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combine like ter hen combine lik e ter Guided Practice In Exercises 9–16, solve each equation for the unknown variable. 9. 4x = 144 11. x 4 = –7 13. –3x + 4 = 19 10. –7x = –7 12. 3 2x = 9 14. 4 – 2x = 18 15. 3x – 2(x – 1) = 2x – 3(x – 4) 16. 3x – 4(x – 1) = 2(x + 9) – 5x Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 7979797979 Independent Practice Solve each of these equations: 1. –2(3x – 5) + 3(x – 1) = –5 2. 4(2a + 1) – 5(a – 2) = 8 3. 5(2x – 1) – 4(x – 2) = –15 4. 2(5m + 7) – 3(3m + 2) = 4m 5. 4(5x + 2) – 5(3x + 1) = 2(x – 1) 6. b – {3 – [b – (2 – b) + 4]} = –2(–b – 3) 7. 4[3x – 2(3x – 1) + 3(2x – 1)] = 2[–2x + 3(x – 1)] – (5x – 1) 8. 30 – 3(m + 7) = –3(2m + 27) 9. 8x – 3(2x – 3) = –4(2 – x) + 3(x – 4) – 1 10. –5x – [4 – (3 – x)] = –(4x + 6) In Exercises 11–17, solve the equations and check your solutions. You don’t need to show all your steps. 11. 4t = 60 13. 15. p 8 y 7 = 1 = 4 17. s −4 = –4 12. x + 21 = 19 14. 7 – y = –11 16. 40 – x = 6 18. Solve –(3m – 8) = 12 – m 19. Denzel takes a two-part math test. In the first part he gets 49 points and in the second part he gets

4 9 of the x points. If his overall grade for the test was 65, find the value of x. 20. Latoya takes 3 science tests. She scores 24%, 43%, and x% in the tests. Write an expression for her average percentage over the 3 tests. Her average percentage is 52. Calculate the value of x. Solve the following equations. Show all your steps and justify them by citing the relevant properties. 21. x + 8 = 13 23. y – 7 = 19 25. 4m = 16 22. 11 + y = 15 24. x 3 = 4 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Solving an equation means isolating the variable. Anything you don’t want on one side of the equation can be “taken over to the other side” by using the inverse operation. You’re always aiming for an expression of the form: “x =...” (or “y =...,” etc.). 8080808080 Section 2.2 Section 2.2 Section 2.2 — Manipulating Equations Section 2.2 Section 2.2 TTTTTopicopicopicopicopic 2.3.12.3.1 2.3.12.3.1 2.3.1 Section 2.3 actions actions ving Frrrrractions ving F ving F RRRRRemoemoemoemoemoving F actions actions ving F ving Frrrrractions RRRRRemoemoemoemoemoving F actions actions ving F ving F actions ving F actions California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as

3(23(23(23(23(2x – 5) + 4( h as sucsucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll use the least common multiple to remove fractional coefficients from equations. Key words: coefficient least common multiple Expressions often seem more complicated if they contain fractions. Getting rid of fractions isn’t too difficult — you just need to use the least common multiple (LCM) again. tions Easier tions Easier es Solving Equa es Solving Equa actions Mak ving Frrrrractions Mak actions Mak ving F RRRRRemoemoemoemoemoving F ving F tions Easier es Solving Equations Easier actions Makes Solving Equa tions Easier es Solving Equa actions Mak ving F To solve equations that contain fractional coefficients, you can get rid of all the fractional coefficients by multiplying both sides of the equation by any common multiple of the denominators of the fractions. You don’t have to remove fractions, but it can make solving the equation a lot easier. The most efficient thing to multiply by is the least common multiple (LCM). If you needed to solve 1 6 common multiple of 4 and 6. To find the LCM, list the prime factors:, you first need to multiply by the least − = Now write each prime factor the greatest number of times it appears in any of the factorizations. The prime factor 2 occurs twice in the factorization of 4, so count two of them. The prime factor 3 occurs only once in the factorization of 6, so just count one of them. Then multiply the factors together to get the LCM: 2 × 2 × 3 = 12 So the LCM of 4 and 6 is 12. Example Example Example Example Example 11111 Solve the equation − = 1 x 1 4 1 6 x. Solution Solution Solution Solution Solution The LCM of the denominators is 12 — so multiply both sides of the equation by 12: × − × = × 1 12 1 x 12 1 12 1 1 4 3x – 12 = 2x 3x – 2x = 12 x = 12 1 6 x Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.

3 8181818181 Guided Practice In Exercises 1–2, find the least common multiple of the denominators: 1. 1 9 x + 13 – In Exercises 3–6, solve the equations for the unknown variable. 3. 1 2 x – 1 = x 5. 1 10. 1 10 x – 3 = 1 4 x 6 or More Fe Fe Fe Fe Frrrrractions actions actions o or Mor k Out the LCM fCM fCM fCM fCM for or or or or TTTTTwwwwwo or Mor o or Mor k Out the L ou Can WWWWWororororork Out the L k Out the L ou Can YYYYYou Can ou Can actions actions o or Mor k Out the L ou Can Example Example Example Example Example 22222 Solve and check the root of Solution Solution Solution Solution Solution LCM of 3, 6, and 2 is 6 × 2x – 1 × 5x – 6 × 3 = 3 × 1x – 6 × 5 4x – 5x – 18 = 3x – 30 –x – 18 = 3x – 30 –x – 3x = –30 + 18 –4x = –12 − − x 12 4 − − 4 4 x = 3 = Checking the solution 15 Check it out: Since the left side of the equation is equal to the right side of the equation, x = 3 is a correct solution of the equation. 5 3 2 − 3 10 2 −11 − = − − − 8282828282 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 Guided Practice In Exercises 7–14, solve your answer for the unknown variable: 7. 9. 2 3 1 8 11. 13. 14 + 22 – 2 3 x n – 1 8 n 10. 8 = – 3 5 (2x – 1 = 17 – 1 2 x – 3 8 x 12. 1 2 + 7 10 x – 1 5 x + 17 = 10 20 + 3 2 x (a – 16) + 3 5 = 7 – 1 15 (a + 4) m + 5 2 (m – 11 + m) Independent Practice In Exercises 1–2, find the least common multiple of the denominators: 1 In Exercises 3–8, solve the equation for the unknown variable: 3. 5. 7 10. 6. 8 21 x – 1

3 x + 6 (m – 2) – 1 5 m = – 1 5 The sum of the measures of the angles of a triangle is 180°. In Exercises 9–11, find the value of x. 9. x 10. 11. 1 2( – 30) x 1 x 4 1 5 x x + 10 + 2 1 x + 5 10 x 2 5 x 2 5 x 12. Qiaofang bought a new car for $27,000. If the value of the car depreciates as 27,000 – 7100 3 n, where n is the number of years since purchase, when will it be worth $12,800? 13. Mary’s weekly allowance increases every year as 2 5 x + $2, where x is Mary’s age. How old is Mary if she gets $5.20 allowance? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up If you multiply one side of an equation by a value, then you need to multiply the other side by the same thing. Make sure that the least common multiple is a multiple of the denominator of every fraction in the equation — otherwise you’ll still end up with annoying fractions in the equation. Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 8383838383 TTTTTopicopicopicopicopic 2.3.22.3.2 2.3.22.3.2 2.3.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as su

csucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll use the least common multiple to remove fractional coefficients from equations. Key words: coefficient least common multiple ficients in ficients in actional Coef actional Coef FFFFFrrrrractional Coef ficients in actional Coefficients in ficients in actional Coef ficients in ficients in actional Coef actional Coef FFFFFrrrrractional Coef ficients in actional Coefficients in ficients in actional Coef essions essions aic Expr aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr essions aic Expressions essions aic Expr AlgAlgAlgAlgAlgeeeeebrbrbrbrbraic Expr essions essions aic Expr aic Expr aic Expressions essions aic Expr essions In the last Topic you dealt with equations with fractional coefficients — but sometimes the coefficients apply to more complicated values. ficients Firststststst ficients Fir ficients Fir actional Coef Eliminate Fte Fte Fte Fte Frrrrractional Coef actional Coef Elimina Elimina actional Coefficients Fir ficients Fir actional Coef Elimina Elimina Some equations contain fractional coefficients — and you can’t isolate the variables until you sort out the fractions. (x – 2) is all divided by 5. You need to deal with the fraction before you can get x on its own 10 The denominators are all different, so multiply the equation by the LCM of all three denominators. If there are fractional coefficients in the equation, multiply both sides of the equation by the least common multiple of the denominators of the fractions to remove the fractional coefficients. You need to keep each numerator (the “x – 2,” “x – 3,” and “3”) as a group. Then you can apply the distributive property to eliminate the grouping symbols. Example Example Example Example Example 11111 Solve − 10. Don’t forget: See Lesson 2.3.1 for more on

the least common multiple. Solution Solution Solution Solution Solution There are 3 different denominators, so you need the LCM. List the prime factors, and use them to work out the LCM. Check it out: Use parentheses to group each numerator before you multiply by the LCM. 5 = 5 6 = 2 × 3 10 = 2 × 5 So LCM = 2 × 3 × 5 = 30 Multiply both sides of the equation by 30 to clear the fractional coefficients. 30 = × 1 3 10 − ) × 30 − × 1 30 1 6(x – 2) – 5(x – 3) = 3 × 3 6x – 12 – 5x + 15 = 9 6x – 5x – 12 + 15 = 9 x + 3 = 9 x = 6 8484848484 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 Check it out 12 = So LCM = 2 × 2 × 3 = 12 Example Example Example Example Example Solve x − 11 12 22222 − 1 6 = + 1 x 4. Solution Solution Solution Solution Solution − 11 12 ( ) 11 12 × − × 12 12 1 1 11 – 2(x – 1) = 3(x + 1) 11 – 2x + 2 = 3x + 3 –5x = –10 x = 2 12 = × 1 ( x + 1 ) 4 Independent Practice Solve each of the following equations: 1. 2 − m = 3 10. 2 5 y ( 5 2 5. 1 + = 10 − ) 1 6 x 6 11 + = 12 +. 8 10 ) 1 = − 7 10 12 ( 10 10. 11. 12 21 ) 5 + 2 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up This Topic is really important, because you can’t simplify an expression fully until you’ve got rid of any fractional coefficients. Luckily, you’ve got the LCM to help you. Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 8585858585 TTTTTopicopicopicopicopic 2.3.32.3.3 2.3.32.3.3 2.3.3 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0

: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as sucsucsucsucsuch as h as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as What it means for you: You’ll remove decimal coefficients from equations by multiplying by a suitable power of 10. Key words: coefficient least common multiple ting Decimal ting Decimal Elimina Elimina ting Decimal Eliminating Decimal ting Decimal Elimina Elimina ting Decimal ting Decimal Elimina Elimina Eliminating Decimal ting Decimal ting Decimal Elimina Elimina ficients ficients CoefCoefCoefCoefCoefficients ficients ficients CoefCoefCoefCoefCoefficients ficients ficients ficients ficients If the coefficients are decimal, you still just need to multiply the equation by a suitable number — in fact, this time it’s slightly easier. er of 10 10 10 10 10 er of y a Py a Pooooowwwwwer of er of y a Py a P ying b ying b y Multipl y Multipl Decimals b Decimals b Get Rid of Get Rid of ying by a P y Multiplying b Decimals by Multipl Get Rid of Decimals b er of ying b y Multipl Decimals b Get Rid of Get Rid of Solving an equation that contains decimals can be made easier if it is first converted to an equivalent equation with integer coefficients. The method is the same as with fractions — you just multiply both sides of the equation by a suitable number. The

idea is to multiply both sides of the equation by a large enough power of 10 to convert all decimals to integer coefficients. For example, 0.35 means “35 hundredths,” so you can write it So multiplying by 100 gives you the integer 35. 35 100. Example Example Example Example Example 11111 Solve the equation 0.35x – 12 = –0.15x. Solution Solution Solution Solution Solution Multiply both sides of the equation by 100: 100(0.35x – 12) = 100(–0.15x) 100 × 0.35x – 100 × 12 = 100 × –0.15x 35x – 1200 = –15x 50x = 1200 x = 24 Example Example Example Example Example 22222 Solve 0.75x – 0.65(13 – x) = 8.35. Solution Solution Solution Solution Solution Multiply both sides of the equation by 100: 100[0.75x – 0.65(13 – x)] = 100(8.35) 100(0.75x) – 100(0.65)(13 – x) = 100(8.35) 75x – 65(13 – x) = 835 75x – 845 + 65x = 835 75x + 65x = 835 + 845 140x = 1680 x = 12 8686868686 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 Guided Practice In Exercises 1–10, solve the equation for the unknown variable: 1. 0.04x – 0.12 = 0.01x 2. –0.01x + 0.3 = 0.03x + 0.18 3. 0.06x – 0.09 = 0.05x + 0.12 – 0.15 4. 0.11 – 0.03x – 0.03 = 0.02x – 0.02 5. 0.16m + 2 = 0.03(5m + 2) 6. 0.01a = 0.45 – (0.04a + 0.15) 7. 0.25(x + 4) + 0.10(x – 2) = 3.60 8. 0.10(x + 100) = 0.08(x – 6) 9. 0.25(x – 50) = 0.2(x – 10)

+ 5 10. 0.25(x + 8) + 0.10(8 – x) + 0.05x = 3.60 y Higher Numbersssss ou Need to Multiply by by by by by Higher Number y Higher Number y Higher Number ou Need to Multipl Sometimes YYYYYou Need to Multipl ou Need to Multipl Sometimes Sometimes y Higher Number ou Need to Multipl Sometimes Sometimes In Examples 1 and 2 you needed to multiply both sides of each equation by 100. You can’t always just multiply by 100, though. If any of the decimals have more than 2 decimal places, you’ll need a higher power of 10. Example Example Example Example Example 33333 Solve 0.015x = 0.2 – 0.025x. Solution Solution Solution Solution Solution Multiply both sides by 1000 this time. 1000(0.015x) = 1000(0.2 – 0.025x) 15x = 200 – 25x 40x = 200 x = 5 If your longest decimal has 3 decimal places, multiply by 103 = 1000. If the longest decimal has 4 decimal places, multiply by 104, and so on. Guided Practice In Exercises 11–20, solve each equation for the unknown variable. 11. 0.3(2x + 7) = 1.8 12. 2.8 = 0.2(4x + 2) 13. 0.016 = 0.002(a – 1) – 0.001a 14. 0.1x + 0.1(3x – 8) = – 230 15. 0.001x + 0.05(2x – 3) = 35.2 16. 0.012p – 0.065p – 0.7 = 0.5 17. 0.003(x + 7) = 0.01x 18. 0.006(x + 3) = 0.002(x + 31) 19. 0.072 – 0.006j = 0.032 + 0.08j 20. 0.003(2x – 0.3) + 0.001x = 0.1979 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 8787878787 Independent Practice Solve each of the equations in Exercises 1–17: 1. 0.11x – 2.3 + 0.

4(2x – 1) = 0.25(3x + 8) 2. 0.5x – 1 = 0.7x + 0.2 3. 0.3v + 4.2v – 11 = 1.5v + 4 4. 1.8(y – 1) = 3.1y + 2.1 5. 0.25(3x – 2) + 0.10(4x + 1) + 0.75(x – 1) = 4.55 6. 0.125(3b – 8) – 0.25(b + 5) – 0.2 = 0.2(2b + 7) 7. 0.4(x + 7) – 0.15(2x – 5) = 0.7(3x – 1) – 0.75(4x – 3) 8. 0.21(x – 1) – 0.25(2x + 1) = 0.5(2x + 1) – 0.6(4x – 15) + 0.03 9. 0.9(2v – 5) + 0.20(–3v – 1) = 0.22(4v – 3) 10. 2.46 – 0.52(x – 10) = 0.35(4x + 8) – 2.82 11. –0.20x – 1.10(3x – 11) = 1.25(2x – 5) – 6.25 12. 16 + 0.50y = 0.60(y + 20) 13. 20 + 1.20m = 1.10(m + 25) 14. 0.03x + 0.30(900 – x) = 72 15. 0.06y + 0.3y – 0.1 = 0.26 16. 0.04k + 0.06(40,000 – k) = 2100 17. –0.02[0.4 – 0.1(2 + 3x)] = 0.004x + 0.005 18. The sides of an equilateral triangle measure 0.2(10x + 90) units each. If the perimeter of the triangle is 612.6 units, find x. 19. The area of a rectangle is 34 units2. If the width is 0.25 units and the length is (x + 10) units, find the value of x. 20.

The cost per minute to make a call is $0.05. If Meimei talks for (x + 20) minutes and the call costs $4.85, what is the value of x? 21. A cell phone plan charges $25 per month plus $0.10 per minute. If your monthly bill is $39.80, write and solve an equation to find out the number of minutes on your bill. 22. A moving van rents for $40 a day plus $0.08 a mile. Ed’s bill is $58.24 and he had the van for one day. Write and solve an equation to find out how many miles he drove. 23. Eylora has x quarters with a value of $0.25x. Emily has dimes that value 0.10(x + 8). If they have a total of $5.00 in coins, how many coins does Eylora have? 24. Michael, William, and Daniel are playing a game. Michael has x points, William has 0.01(x + 18,000) points, and Daniel has 0.02(x – 800) points. If together they have earned 11,288 points, how many points does Michael have? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Whether you’re dealing with fractional or decimal coefficients, the method’s essentially the same — you multiply everything by a number that will make the algebra easier and mistakes less likely. Then you can start isolating the variable. 8888888888 Section 2.3 Section 2.3 Section 2.3 — More Equations Section 2.3 Section 2.3 TTTTTopicopicopicopicopic 2.4.12.4.1 2.4.12.4.1 2.4.1 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele

,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as h as sucsucsucsucsuch as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solveeeee Students solv Students solv ultiste mmmmmultiste lems,,,,, lems lems p prp proboboboboblems p prp pr ultiste ultistep pr lems ultiste d prd proboboboboblems incincincincincluding w lems,,,,, lems lems d prd pr luding wororororord pr luding w luding w luding w lems tions tions olving linear equa olving linear equa inininininvvvvvolving linear equa tions olving linear equations tions olving linear equa in one in one and linear inequalities in one in one in one vvvvvariaariaariaariaariabbbbble and pr le and prooooovidevidevidevidevide le and pr le and pr le and pr h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll set up and solve equations that model real-life situations. Key words: application linear equation Check it out: Look out for words that give math clues — for example, “twice” means two times and the word “is” means equals. tions of tions of pplica pplica AAAAApplica tions of pplications of tions of pplica AAAAApplica tions of tions of pplica pplica pplications of tions of pplica tions of tions tions Linear Equa Linear Equa tions Linear Equations tions Linear Equa Linear Equa tions tions

Linear Equa Linear Equa Linear Equations tions tions Linear Equa Linear Equa “Applications” are just “real-life” tasks. In this Topic, linear equations start to become really useful. asksasks eal-Life”e”e”e”e” TTTTTasks asksasks eal-Lif tions are “Re “Re “Re “Re “Real-Lif eal-Lif tions ar tions ar Equa Equa tions of tions of pplica AAAAApplica pplica Equations ar tions of Equa pplications of eal-Lif tions ar Equa tions of pplica Applications questions are word problems that require you to set up and solve an equation. First decide how you will label the variables......then write the task out as an equation......making sure you include all the information given......then you can solve your equation. Example Example Example Example Example 11111 The sum of twice a number c and 7 is 21. Set up and solve an equation to find c. Solution Solution Solution Solution Solution You’re given the label “c” in the question — so just write out the equation. 2c + 7 = 21 2c = 14 c = 7 Guided Practice 1. Twice a number c plus 17 is 31. Find the value of c. 2. Three times a number k minus 8 is 43. Find the number k. 3. The sum of four times a number m and 17 is the same as 7 less than six times the number m. Find the number m. 4. Seven minus five times the number x is equal to the sum of four times the number x and 25. Find the number x. Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 8989898989 Check it out: “4v + 15” is just the phrase “15 more than four times v” written in math-speak. en the VVVVVariaariaariaariaariabbbbbleslesleslesles en the ys Be Givvvvven the en the ys Be Gi YYYYYou ou ou ou ou WWWWWononononon’’’’’t t t t t AlAl

AlAlAlwwwwwaaaaays Be Gi ys Be Gi en the ys Be Gi Sometimes you’ll have to work out for yourself what the variables are, and decide on suitable labels for them. Example Example Example Example Example 22222 Juanita’s age is 15 more than four times Vanessa’s age. The sum of their ages is 45. Set up and solve an equation to find their ages. Solution Solution Solution Solution Solution This time you have to decide for yourself how to label each term. Let v = Vanessa’s age 4v + 15 = Juanita’s age v + (4v + 15) = 45 v + 4v + 15 = 45 5v + 15 = 45 5v = 30 v = 6 es is 45 es is 45 their aggggges is 45 their a their a he sum of he sum of TTTTThe sum of es is 45 he sum of their a es is 45 their a he sum of Plug in the value for v to get Juanita’s age: 4v + 15 = 4 × 6 + 15 = 24 + 15 = 39 So Vanessa is 6 years old and Juanita is 39 years old. Guided Practice 5. The length of a rectangular garden is 3 meters more than seven times its width. Find the length and width of the garden if the perimeter of the garden is 70 meters. 6. A rectangle is 4 meters longer than it is wide. The perimeter is 44 meters. What are the dimensions and area of the rectangle? 7. Abraham’s age is 4 less than half of Dominique’s age. Dominique’s age is 6 more than three times Juan’s age. The sum of their ages is 104. Find the age of each person. 8. The sides of an isosceles triangle are each 2 inches longer than the base. If the perimeter of the triangle is 97 inches, what are the lengths of the base and sides of the triangle? 9. The sum of two consecutive integers is 117. What are the integers? 9090909090 Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 Independent Practice 1. Find the value of x, if the line segment shown on the right is 21 cm long. Also, find the length of each part of the line segment. (4x – 5) (3x – 2) cm 21 cm cm 2. Point M

is the midpoint of the line segment shown. Find the value of x and the length of the entire line segment. 6x – 11 4x + 23 M 3. The perimeter of the rectangular plot shown below is 142 feet. Find the dimensions of the plot. (6 – 1) ft x (4 + 2) ft x 4. The sum of the interior angles of a triangle is 180°. Find the size of each angle in the triangle shown below. 4x 3x 5x 5. The sum of one exterior angle at each vertex of any convex polygon is 360°. Find the size of each exterior angle shown around the triangle below. 3 + 10 y 2 + 45 y 4 – 10 y 6. The interior angles of a triangle sum to 180°. Find the size of each angle in the triangle sketched on the right. 3x 3x + 5 x 7. A rectangular garden has a length that is five meters less than three times its width. If the length is reduced by three meters and the width is reduced by one meter, the perimeter will be 62 meters. Find the dimensions of the garden. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The thing to do with any word problem is to write out a math equation that describes the same situation. Then you can use all the techniques you’ve learned to solve the equation. Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 9191919191 TTTTTopicopicopicopicopic 2.4.22.4.2 2.4.22.4.2 2.4.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4

e Linear Equations About Mone tions e Linear Equa asks In Coin If you have a collection of coins, there are two quantities you can use to describe it — how many coins there are, and how much they are worth. In coin tasks you’ll have to use both quantities. Example Example Example Example Example 11111 Jazelle has a coin collection worth $3.50. She only has nickels, dimes, and quarters. If she has four more dimes than quarters and twice as many nickels as she has dimes, how many coins of each kind does she have in her collection? Solution Solution Solution Solution Solution The first thing to do is to write expressions for how many of each type of coin she has. Call the number of quarters q. Then the number of dimes is q + 4. And the number of nickels is 2(q + 4) = 2q + 8. Then write expressions for the total value (in cents) of each type of coin. There are q quarters, so the total value of the quarters is 25q. There are q + 4 dimes, so the total value of the dimes is 10(q + 4) = 10q + 40. There are 2q + 8 nickels, so the total value of the nickels is 5(2q + 8) = 10q + 40. Write an equation to represent the fact that the collection is worth $3.50. 25q + (10q + 40) + (10q + 40) = 100(3.50) Solving for q gives: 45q + 80 = 350 45q = 270 q = 6 So Jazelle has: q = 6 quarters, q + 4 = 6 + 4 = 10 dimes, 2(q + 4) = 2(6 + 4) = 20 nickels. 9292929292 Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 Example Example Example Example Example 22222 Tickets to a puppet show sell at $2.50 for children and $4.50 for adults. There are five times as many children at a performance as there are adults and the show raises $3009. How many adult and children’s tickets were sold for the show? Solution Solution Solution Solution Solution Let x = adult tickets sold. Then 5x = children’s tickets sold. The value is the number of

tickets multiplied by the cost of each ticket. So 4.50x = amount raised from adults (in dollars) 5x(2.50) = amount raised from children Write an equation showing that the sum of these amounts is $3009: 4.50x + 5x(2.50) = 3009 17x = 3009 x = 177 So 177 adult tickets and 5 × 177 = 885 children’s tickets were sold. Independent Practice 1. The total cost of buying some music CDs and having them shipped to Charles was $211.50. If the CDs cost $11.50 each and the shipping for the box of CDs was $4.50, how many CDs did Charles receive? 2. Jerome bought 12 CDs. Some of the CDs cost $7.50 each and the rest cost $6.50 each. How many CDs were bought at each price if Jerome spent a total of $82? 3. Rajan’s coin collection is valued at $22.70. He has one fewer halfdollar coin than twice the number of dimes and four more quarters than three times the number of dimes. How many dimes, quarters, and half-dollars does Rajan have, assuming he has no other coins? 4. Liza has 60 coins in her collection. The coin collection consists of nickels, dimes, and quarters. She has five fewer quarters than nickels and ten more dimes than quarters. How many coins of each kind does Liza have? 5. A school cafeteria cashier has collected $243 in one-dollar, fivedollar, and ten-dollar bills. The number of one-dollar bills is eight more than 20 times the number of ten-dollar bills. The cashier also has seven more than twice the number of ten-dollar bills in five-dollar bills. How many bills of each value does the cashier have? Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 9393939393 6. Dan’s algebra class is planning a summer afternoon get-together. Dan is supposed to bring some melons at $1.25 each, juice boxes at $0.50 each, and granola bars for $0.75 each. If he buys nine more than ten times the number of melons in juice boxes and seven more than five times the number of melons in granola bars, how many items of

each kind did he buy with $29.75? 7. Martha bought some baseball uniforms for $313 and had them shipped to her. If the baseball uniforms cost $23.75 each and the shipping was $4.25 for the whole order, how many baseball uniforms did Martha buy? 8. Fifteen children’s books cost $51.25. Some were priced at $2.25 each, and the rest of the books were sold at $4.75 each. How many books were purchased at each price? 9. Dwight purchased various stamps for $16.35. He purchased 12 more 25¢ stamps than 35¢ stamps. The number of 30¢ stamps was four times the number of 35¢ stamps. Finally, he bought five 15¢ stamps. How many of each kind of stamp did Dwight buy? 10. A waiter has collected 150 coins from the tips he receives from his customers. The coins consist of nickels, dimes, and quarters. He has five more than twice the nickels in dimes and five more than four times the nickels in quarters. i) How many coins of each kind does the waiter have? ii) How much money does the waiter have? 11. Jessica has two more nickels than dimes, and three more quarters than nickels, but no other coins. If she has a total of $5.35, how many coins of each kind does she have? 12. A grocer’s deposit box contains 150 coins worth $12.50. They are all nickels and dimes. Find the number of each coin in the box. 13. A store sells nineteen different video games. Several games are priced at $19.99, while half that number are priced at $39.99, and 4 are priced at $49.99. How many games are priced at $19.99? 14. Mark bought packets of popcorn, drinks, and bags of nuts for him and some friends at the movie theater. Everyone got one of each. Drinks cost $4.50 each, packets of popcorn cost $3.75 each, and bags of nuts cost $2.00 each. If he spent $41, how many friends did Mark have with him? 15. John has quarters, nickels, and dimes. He has 4 more nickels than quarters and twice as many dimes and nickels. If he has $6.00, how many quarters does he have?

ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up To answer these questions, you need to use (i) the number of items, and (ii) their value. Then, when you have set up your equation and solved it, be sure to give your final answer in the form asked for in the problem. 9494949494 Section 2.4 Section 2.4 Section 2.4 — Using Equations Section 2.4 Section 2.4 TTTTTopicopicopicopicopic 2.5.12.5.1 2.5.12.5.1 2.5.1 Section 2.5 asksasks asksasks e Integgggger er er er er TTTTTasks e Inte e Inte Consecutivvvvve Inte Consecuti Consecuti e Inte Consecuti Consecuti e Integgggger er er er er TTTTTasks Consecutivvvvve Inte asksasks asksasks e Inte e Inte Consecuti Consecuti e Inte Consecuti Consecuti California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as h as sucsucsucsucsuch as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solveeeee Students solv Students solv ultiste mmmmmultiste lems,,,,, lems le

ms p prp proboboboboblems p prp pr ultiste ultistep pr lems ultiste d prd proboboboboblems incincincincincluding w lems,,,,, lems lems d prd pr luding wororororord pr luding w luding w luding w lems tions tions olving linear equa olving linear equa inininininvvvvvolving linear equa tions olving linear equations tions olving linear equa in one in one and linear inequalities in one in one in one vvvvvariaariaariaariaariabbbbble and pr le and prooooovidevidevidevidevide le and pr le and pr le and pr h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve word problems that refer to patterns of integers. Key words: consecutive integer common difference When you’re solving word problems, the most important thing to do is to write down what you know. Then create an equation that represents the relationship between the known and unknown quantities. Finally, you have to solve that equation. Sequences with a Common Difffffferererererence ence ence Sequences with a Common Dif Sequences with a Common Dif ence ence Sequences with a Common Dif Sequences with a Common Dif A sequence of integers with a common difference is a set of integers that increase by a fixed amount as you move from term to term 10 Common difference = 2 1 2 3 4 5 6 87 9 10 3, 3, 6 3, 3 9, Common difference = 3 Consecutive EVEN integers (for example, 2, 4, and 6) form a sequence with common difference 2. Consecutive ODD integers (for example, 1, 3, and 5) also form a sequence with common difference 2. So if you are given an even integer (or an odd integer) and you are asked to find the next even (or odd) integer, just add two. Example Example Example Example Example 11111 Check it out: Add 2 to x to find the second even integer, and another 2 to find the third. Find the three consecutive even integers whose sum is 48. Solution Solution

Solution Solution Solution Call the first (smallest) even integer x. Then you can write down an expression for the other even integers in terms of x. 1st even integer = x 2nd even integer = x + 2 3rd even integer = x + 4 But the sum of the three even integers is 48, so x + (x + 2) + (x + 4) = 48. Now you can rearrange and simplify this formula, and then solve for x. x + (x + 2) + (x + 4) = 48 (x + x + x) + (2 + 4) = 48 3x + 6 = 48 3x = 42 x = 14 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 9595959595 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin So the smallest of the three even integers is 14. Now just add 2 to find the second even integer, and add another 2 to find the third. So the three consecutive even integers are: x = 14, x + 2 = 16, and x + 4 = 18. wns Firststststst wns Fir wns Fir or the Unkno or the Unkno essions f essions f rite Expr WWWWWrite Expr rite Expr or the Unknowns Fir essions for the Unkno rite Expressions f wns Fir or the Unkno essions f rite Expr The most important thing with any word problem is to first write it out again in math-speak. Example Example Example Example Example 22222 Find a sequence of four integers with a common difference of 4 whose sum is 92. Solution Solution Solution Solution Solution Call the first (smallest) integer v, for example. Then you can write the other three in terms of v. 1st integer = v 2nd integer = v + 4 3rd integer = v + 8 4th integer = v + 12 The sum of the four integers is 92, so: v + (v + 4) + (v + 8) + (v + 12) = 92 Now you can solve for v: (v + v + v + v) + (4 + 8 + 12) = 92 4v + 24 = 92 4v = 68 v = 17 So the smallest of the integers is 17. Since they differ by 4, the others

must be 21, 25, and 29. Once you’ve completed the problem, do a quick answer check — add the integers together and see if you get what you want: 1st integer = v = 17 2nd integer = v + 4 = 21 3rd integer = v + 8 = 25 4th integer = v + 12 = 29 92 9696969696 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 ✓ Example Example Example Example Example 33333 Find a sequence of four integers with a common difference of 2 whose sum is 944. Decide whether these integers are even or odd. Solution Solution Solution Solution Solution Call the first integer x, then write expressions for the others: Check it out: The difference between the integers is 2, whether they are odd or even. 1st integer = x 2nd integer = x + 2 3rd integer = x + 4 4th integer = x + 6 The sum of the four integers is 944, so: x + (x + 2) + (x + 4) + (x + 6) = 944 Now solve for x: 4x + 12 = 944 4x = 932 x = 233 So the smallest of the integers is 233, which means the others must be 235, 237, and 239. These are consecutive odd integers. Now do a quick answer check — add the integers together and see if you get what you want: 1st integer = x = 233 2nd integer = x + 2 = 235 3rd integer = x + 4 = 237 4th integer = x + 6 = 239 944 Guided Practice Find the unknown numbers in each of the following cases. 1. The sum of two consecutive integers is 103. 2. The sum of three consecutive integers is –138. 3. The sum of two consecutive even integers is 194. 4. The sum of three consecutive odd integers is –105. 5. The sum of a number and its double is 117. 6. Two integers have a difference of three and their sum is 115. 7. Three integers have a common difference of 3 and their sum is –78. 8. Four rational numbers have a common difference of 5 and a sum of 0. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 9797979797 ✓ Independent

Practice 1. Three consecutive integers have a sum of 90. Find the numbers. 2. Find the four consecutive integers whose sum is 318. 3. Find three consecutive integers such that the difference between three times the largest and two times the smallest integer is 30. 4. Find three consecutive integers such that the sum of the first two integers is equal to three times the highest integer. 5. Two numbers have a sum of 65. Four times the smaller number is equal to 10 more than the larger number. Find the numbers. 6. Four consecutive even integers have a sum of 140. What are the integers? 7. Find three consecutive even integers such that six more than three times the smallest integer is 54. 8. Find three consecutive odd integers whose sum is 273. 9. Find four consecutive odd integers such that 12 more than four times the smallest integer is 144. 10. Find three consecutive even integers such that six more than twice the first number is 94. 11. Find three consecutive even integers such that the product of 16 and the third integer is the same as the product of 20 and the second integer. 12. A 36-foot pole is cut into two parts such that the longer part is 11 feet longer than 4 times the shorter part. How long is each piece of the pole? 13. Find three consecutive odd integers such that four times the largest is one more than nine times the smallest integer. 14. Ten thousand people attended a three-day outdoor music festival. If there were 800 more girls than boys, and 1999 fewer adults than boys, how many people of each group attended the festival? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Consecutive integer tasks are a strange application of math equations — but they appear a lot in Algebra I. Always make sure you’ve answered the question — you’ve always got to remember that your solution isn’t complete until you’ve stated what the integers actually are. 9898989898 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 TTTTTopicopicopicopicopic 2.5.22.5.2 2.5.22.5.2 2.5.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.

0: Students simplify 4.0: Students simplify essions befororororore solving eeeeexprxprxprxprxpressions bef e solving e solving essions bef essions bef e solving e solving essions bef tions tions linear equa linear equa tions and linear equations tions linear equa linear equa in one variaariaariaariaariabbbbblelelelele,,,,, in one v in one v inequalities in one v in one v – 2) = 12. – 2) = 12. – 5) + 4( – 5) + 4( h as 3(23(23(23(23(2x – 5) + 4( h as h as sucsucsucsucsuch as – 2) = 12. – 5) + 4(x – 2) = 12. – 2) = 12. – 5) + 4( h as 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solveeeee Students solv Students solv ultiste mmmmmultiste lems,,,,, lems lems p prp proboboboboblems p prp pr ultiste ultistep pr lems ultiste d prd proboboboboblems incincincincincluding w lems,,,,, lems lems d prd pr luding wororororord pr luding w luding w luding w lems tions tions olving linear equa olving linear equa inininininvvvvvolving linear equa tions olving linear equations tions olving linear equa in one in one and linear inequalities in one in one in one vvvvvariaariaariaariaariabbbbble and pr le and prooooovidevidevidevidevide le and pr le and pr le and pr h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve equation problems that refer to people’s ages. Key words: age task asksasks asksasks ted TTTTTasks ted ted AgAgAgAgAge-Re-Re-Re-Re-Relaelaelaelaelated ted

ted TTTTTasks AgAgAgAgAge-Re-Re-Re-Re-Relaelaelaelaelated asksasks asksasks ted ted ted Age tasks relate ages of different people at various time periods — in the past, the present, or the future. asksasks ted TTTTTasks asksasks ted AgAgAgAgAge-Re-Re-Re-Re-Relaelaelaelaelated ted ted Your age at any point in your life can always be written as your current age plus or minus a certain number of years. If your current age is x years......then 5 years ago, your age was 5 fewer than x......and in 5 years’ time, your age will be 5 more than x. More generally, your age c years ago was: And your age in c years will be In much the same way, anybody’s age can always be related to someone else’s by adding or subtracting a certain number of years. wn Quantities wn Quantities or the Unkno or the Unkno essions f essions f rite Expr WWWWWrite Expr rite Expr wn Quantities or the Unknown Quantities essions for the Unkno rite Expressions f wn Quantities or the Unkno essions f rite Expr Solving an age task is pretty similar to solving a consecutive integer task. You need to write down expressions for the unknown quantities in terms of one variable (like x). Then you can use the information in the question to write an equation that you can go on to solve. Example Example Example Example Example 11111 Charles is 7 years older than Jorge. In 20 years’ time, the sum of their ages will be 81 years. How old is each one now? Solution Solution Solution Solution Solution The first thing to do is write expressions relating all the ages to one another. Present You’re not told Jorge’s age, so call it x. Charles is currently 7 years older — that is, 7 more than x. Jorge’s age = x Charles’s age = x + 7 Future (in 20 years) Jorge will be 20 years older than at present. Charles will be 20 years older than at present. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 9999999999 Example 1 continueduedueduedued Example 1 contin Example 1

contin Example 1 contin Example 1 contin Jorge’s future age = x + 20 Charles’s future age = (x + 7) + 20 = x + 27 Now use the information from the question to combine these expressions into an equation. The sum of their ages in 20 years will be 81, so: (x + 20) + (x + 27) = 81 Now solve your equation for x: (x + 20) + (x + 27) = 81 2x + 47 = 81 2x = 34 x = 17 So Jorge is currently 17 years old. That means that Charles is 17 + 7 = 24 years old. Example Example Example Example Example 22222 Juanita is twice as old as Vanessa. If 5 years were subtracted from Vanessa’s age and 2 years added to Juanita’s age, then Juanita’s age would be five times Vanessa’s. How old are the girls now? Solution Solution Solution Solution Solution As before, start by writing down mathematical expressions for the ages mentioned in the question. Present You’re not given Vanessa’s age, so call it v. Juanita is twice as old — that is, twice v. Vanessa’s age = v Juanita’s age = 2v “Adjusted” ages “If 5 years is subtracted from Vanessa’s age...” = v – 5 “If 2 years is added to Juanita’s age...” = 2v + 2 Now you can write an equation. Remember that Juanita’s “adjusted” age is five times as big as Vanessa’s. 2v + 2 = 5(v – 5) Now solve for v to find Vanessa’s current age: 2v + 2 = 5(v – 5) 2v + 2 = 5v – 25 –3v = –27 v = 9 So Vanessa is 9 years old. And since Juanita is twice as old, Juanita is 2 × 9 = 18 years old. Check it out: Watch out — you haven’t answered the question unless you say what v = 9 means for the people’s ages. 100100100100100 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 Example Example Example Example Example 33333 A father is three times as old as his daughter.

In 15 years’ time, the father will be twice as old as his daughter. What are their current ages? Solution Solution Solution Solution Solution Present You are not given the daughter’s age, so call it x. The father is three times as old — three times x. Daughter’s age = x Father’s age = 3x Future (in 15 years) The daughter and father will both be 15 years older. Daughter’s future age = x + 15 Father’s future age = 3x + 15 Don’t forget: Remember to say what your value of x means in the context of the question. In 15 years, the father’s age will be twice as great as his daughter’s. Write this as an equation, and solve: 3x + 15 = 2(x + 15) 3x + 15 = 2x + 30 x = 15 So the daughter is 15 years old, and the father is 3 × 15 = 45 years old. Guided Practice 1. Eylora is 4 times as old as Leo. If the sum of their ages is 5, how old is Eylora? 2. Clarence is 8 years older than Maria. In 24 years, the sum of their ages will be 100. How old is Clarence? 3. Tyler, Nick, and Sid are brothers. The sum of their ages is 54. The oldest brother, Nick, is 2 years older than Sid, and Sid is 2 years older than Tyler. How old is Sid? 4. A father is 9 times older than his daughter and 2 years older than his wife. If the sum of their ages is 74, how old is the father? 5. Ruby and Emily are twins. Rebecca is 6 more than 2 times Ruby and Emily's age. Altogether the sum of their three ages is 50. How old are Emily and Ruby? 6. Santos is 30 years older than his daughter Julia. If their ages are increased by 10% and added together the sum is 77. How old is Santos? 7. James is 4 years less than 7 times the age of his daughter, who is 4 more than half of her brother's age. The sum of their ages is 38. How old is James? Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 101101101101101 Independent Practice 1. Keisha, Juan, and Jose are friends

who are all different ages. There is a 2 year difference in age between the oldest and youngest. Juan is not as old as Jose, but he is older than Keisha. If the sum of their ages is 36, how old is the oldest child? 2. Sonita’s father is three times as old as she is now. In ten years, her father will be twice as old as Sonita will be then. How old are Sonita and her father now? 3. Kadeeja is four times as old as her niece. In three years, Kadeeja will be three times as old as her niece. How old is each of them now? 4. Sally is twice as old as Daniel. Ten years ago, the sum of their ages was 70 years. How old is each one of them now? 5. Andy is four times as old as Alejandro. Five years ago, Andy was nine times as old as Alejandro. How old is each one now? 6. Chris is 40 years younger than his uncle. In ten years’ time the sum of their ages will be 80 years. How old are they now? 7. Jorge is three times Martha’s age. If 30 years is added to Martha’s age and 30 years is subtracted from Jorge’s age, their ages will be equal. How old is each person now? 8. Mia’s age in 20 years will be the same as Simon’s age is now. Ten years from now, Simon’s age will be twice Mia’s age. How old is each one now? 9. Paula is three times as old as Duncan. If four is subtracted from Duncan’s age and six is added to Paula’s age, Paula will then be four times as old as Duncan. How old are they now? 10. If you decrease Marvin's age by 25%, you will find his age 4 years ago. How old is Marvin now? 11. If you increase Qin's age by 75% then you will find his age 6 years from now. How old will Qin be in 6 years? 12. Jaya is 10 years younger than Sid. If you increase both of their ages by 20%, the difference between their ages is 18 less than Jaya's current age. How old is Sid? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up

ound Up ound Up Age tasks are just another example of real-life equations. As always, you have to set up an equation from the information you’re given, then solve the equation. Your answer is only complete when you include the actual ages of the people involved. 102102102102102 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 TTTTTopicopicopicopicopic 2.5.32.5.3 2.5.32.5.3 2.5.3 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec aic techniques to hniques to hniques to aic tec solvsolvsolvsolvsolve re re re re raaaaate pr lems lems te proboboboboblems te pr te pr lems, work lems te pr problems, and percent mixture problems. What it means for you: You’ll learn the formulas for rate, distance, and time, and use them to set up equations to solve real-life problems involving rates. Key words: rate speed distance time RRRRRaaaaatetetetete,,,,, TimeTimeTimeTimeTime,,,,, RRRRRaaaaatetetetete,,,,, TimeTimeTimeTimeTime,,,,, asksasks asksasks and Distance TTTTTasks and Distance and Distance and Distance and Distance and Distance TTTTTasks asksasks asksasks and Distance and Distance and Distance and Distance The greater an object’s speed, the greater the distance it travels in a given amount of time. Rate, time, and distance tasks are a little different to the ones you’ve seen in this Section so far because they have particular formulas that you need to learn. y a Fy a Fororororormmmmmulaulaulaulaula y a Fy a F ted b and Distance are Re Re Re Re Relaelaelaelaelated b ted b and Distance ar Speed, TimeTimeTimeTimeTime,,,,, and Distance ar and Distance ar Speed, Speed, ted by a F ted b and Distance ar Speed,

Speed, The quantities of distance, time, and speed are related by a formula. Speed is the distance traveled per unit of time — for instance, the distance traveled in one second, one hour, etc. speed = distance time The units of speed depend on the units used for the distance and time. For example, if the distance is in miles and the time is in hours, then the speed will be in miles per hour. Time Fororororormmmmmulaulaulaulaula Time F Distance,,,,, Time F Time F Distance Distance e the Speed, ou Can Rearearearearearrrrrrangangangangange the Speed, e the Speed, ou Can R YYYYYou Can R ou Can R e the Speed, Distance Time F Distance e the Speed, ou Can R The above formula can be rearranged to give these important formulas for distance and time: distance = speed × time time = distance speed Guided Practice Use the speed, distance, and time formulas to work out the following: 1. Find the average speed if distance = 116 miles and time = 2 hours. 2. Find the average speed if distance = 349 km and time = 5 h. 3. Find the distance if speed = 75 mph and time = 2.5 h. 4. A car completes a 125 mile journey traveling at an average speed of 50 miles per hour. Work out the time taken. 5. A train travels a distance of 412.5 km at an average speed of 120 km per hour. How long did the journey take? 6. A train travels at an average speed of 140 mph. If it takes 4 hours to reach its destination, how far did it travel? 7. A long-distance runner completes a half marathon (13.1 miles) in a time of 1 hour 45 minutes. Find the runner’s average speed. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 103103103103103 e an Equationtiontiontiontion e an Equa e an Equa hen Solv ou Knowwwww,,,,, TTTTThen Solv hen Solv ou Kno rite Down wn wn wn wn WWWWWhahahahahat t t t t YYYYYou Kno ou Kno rite Do WWWWWrite Do rite Do hen Solve an Equa e an Equa hen

Solv ou Kno rite Do Motion tasks normally involve two objects with different speeds. Example Example Example Example Example 11111 Tim drives along a road at 70 km/h. Josh leaves from the same point an hour later and follows exactly the same route. If Josh drives at 90 km/h, how long will it take for Josh to catch up with Tim? Solution Solution Solution Solution Solution As always, first write down what you know — and use sketches, arrows, and tables if they help you visualize what is happening. Drawing arrows helps remind you which way each person is traveling: Tim at 70 km/h Josh at 90 km/h Tim left first......but Josh is traveling faster (and in the same direction), so he will catch up. Suppose Josh catches Tim x hours after Josh left. At that time, Tim will have been traveling for (x + 1) hours. emiT )sruohni( deepS )h/mkni( ecnatsiD )mkni( hsoJ x miT x 1+ 09 07 09 x (07 x )1+ = speed × time Time is in hours and speed is in kilometers per hour — so all distances will be in kilometers. Josh catches Tim when they have both traveled the same distance. So you need to solve: 90x = 70(x + 1) 90x = 70x + 70 20x = 70 x = 3.5 So it will take Josh 3.5 hours to catch up with Tim. 104104104104104 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 Example Example Example Example Example 22222 Lorraine is driving to a theme park at 45 miles per hour. Twenty minutes after Lorraine leaves, Rachel sets off along the same freeway. If Rachel is traveling at 55 miles per hour, how long does it take her to catch up with Lorraine? Solution Solution Solution Solution Solution Lorraine at 45 mph Rachel at 55 mph Both vehicles are moving in the same direction......but at different speeds. Lorraine left first, but she’s traveling slower. Rachel left afterwards but is traveling faster — so at some point Rachel will catch up. If you call Rachel’s travel time x (hours), then Lorraine’s travel time will be x +( (since Lorraine left 20 minutes (= of an hour) earlier). )1 1 3 3 emiT

)sruohni( deepS )hpmni( ecnatsiD )selimni( = speed × time lehcaR eniarroL x x+1 3 55 54 55 x x+⎛ ⎜⎜⎜ 45 ⎝ ⎞ ⎟⎟⎟⎟ ⎠ 1 3 When Rachel catches up, Rachel and Lorraine have traveled equal x+⎛ ⎜⎜⎜ distances, so 55x = 45 ⎝ 1 3 ⎞ ⎟⎟⎟⎟. So solve this equation to find x: ⎠ 55 x = 45 x + 45 3 10 = x x = 15 3 2 x = 1 1 2 So, Rachel catches up with Lorraine 1 1 2 hours after Rachel left. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 105105105105105 Guided Practice 8. Felipe sets off from Phoenix for Los Angeles, driving at 60 mph. Thirty minutes later an emergency vehicle takes off after Felipe, on the same route, at an average speed of 80 mph. How long will it take the emergency vehicle to overtake Felipe? 9. Lavasha and Keisha travel separately to their grandmother's house, 114 miles away. Keisha leaves from the same place 30 minutes after Lavasha. If Lavasha is traveling at 40 mph and Keisha travels at 55 mph, how long will it take Keisha to catch up with Lavasha? 10. An express train leaves Boston for Washington D.C., traveling at 110 mph. Two hours later, a plane leaves Boston for Washington D.C., traveling at a speed of 550 mph. If the plane flies above the train route, how long will it take the plane to pass the express train? ections ections ving in Opposite Dir ving in Opposite Dir hings Mo h Out for or or or or TTTTThings Mo hings Mo h Out f WWWWWaaaaatctctctctch Out f h Out f ections ving in Opposite Directions hings Moving in Opposite Dir ections ving in Opposite Dir hings Mo h Out f When you’re solving motion questions, you always need to work out which direction each of the objects is moving in. If

things are traveling in opposite directions, you have to think carefully about the equation. Example Example Example Example Example 33333 Bus 1 leaves Bulawayo at 8 a.m. at a speed of 80 km/h. It is bound for Harare, 680 km away. Bus 2 leaves the Harare depot at 8 a.m., heading for Bulawayo along the same highway at a speed of 90 km/h. Calculate at what time the buses pass each other. Solution Solution Solution Solution Solution Bus 1 at 80 km/h Bus 2 at 90 km/h This time the buses are traveling in opposite directions. emiT )sruohni( deepS )h/mkni( ecnatsiD )mkni( 1suB 2suB x x 08 09 08 x 09 x Check it out: Careful — the question asked for the time the buses pass, not for how long they will have been traveling. The initial distance between the two buses is 680 km, and they pass when the distance between them is zero. This means they must pass when the distances they have traveled sum to 680 km. So solve 80x + 90x = 680. This gives 170x = 680, or x = 4. Now you have to say what this value of x means. Since they pass 4 hours after their departure time of 8 a.m., they must pass at noon. 106106106106106 Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 Guided Practice 11. An express train leaves Chicago for Atlanta at 100 mph. At the same time a freight train leaves Atlanta for Chicago at 80 mph. If the distance between the two cities is 720 miles, how long will it take the trains to pass each other? 12. Ernesto, a resident of Los Angeles, sets off to visit his uncle in San Francisco. At the same time, Chang, a resident of San Francisco, sets off to visit a friend in Los Angeles. The road between San Francisco and Los Angeles is 390 miles long. If Ernesto drives at 60 mph while Chang drives at 70 mph, how long will it take for the drivers to pass each other? 13. Mr. and Mrs. Ding leave their home traveling in opposite directions on a straight road. Mrs. Ding drives 10 mph faster than Mr. Ding. After 2 hours, they are 200 miles apart. Find the rates that Mr. and

Mrs. Ding are traveling. Independent Practice 1. A plane leaves Miami for Seattle at 9:00 a.m., traveling at 450 mph. At the same time another plane leaves Seattle for Miami, flying at 550 mph. At what time will the planes pass if the distance between Miami and Seattle is 3300 miles? 2. A jet leaves the airport traveling at a speed of 560 km/h. Another jet, leaving the same airport and traveling in the same direction, leaves 45 minutes later traveling at 750 km/h. About how long will it take for the second jet to overtake the first jet? 3. Two planes depart Los Angeles at the same time. One plane flies due east at 500 mph while the other plane flies due west at 600 mph. How long will it be before the planes are 3300 miles apart? 4. Two planes leave from the same city at the same time. One plane flies west at 475 mph and the other plane flies east at 550 mph. How long will it be before the planes are 4100 miles apart? 5. Two boaters leave a boat ramp traveling in opposite directions. The first boat travels 15 mph faster than the second one. If after 4 hours, the boats are 220 miles apart, how fast are the boats traveling? 6. Bus A leaves Eastport at 7 a.m. at a speed of 40 km/h and is bound for Westport. Bus B leaves Eastport at 7.45 a.m. and drives along the same freeway at 50 km/h. Bus C leaves Westport at 7.45 a.m. traveling towards Eastport along the same freeway at 60 km/h. Westport and Eastport are 220 km apart. Will bus B or bus C pass bus A first? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up As long as you learn the formulas for speed, distance, and time, then these problems are just like solving the rest of the real-life problems in this Section. As always, first set up an equation, then solve. Section 2.5 Section 2.5 Section 2.5 — Consecutive Integer Tasks, Time and Rate Tasks Section 2.5 Section 2.5 107107107107107 TTTTTopicopicopicopicopic 2.6.12.6.1 2.6.12.6.1 2.6.1 Section 2.6

asksasks asksasks estment TTTTTasks estment estment InInInInInvvvvvestment estment estment TTTTTasks InInInInInvvvvvestment asksasks asksasks estment estment estment California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec hniques to aic techniques to hniques to aic tec solvsolvsolvsolvsolve e e e e rate problems, work cent cent per per problems, and per cent percent cent per lems..... lems lems e pre proboboboboblems e pre pr mixtur mixtur mixture pr mixtur lems mixtur What it means for you: You’ll learn what annual interest is, and you’ll solve real-life problems involving interest on investment. Key words: annual interest rate investment Interest is the money you earn by investing — that’s why it’s also known as “return on investment.” Money can be invested in lots of ways (for example, savings accounts, stocks, property, etc.), and each can have a different rate of interest. After One YYYYYearearearearear After One est is the Returetureturetureturn n n n n After One After One est is the R est is the R ual Inter AnnAnnAnnAnnAnnual Inter ual Inter ual Interest is the R After One est is the R ual Inter If your money is invested in a plan that pays interest at a rate of r per year, then the compound interest formula tells you the value of your savings after t years. A = amount including interest A = p(1 + r)t t = length of investment in years p = initial amount invested r = annual interest rate Putting t = 1 shows that after one year, you will have A = p + pr. But p is the original amount you invest (called the principal), and so the amount of interest earned in a single year (I) is equal to pr. Annual interest: I = pr Example Example Example Example Example 11111 Maria invests $5000 in a savings account with an annual

interest rate of 10%. What is the return on her investment at the end of one year? Solution Solution Solution Solution Solution The formula you need is I = pr, where p = $5000. Convert the percent to a decimal: r = 10 100 = 0.10 Substitute in your values for p and r: I = pr = 5000 × 0.10 = 500 So Maria makes $500 in interest on her investment. 108108108108108 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 1. A banker invested $7000 at an annual interest rate of 8%. What would be the return on the investment at the end of the year? 2. A banker invested $5000 at an annual rate of 2.5%. What would be the return on the investment at the end of one year? 3. A banker invested $10,000 at an annual rate of 13.25%. What would be the return on the investment at the end of one year? 4. At the end of one year an investment at 6% earned $795.00 in interest. What amount was invested? 5. At the end of one year an investment at 3.5% earned $28.00 in interest. What amount was invested? 6. At the end of one year an investment of $1250 earned $31.25 in interest. What was the interest rate? 7. At the end of one year an investment of $6000 earned $630 in interest. What was the interest rate? t Once t Once hemes a est in TTTTTwwwwwo Sco Sco Sco Sco Schemes a hemes a est in ou Could Invvvvvest in est in ou Could In YYYYYou Could In ou Could In t Once hemes at Once t Once hemes a est in ou Could In If you invest your money in two different schemes, the total interest over one year is the sum of the interest earned from each of the different schemes. where p1 is the amount invested in Scheme 1 at an interest rate of r1, and p2 is the amount invested in Scheme 2 at an interest rate of r2. I = p1r1 + p2r2 Example Example Example Example Example 22222 Francis has $10,000 to invest for one year. He plans to invest $6000 in stocks, and put the rest in a savings account. If the stocks

pay 10% annually and the savings account pays 8%, how much interest will he make over the year? Solution Solution Solution Solution Solution I = p1r1 + p2r2 where p1 = 6000, r1 = 0.1 (= 10%) p2 = 10,000 – 6000 = 4000, r2 = 0.08 (= 8%) I = p1r1 + p2r2 = (6000 × 0.10) + (4000 × 0.08) = 600 + 320 = 920 So Francis will earn $920 over the year. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 109109109109109 Guided Practice 8. Tyler invested $12,000 in two different savings accounts for one year. One account had $8000 and paid a 10% return annually and the other had $4000 and paid a 12.5% return. How much interest did he earn over the year? 9. Maria invested some money in two different stock accounts for one year. If she invested $5000 in an account with an annual return of 2.5%, how much would she have had to invest at 3% in order to receive $260 interest for the year? 10. Yang is investing the same amount of money into two different savings accounts. He invests in an account returning 10% and an account returning 12% for one year. If he earned $1226.50 in interest, how much did he invest in each account? k Backwkwkwkwkwararararardsdsdsdsds k Bac ou to WWWWWororororork Bac k Bac ou to equire e e e e YYYYYou to ou to equir equir lems R Some Proboboboboblems R lems R Some Pr Some Pr lems Requir k Bac ou to equir lems R Some Pr Some Pr Sometimes you’re given information about total return on investment — and you have to work backwards to figure out how money was invested in the first place. Example Example Example Example Example 33333 A banker invested $12,000 for one year. He invested some of this money in an account with an interest rate of 5%, and the rest of the money in stocks which paid 9% interest annually. How much money did he invest in each plan if the total return from the investments was $700? Solution Solution Solution Solution Solution Call p1 the amount invested

at 5%, and p2 the amount invested at 9%. Let p1 = x, then p2 = 12,000 – x. Also, r1 = 0.05 (= 5%), and r2 = 0.09 (= 9%). Substitute these values into the annual interest formula: I = p1 r1 + p2r2 = 0.05x + 0.09(12,000 – x) = 700 Now you can solve for x: 0.05x + 0.09(12,000 – x) = 700 5x + 9(12,000 – x) = 70,000 Get rid of ficients ficients decimal coef decimal coef Get rid of Get rid of ficients decimal coefficients Get rid of decimal coef ficients decimal coef Get rid of 5x + 108,000 – 9x = 70,000 5x – 9x = 70,000 – 108,000 Gr e termsmsmsmsms e ter e ter oup lik oup lik Gr Gr oup like ter Group lik e ter oup lik Gr Simplify –4x = –38,000 Simplify Simplify Simplify Simplify x = 9500 So he invested $9500 in the account paying 5%, and 12,000 – 9500 = $2500 in stocks. 110110110110110 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 11. Latisha invested a total of $60,000. She invested a part of her money at an annual interest rate of 6% and the rest at 10%. If the total return at the end of the year was $5200, how much was invested at each rate? 12. Robin invested $80,000. He invested some of his money at an annual interest rate at 10% and the rest at 12%. If the total interest earned at the end of one year was $8300, how much was invested at each rate? 13. Leon invests $15,000 in two different accounts. He invests some at a 3% interest rate and the remaining at 4.5%. If he earns $621.00 in interest for one year, how much did he invest at each rate? 14. Fedder invested some money in three different savings accounts. If he invested $7500 in an account with an annual return of 8% and $4000 in an account earning 8.9%,

how much money, to the nearest dollar, did he invest at 8.25% if he earned $1166.30 total interest in a year? est are Re Re Re Re Relaelaelaelaelatedtedtedtedted est ar est ar Inter Inter Amounts of Sometimes TTTTTwwwwwo o o o o Amounts of Amounts of Sometimes Sometimes Interest ar Amounts of Inter est ar Inter Amounts of Sometimes Sometimes In Example 4, instead of relating the sum of the individual amounts of interest to a given total, you have to relate them to each other. Example Example Example Example Example 44444 A banker had $80,000 to invest. She put some of the money in a deposit account paying 10% a year, and invested the rest at 8%. The annual return on the money invested at 8% was $100 more than the return on the 10% investment. How much money did she invest at each rate? Solution Solution Solution Solution Solution Call p1 the amount invested in the deposit account, and p2 the amount invested at 8%. Let p1 = x, then p2 = 80,000 – x. Also r1 = 0.1 (= 10%), and r2 = 0.08 (= 8%). This time, you know the return from the 8% plan was $100 more than the return from the 10% plan. Writing this as an equation gives: p1r1 + 100 = p2r2 Substitute all the information into this formula, then solve for x: 0.1x + 100 = 0.08(80,000 – x) 10x + 10,000 = 8(80,000 – x) 10x + 10,000 = 640,000 – 8x 18x = 630,000 x = 35,000 So she invested $35,000 at a rate of 10%, and 80,000 – 35,000 = $45,000 at a rate of 8%. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 111111111111111 Guided Practice 15. Dorothy divided $14,500 among two accounts paying 11% and 8% interest annually. The interest earned after 1 year in the 8% account was $455 less than that earned in the 11% account. How much money was invested in each account? 16. Michael invested $18,000 in two different accounts. The account paying 14% annually earned $1357.10 more than

the one earning 15% interest. How much was invested in each account? 17. Lavasha invested $16,000 among two different accounts paying 10% and 12% in one year. If she earned twice as much interest in the account paying 12%, how much did she invest in each account? Independent Practice 1. In one year, an investment at 8% interest earned $1060. How much money was invested? 2. Louise invested $25,000 in two different accounts. She invested some money at a 4.5% interest rate and the remaining amount at a 3.25% interest rate. If she earned $1062.50 in interest in a year, how much did she invest at each interest rate? 3. Mackey invested some money in an account paying 6% interest for one year. She invested $1000 more than this amount in an account paying 8.5%. How much did she invest in total if the total interest earned in the year was $737.50? 4. Maxwell invested a total of $100,000. He invested his money in three different accounts. He invested $35,000 at 10% annual interest rate, $42,000 at 8% annual interest rate, and the remainder at 14% interest rate. How much interest did Maxwell receive in one year? 5. Garrett invested $12,000 at an annual interest rate of 6%. How much money would Garrett have had to invest at 10% so that the combined interest rate for both investments was 7% over the year? 6. Two business partners decided to borrow $30,000 start-up money for their business. One borrowed a certain amount at a 16.5% interest rate and the other one borrowed the rest of the money at a 12.5% interest rate. How much money did each business partner borrow if the total interest amount at the end of the year was $4150? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Any problems that involve adding together investments with different interest rates are actually a type of mixture problem. In the next Topic you’ll deal with mixture problems not involving money. 112112112112112 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 TTTTTopicopicopicopicopic 2.6.22.6.2 2.6.22.

6.2 2.6.2 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec hniques to aic techniques to hniques to aic tec solvsolvsolvsolvsolve e e e e rate problems, work cent cent per per problems, and per cent percent cent per lems..... lems lems e pre proboboboboblems e pre pr mixtur mixtur mixture pr mixtur lems mixtur What it means for you: You’ll set up and solve equations involving mixtures. Key words: mixture task volume mass concentration ingredients asksasks asksasks Mixture e e e e TTTTTasks Mixtur Mixtur Mixtur Mixtur Mixture e e e e TTTTTasks asksasks asksasks Mixtur Mixtur Mixtur Mixtur The interest problems in Topic 2.6.1 were actually mixture problems because you had to add together returns from investments with different interest rates. In this Topic you’ll see some mixture problems that don’t involve money. olume” olume” Amount per Unit Mass/V Amount per Unit Mass/V asks Use “ Some TTTTTasks Use “ asks Use “ Some Some olume” Amount per Unit Mass/Volume” asks Use “Amount per Unit Mass/V olume” Amount per Unit Mass/V asks Use “ Some Some Mixtures are made by combining ingredients. In math, mixture problems involve using algebra to work out the precise amounts of each ingredient. To do this, you have to make use of an equation of the form: concentration = amount of substance total volume (or total mass) Check it out: Concentration means “amount per liter,” or “amount per kilogram,” etc. The “amount of substance” could be given to you as a volume or as a mass — but it doesn’t make any difference to the math. Just use whatever units they give you in the question. Concentration can also be described as a percent: percent of ingredient = amount of ingredient total weight or volume

of mixture × 100 The following example doesn’t involve a mixture, but it shows how the above formula can be used. Example Example Example Example Example 11111 If a 1 kg bag of granola consists of 15% raisins (by mass), how many grams of raisins are there in the bag? Solution Solution Solution Solution Solution The formula tells you: Percent of raisins = total amount of raisins total amount of granola 0.15 = total amount of raisins 1000 So total mass of raisins = 1000 × 0.15 = 150 grams Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 113113113113113 Guided Practice 1. A 19-ounce can of a disinfectant spray consists of 79% ethanol by weight. How many ounces of pure ethanol does the spray contain? 2. A 2 kg bag of mixed nuts contains 20% walnuts (by mass). How many grams of walnuts are in the bag? 3. What is the percent of juice if a 2 liter bottle of juice drink has 250 ml of juice? 4. A 142 g bottle of baby powder contains 15% zinc oxide. How many grams of zinc oxide are in the powder? 5. What is the concentration of a hydrogen peroxide solution if 500 ml of the solution contains 15 ml of hydrogen peroxide? tion as a Pererererercent cent cent tion as a P te the Concentraaaaation as a P tion as a P te the Concentr te the Concentr ou Can Calcula YYYYYou Can Calcula ou Can Calcula cent ou Can Calculate the Concentr cent tion as a P te the Concentr ou Can Calcula This example shows how you can apply the formula to mixtures of various concentrations. Example Example Example Example Example 22222 If you combine 5 liters of a 10% fruit juice drink and 15 liters of a 20% fruit juice drink, what percent of fruit juice do you have? Solution Solution Solution Solution Solution Here, you have to use your formula more than once, but the principle is the same. Concentration of fruit juice = volume of fruit juice volume of fruit drink For the 10% fruit drink, concentration = 0.1 = x 5, where x is the volume of fruit juice. The volume of fruit juice in the 10% solution is x = 0.1 × 5 = 0.5 liters. And for the

20% fruit drink, concentration = 0.2 = y 15, where y is the volume of fruit juice. The amount of fruit juice in the 20% solution is y = 0.2 × 15 = 3 liters. Also, the total volume of the combined fruit drink is 5 + 15 = 20 liters. The total amount of fruit juice in the mixture is x + y = 0.5 + 3 = 3.5 liters. Therefore, the concentration of the final mixture is: Concentration = volume of fruit juice volume of fruit drink =. 3 5 20 = 0.175 = 17.5% 114114114114114 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 6. If you combine 10 liters of juice cocktail made with 100% fruit juice with 5 liters of another juice cocktail containing 10% fruit juice, what percent of the 15 liters of juice cocktail will be fruit juice? 7. There is 4% hydrogen peroxide in a 475 ml hydrogen peroxide solution. The 4% solution is mixed with 375 ml of a hydrogen peroxide solution containing 15% hydrogen peroxide. What percent of hydrogen peroxide is in the combined solution? 8. Tina mixes 4.75 mg of a lotion containing 1% vitamin E with 2 mg of a lotion containing 7% vitamin E. What is the percentage of vitamin E in the combined lotion? Independent Practice 1. A 12 fl. oz. bottle of lotion contains 2.5% lavender. How many fluid ounces of lavender are contained in the bottle of lotion? 2. A 2.5 fl. oz. bottle of nose spray contains 0.65% sodium chloride. What volume of sodium chloride does the spray contain? 3. A 5-gallon car radiator should contain a mixture of 40% antifreeze and 60% water. What volumes of water and antifreeze should the radiator contain? 4. A 1.75 ml bottle of medicine contains 80 mg of an active ingredient per 0.8 ml of medicine. How many milligrams of active ingredient does the bottle contain? 5. A 473 ml bottle of rubbing alcohol contains 331.1 ml of isopropyl alcohol. What percentage of the bottle is isopropyl alcohol? 6. Two liters of 100% pure pineapple juice is mixed with 2 liters of soda water and 1 liter of orange juice to make

a party punch. What percentage of the party punch is pineapple juice? 7. A 2 kg bag of walnuts is mixed with 3 kg of pecans, 4 kg of hazelnuts, and 1 kg of brazil nuts to make a 10 kg bag of mixed nuts. What percentage of the 10 kg bag of nuts is pecans and hazelnuts? 8. 473 ml of 70% rubbing alcohol is combined with 473 ml of 90% rubbing alcohol and 473 ml of water. What is the percentage of rubbing alcohol in the mixture? 9. Two liters of juice cocktail are combined with 6 liters of a different juice cocktail to make 8 liters of juice cocktail containing 50% juice. If the 6 liters of juice cocktail contains 33 percentage of fruit juice is the 2 liters of juice cocktail? 1 3 % fruit juice, what ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up It’s a good idea to memorize the two formulas at the start of this Topic — knowing them will make it a lot easier to deal with any mixture tasks involving liquids or ingredients. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 115115115115115 TTTTTopicopicopicopicopic 2.6.32.6.3 2.6.32.6.3 2.6.3 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec hniques to aic techniques to hniques to aic tec solvsolvsolvsolvsolve e e e e rate problems, work cent cent per per problems, and per cent percent cent per lems..... lems lems e pre proboboboboblems e pre pr mixtur mixtur mixture pr mixtur lems mixtur What it means for you: You’ll set up and solve equations involving mixtures. Key words: mixture task volume mass concentration ingredients asksasks asksasks e Mixture e e e e TTTTTasks e Mixt

ur e Mixtur MorMorMorMorMore Mixtur e Mixtur e Mixture e e e e TTTTTasks MorMorMorMorMore Mixtur asksasks asksasks e Mixtur e Mixtur e Mixtur Remember that a mixture is made up of only the original ingredients. Nothing can mysteriously appear in the final mixture that was not part of the original ingredients, and nothing can disappear either. t Changeeeee t Chang edient Doesn’’’’’t Chang t Chang edient Doesn h Ingrrrrredient Doesn edient Doesn h Ing h Ing Eac Eac Amount of Amount of otal TTTTThe he he he he TTTTTotal otal Each Ing Amount of Eac otal Amount of t Chang edient Doesn h Ing Eac Amount of otal total amount of each substance in the ingredients = amount of that substance in the final mixture So if you know the amount of a substance in each of the ingredients you are mixing, you can always calculate the amount of that substance in the final mixture. Similarly, if you know the amount of a substance in one ingredient and in the final mixture, you can find out how much was in the other ingredient(s). Example Example Example Example Example 11111 There are 1000 gallons of water in a wading pool. The water is 5% chlorine. What quantity of a 65% chlorine solution would need to be added to bring the pool’s chlorine concentration up to 15%? Solution Solution Solution Solution Solution Ingredient 1: You know the volume and concentration of the water in the wading pool. Volume = 1000 gallons. Concentration of chlorine = 0.05 (= 5%). So the amount of chlorine in the wading pool to begin with is given by: Original amount of chlorine = volume × concentration = 1000 × 0.05 = 50 Ingredient 2: You also know that the concentration of the chlorine solution to be added is 0.65 (or 65%). So if you call the volume of the added chlorine solution x, then: The amount of chlorine added is 0.65x (= volume × concentration) Mixture: This means the final volume of water in the wading pool is 1000 + x, and so the final amount of chlorine in the pool is 0.15(1000 + x). Therefore 50 + 0.65x = 0.15(1000 + x). Now solve for x: 50 + 0.65x = 150 + 0.

15x 0.5x = 100 x = 200 Therefore 200 gallons of the 65% solution are required. 116116116116116 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 1. A pest control company has 200 gallons of 60% pure insecticide. To create a 70% pure insecticide solution, the company must mix the 60% pure insecticide solution with a 90% pure insecticide solution. How many gallons of the 90% pure insecticide solution should the company mix with the existing solution? 2. A chemist needs to dilute a 60% citric acid solution to a 20% citric acid solution. She needs 30 liters of the 20% solution. How many liters of the 60% solution and water should be used? (Hint: water has 0% acid.) 3. An alloy containing 40% gold is mixed with an 80% gold alloy to get 1000 kilograms of an alloy that is 50% gold. How many kilograms of each alloy are used? h Substance h Substance Eac Eac Amount of Amount of or the or the ula f s a Fororororormmmmmula f ula f s a F TTTTTherherherherhere’e’e’e’e’s a F s a F h Substance Each Substance Amount of Eac or the Amount of ula for the h Substance Eac Amount of or the ula f s a F Multiplying the volume by the concentration gives you the amount of substance in each solution. And since the total amount of substance is the same before and after, you can use this handy formula: c1v1 + c2v2 = cv where c1 and v1 are the concentration and volume of the first ingredient, c2 and v2 are the concentration and volume of the second ingredient, c and v are the concentration and volume of the mixture of the two. and The next example is quite similar to Example 1, but it uses the formula shown above. Example Example Example Example Example 22222 A pharmacist has 500 cm3 of a 30% acid solution. He replaced x cm3 of the 30% solution with a 70% acid solution to get 500 cm3 of a new 40% acid solution. What volume of the 30% acid solution did he replace with the 70% acid solution? Solution Solution Solution Solution Solution Assume that the pharmacist poured

away x cm3 of the 30% solution and replaced it with 70% solution. Then 0.3(500 – x) + 0.7x = 0.4(500) 150 + 0.4x = 200 x = 125 So the pharmacist replaced 125 cm3 of the 30% acid solution. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 117117117117117 Guided Practice 4. A doctor prescribes 20 grams of a 62% solution of a generic medicine. The pharmacist has bottles of 50% and 70% solutions in stock. How many grams of each solution can the pharmacist use to fill the prescription for the patient? 5. Fifty pounds of special nuts costing $4.50 per pound were mixed with 120 pounds of generic nuts that cost $2 per pound. What is the value of each pound of the nut mixture? 6. Cherry juice that costs $5.50 per liter is to be mixed with 50 liters of orange juice that costs $2.50 per liter. How much cherry juice should be used if the value of the mixture is to be $3.50 per liter? 7. Sixty liters of a syrup that costs $8.50 per liter were mixed with honey that costs $4.75 per liter. How many liters of honey were used if the value of the mixture is $6 per liter? After” TTTTTotals otals otals After” After” and “ te the “Befororororore”e”e”e”e” and “ and “ te the “Bef AlAlAlAlAlwwwwwaaaaays Rys Rys Rys Rys Relaelaelaelaelate the “Bef te the “Bef otals and “After” otals After” and “ te the “Bef Example Example Example Example Example 33333 Marie ordered 40 pounds of walnuts at $1.50 per pound. She mixed this with hazelnuts costing $1.00 per pound, and sold the mixed nuts at $1.25 per pound. How many pounds of hazelnuts did she use, if she broke even? Solution Solution Solution Solution Solution Here, your “amount per unit weight” formula is: Price per pound (p) = v total value ( ) ) weight

( w Check it out: Notice that p1w1 + p2w2 = pw. Compare this to the formula on the previous page — it’s essentially the same. First write down what you know about the original ingredients and the final mixture (calling the quantity you need to find x, for example). Then relate a “before” total to the “after” total — so the total value of the ingredients is the same as the total value of the mixture — v1 + v2 = v. $1.50-per-pound walnuts: weight = 40 lb. Total value = 1.50 × 40 = 60. $1.00-per-pound hazelnuts: weight = x lb. Total value = 1.00 × x = x. Mixed ($1.25) nuts: weight = (40 + x) lb. Total value = 1.25(40 + x). To break even, the value of the mixed nuts must equal the sum of the values of the original nuts. That is, x + 60 = 1.25(x + 40). Solve this to find x: x + 60 = 1.25(x + 40) x + 60 = 1.25x + 50 –0.25x = –10 x = 40 Therefore she used 40 pounds of hazelnuts. 118118118118118 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 Guided Practice 8. Mrs. Roberts owns a pet store and wishes to mix 4 pounds of cat food worth $2.00 per pound with another cat food costing $3.00 per pound. How much of the $3.00 per pound cat food should be used if the mixture is to have costed $2.75 per pound? 9. A local garden center makes a medium grade plant seed by mixing a low grade plant seed bought for $2.50 a pound with 14 pounds of superior quality seed bought for $5.00 a pound. How much of the low grade plant seed needs to be mixed with the superior quality seed if it is to be worth $3.50 per pound? 10. A tea blend is made by using 2 kg of $2.00 per kg tea leaves and another tea leaf costing $4.00 per kilogram. How many kilograms of the $4.00 tea is needed to make a tea blend worth $3.75 per

kilogram? 11. A landscaper wants to make a blend of grass seed using 300 pounds of $0.40 per pound grass seed and another seed costing $0.75 per pound. How much of the $0.75 seed does the landscaper need to make a $0.60 per pound blend? ut the Same Maththththth ut the Same Ma ut the Same Ma xt — b xt — b ent Conte A Difffffferererererent Conte ent Conte A Dif A DifA Dif xt — but the Same Ma ent Context — b ut the Same Ma xt — b ent Conte A Dif Example Example Example Example Example 44444 A retailer mixed two fruit drinks. He mixed 30 liters of a fruit drink that cost him $1.90 per liter with an unknown amount of a fruit drink that cost him $2.50 per liter. If the mixture’s ingredients cost $2.40 per liter, what volume of the $2.50-per-liter drink did he use? Solution Solution Solution Solution Solution This time, use: price per liter (p) = c total cost ( ) v volume ( ) $1.90-per-liter drink: volume = 30 liters. Total cost = 1.90 × 30 = 57. $2.50-per-liter drink: volume = x liters. Total cost = 2.50x. Mixed ($2.40) drink: volume = (30 + x) liters. Total cost = 2.40(30 + x). But the cost of the mixture is the sum of the ingredients’ costs, so 57 + 2.50x = 2.40(30 + x). Solving this for x gives: 57 + 2.5x = 72 + 2.4x 0.1x = 15 x = 150 So he used 150 liters of the $2.50-per-liter fruit drink. Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 119119119119119 Guided Practice 12. A pet store owner mixed two different bird seeds. She mixed 20 pounds of a seed that costs $2.25 per pound with 50 pounds of another seed. If the mixture of seeds cost $2.75 per pound, how much per pound did the 50 pounds of seed cost? 13. Apples cost $0.80

per pound and grapes cost $1.10 per pound. Michael wishes to make a fruit tray using only apples and grapes that costs $1.00 per pound. If Michael has 8 lbs of apples, how many pounds of grapes are needed? 14. A roast coffee blend costing $6.60 per pound is made by mixing a bean that costs $2.00 per pound with another one that costs $7.00 per pound. If 5 pounds of the $2.00 bean are used, how many pounds of the $6.60 beans would be produced? Independent Practice 1. Dave and his parents went to the movies. Adult tickets cost $7.00 and children’s tickets cost $3.00. For the screening of one movie, 500 tickets were sold for a total of $2000. Find the number of each kind of ticket that was sold for the movie. 2. A 70% salt solution was diluted with purified water to produce about 50 liters of 55% salt solution. Approximately how much purified water was used? 3. A chemist wants to dilute a 60% boric acid solution to a 15% solution. He needs 30 liters of the 15% solution. How many liters of the 60% solution and water must the chemist use? 4. A student wants to dilute 35 liters of a 35% salt solution to a 16% solution. How many liters of distilled water does the student need to add to the 35% salt solution to obtain the 16% salt solution? 5. A tea blend is made by mixing 2 kg of a $2.00 per kg tea with 5 kg of another tea. If the total cost of the ingredients in the blend is $24.00, what is the price per kg of the 5 kg of tea? 6. How many liters of a 100% alcohol solution must be mixed with 20 liters of a 50% solution to get a 70% solution? 7. A chemist has 8 liters of a 30% solution of a compound. How much of a 100% solution of the compound must be added to get a 50% solution? 8. Milk that contains 2% fat is mixed with milk containing 0.5% fat. How much 0.5% fat milk is needed to be added to 10 gallons of 2% fat milk to obtain a mixture of milk containing 1% fat? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRR

ound Up ound Up ound Up Over the last couple of Topics you’ve seen lots of examples of mixture tasks. The most difficult part of a mixture task is setting up the equation — once you’ve done that it’s all a lot easier. 120120120120120 Section 2.6 Section 2.6 Section 2.6 — Investment and Mixture Tasks Section 2.6 Section 2.6 TTTTTopicopicopicopicopic 2.7.12.7.1 2.7.12.7.1 2.7.1 Section 2.7 WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec aic techniques to hniques to hniques to aic tec solvsolvsolvsolvsolve re re re re raaaaate pr lems,,,,, w w w w worororororkkkkk lems lems te proboboboboblems te pr te pr lems te pr prprprprproboboboboblems lems,,,,, and percent lems lems lems mixture problems..... What it means for you: You’ll solve combined work rate problems by calculating each work rate in turn. Key words: work rate combined work rate Check it out: Problems like this usually assume that both people are working all the time (rather than one of them stopping early). Work problems are similar to the mixture problems you saw in Section 2.6. Again, the only new thing is that there are a few formulas that you need to know in order to set up the equations. About Speeds of WWWWWorororororkingkingkingkingking About Speeds of asks are e e e e About Speeds of About Speeds of asks ar ted TTTTTasks ar asks ar ted WWWWWororororork-Rk-Rk-Rk-Rk-Relaelaelaelaelated ted About

Speeds of asks ar ted This area of math is concerned with calculating how long certain jobs will take if the people doing the job are working at different rates. Example Example Example Example Example 11111 John takes 1 hour to deliver 100 newspapers, and David takes 90 minutes to deliver 100 newspapers. How long would it take them to deliver 100 newspapers between them? Assume that they work independently, that they both start at the same time, and that they are both working the whole time. Solution Solution Solution Solution Solution It’s tempting just to work out how long it takes John and David to deliver half the papers each. But that does not take into account the fact that they are working as a team — and since John works faster than David, he will deliver more papers. Instead, you have to figure out how quickly they work as a team — not just as two individuals. The problem is that you don’t know how many papers each of them delivers — you only have the information given in the question, which is shown in this diagram: John David 0 30 100 newspapers 60 90 mins 100 newspapers You need to work out their rate of delivering the papers. Based on how long it takes each person to deliver 100 papers, you can calculate how many papers are delivered every minute in total. If John can deliver 100 newspapers in 1 hour (60 minutes), he can deliver 100 ÷ 60 = 5 3 newspapers in 1 minute. If David can deliver 100 newspapers in 90 minutes, 10 9 10 9 he can deliver 100 ÷ 90 = newspapers in 1 minute. So in total 5 3 15 + = + = 9 10 9 25 9 newspapers are delivered each minute. This means that 100 papers will take 100 ÷ 25 9 = 36 minutes to deliver. Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 121121121121121 Guided Practice 1. Akemi can weed the garden in 4 hours. Keira can weed the garden in 12 hours. How long would it take the two of them to weed the garden together? 2. Martha can clean a statue in 15 hours. Chogan can clean the same statue in 9 hours. If they clean the statue together, how long would it take them to finish? 3. A carpenter can build a cabinet in 10 hours. Her assistant can build the same cabinet in 15 hours. How long would it take them to build the cabinet together, assuming they can work independently? k Done �

� TimeTimeTimeTimeTime k Done ÷ te = WWWWWororororork Done ÷ k Done ÷ te = WWWWWororororork Rk Rk Rk Rk Raaaaate = te = k Done ÷ te = Work rate is the amount of work carried out per unit time. The work completed can be given as a fraction. For example, if only half the task is completed, write 1 2. If the whole task is completed, write 1. Work rate = work completed time When you are solving a problem like this, you need to start by identifying the “work completed” and the time that it took. The “work completed” part in the previous example was quite straightforward. The example below is not so simple — make sure you understand each step. Example Example Example Example Example 22222 An inlet pump can fill a water tank in 10 hours. However, an outlet pump can empty the tank in 12 hours. An engineer turns on the inlet pump but forgets to switch off the outlet pump. With both pumps running, how long does it take to fill the tank? Solution Solution Solution Solution Solution Inlet pump’s work rate = work completed time = 1 10 Outlet pump’s work rate = work completed time = 1 12 The two pumps are working in opposite directions. The combined work rate is the difference between the two rates. So the combined rate = 1 10 1 − = 12 ⋅ 6 1 60 − ⋅ 5 1 60 = − 6 5 60 = 1 60 So after 60 hours, the net amount of water that has gone in is 1 tank’s worth — which means the tank will be full. So the answer is 60 hours. Check it out: The inlet pump’s “work completed” is 1 tank of water ininininin, and the outlet pump’s “work completed” is 1 tank of water outoutoutoutout. Check it out: The fractions have been subtracted using the least common multiple (LCM) of 10 and 12, which is 60. See Topic 2.3.1 for more on LCMs. 122122122122122 Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 Guided Practice 4. Megan, Margarita, and James work in a fast-food restaurant after school. It takes

Megan 6 hours, Margarita 3 hours, and James 4 hours to clean the utensils individually. How long would it take the three of them to clean the utensils if they worked together? 5. A bathtub can be filled from a faucet in 10 minutes. However, a pump can empty the bathtub in 15 minutes. If the faucet and the pump are on at the same time, how long will it take to fill the bathtub? 6. Isabel can fence the family three-acre lot in 8 days. If it would take José 6 days and Marvin 12 days to fence the same lot, how long would it take the three people to fence the lot together? Independent Practice 1. Bill can paint a house in 5 days and Samantha can paint the same house in 7 days. How long will it take them, working together, to paint the house? 2. There are two drains in a tub filled with water. One drain would empty the tub in 3 hours, if opened. The other drain would empty the tub in 4 hours. If both drains are opened at the same time, how long will it take to empty the tub? 3. A tub has 2 drains. One drain can empty the full tub in 20 minutes. The other drain can empty half the tub in 30 minutes. How long will it take to empty the full tub if both drains are opened together? 4. Jose can paint a house in 4 days. Leroy can paint the same house in 5 days. How long will it take Jose and Leroy, working together, to paint the house? 5. Sam, Doris, and Alice are weeding a yard. Working by himself, Sam could weed the yard in 1½ hours. Doris could do the same yard by herself in 1 hour and Alice could do it in only 45 minutes. How long will it take the three of them working together to weed the yard? 6. A barrel has two filling pipes and one draining faucet. One pipe could fill the barrel in 2 hours and the other could fill it in 2½ hours. The faucet could empty the barrel in 4 hours. How long will it take to fill the barrel if both pipes are filling and the draining faucet is opened? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The very first thing you need to do with a work rate problem

is look through the word problem to identify the work completed and the time taken. Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 123123123123123 TTTTTopicopicopicopicopic 2.7.22.7.2 2.7.22.7.2 2.7.2 California Standards: Students applpplpplpplpplyyyyy Students a Students a 15.0: 15.0: 15.0: Students a 15.0: Students a 15.0: algalgalgalgalgeeeeebrbrbrbrbraic tec hniques to hniques to aic tec aic tec aic techniques to hniques to hniques to aic tec solvsolvsolvsolvsolve re re re re raaaaate pr lems,,,,, w w w w worororororkkkkk lems lems te proboboboboblems te pr te pr lems te pr prprprprproboboboboblems lems,,,,, and percent lems lems lems mixture problems..... What it means for you: You’ll solve combined work rate problems by first calculating the combined work completed. Key words: work rate combined work rate Check it out: The fractions have been multiplied by the LCM of 10, 12, and x, to get rid of the fractional coefficients. Combined WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete Combined Combined Combined Combined Combined WWWWWororororork Rk Rk Rk Rk Raaaaatetetetete Combined Combined Combined Combined In Topic 2.7.1 you calculated each work rate in turn, then added or subtracted them. This Topic contains a method for combining each piece of information directly into one equation — which can make the whole calculation a lot more straightforward. All the Infororororormamamamamationtiontiontiontion All the Inf All the Inf lude lude tes Inc Combined WWWWWororororork Rk Rk Rk Rk Raaaaates Inc tes Inc Combined Combined lude All the Inf tes Include All the Inf lude tes Inc Combined Combined Another way to approach work rate problems is to put all the information from the question directly into this equation, which can then be solved:

Combined work rate = combined work completed combined work completed total time total time Here’s the same problem that you saw in Example 2 in Lesson 2.7.1 — but now using the new method. Example Example Example Example Example 11111 An inlet pump can fill a water tank in 10 hours. However, an outlet pump can empty the tank in 12 hours. An engineer turns on the inlet pump but forgets to switch off the outlet pump. With both pumps running, how long does it take to fill the tank? Solution Solution Solution Solution Solution Combined work rate = total water in the tank time taken to fill the tank Let x = number of hours to fill the tank, and then substitute everything you know into the formula: 1 10 1 − = 12 1 x 1 tank’s worth of water total time (left-hand side of equation = combined rate) Then rearrange to solve: 60 x ⎛ ⎜⎜⎜ ⎝ 1 10 x 60 10 − − ⎞ ⎟⎟⎟ = ⎠ = 1 12 x 60 12 x x 60 x 60 x 6x – 5x = 60 x = 60 So it takes 60 hours to fill the tank. 124124124124124 Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 Example Example Example Example Example 22222 Jesse can paint a wall in 6 hours. Melinda can paint the same wall in 4 hours. How long would it take the two of them to paint the wall together if they worked independently and started at the same time? Solution Solution Solution Solution Solution Jesse can complete the whole task in 6 hours, so her work rate = 1 6. Melinda can complete the whole task in 4 hours, so her work rate = 1 4. You need to find out how many hours they would take to do the work together. They are working on the same wall, so add the contribution from each person: Combined work rate = 1 6 + 1 4 Use x for the hours it takes in total, and write out the equation for combined work rate. 1 x 1 6 1 + = 4 ⎞ ⎟⎟⎟ = ⎠ ⎛ ⎜⎜⎜ ⎝ x + 12 1 4 1 6 12 x 2x + 3x = 12 5x = 12 x and x Multiply by by

by by by 12y 12y 12y 12y 12x,,,,, the L Multipl Multipl the L the L CM of CM of 6, 6, 4, 4, and and the LCM of CM of 6, 6, 4, 4, and Multipl Multipl the L CM of 6, 4, and hen solve fe fe fe fe for or or or or x TTTTThen solv hen solv hen solv hen solv x = 2 2 5 Therefore painting the wall would take Jesse and Melinda 2 2 5 hours = 2 hours and 24 minutes. Example Example Example Example Example 33333 Liza can dig a garden in 7 hours alone. If Marisa helps her, they finish all the digging in just 3 hours, working independently. How long would it take Marisa to dig the garden alone? Solution Solution Solution Solution Solution Liza can complete the whole task in 7 hours, so her work rate = 1 7. Together they can complete the whole task in 3 hours, so their combined work rate = 1 3. Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 125125125125125 Example 3 continueduedueduedued Example 3 contin Example 3 contin Example 3 contin Example 3 contin You need to find out how fast Marisa could do the work on her own. Use x for the hours it takes Marisa, so her work rate = 1 x. This time you know the combined work rate, but not Marisa’s work rate. The formula is still: combined work rate = combined work completed total time. 1 3 rite out the equationtiontiontiontion rite out the equa rite out the equa WWWWWrite out the equa rite out the equa 7, x,,,,, and 3 Multiply by by by by by 21y 21y 21y 21y 21x,,,,, the L 7, 7, CM of CM of and 3 and 3 the L the L Multipl Multipl CM of 7, and 3 the LCM of 7, CM of and 3 the L Multipl Multipl hen solve fe fe fe fe for or or or or x hen solv hen solv TTTTThen solv hen solv 1 7 + =1 x ⎞ ⎟⎟⎟ = ⎠ ⎛ ⎜⎜⎜ ⎝ + 1 x 1 7 x 21 x 21 3

3x + 21 = 7x 4x = 21 x = 5 1 4 So Marisa can dig the garden in 5 1 4 hours = 5 hours and 15 minutes. Guided Practice 1. A pump can fill a fuel tank in 30 minutes. A second pump can fill the same tank in 60 minutes. How long would it take to fill the fuel tank if both pumps were filling the tank together? 2. Machine A can pack 50 crates of canned dog food in 15 minutes. When Machine A and Machine B are working at the same time, they can pack 50 crates of canned dog food in 9 minutes. How long would it take Machine B to pack 50 crates of canned dog food by itself? 3. Three student-service workers are cataloging books in a school library. Joaquin can catalog the books in 8 hours, Caroline can do the same job in 4 hours, and Joshua can do it in 6 hours. If the three students work together, how long will it take them to finish cataloging the books? 4. A central heater can warm a house in 24 minutes. When the central heater and a floor heater are used together they can warm the house in 16 minutes. How long would the floor heater take to warm the house alone? 5. When Dionne’s cell phone has been fully charged, it can be used for 5 hours before its battery runs out. When its battery runs out, the phone can be fully charged in 3 hours if it is not in use. Dionne’s phone’s battery has run out. If she sets it to recharge, and uses the phone constantly while it is recharging, how long will it take to fully charge her phone? 126126126126126 Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 Independent Practice 1. A window cleaner can clean the windows of a house in 5 minutes and his trainee can clean the same windows in 20 minutes. How long would it take the window cleaner and his trainee to clean the windows together? 2. Susan and John are hired to stuff envelopes with parent notices about an upcoming school event. If the task would take Susan 120 minutes and John 90 minutes individually, how long would it take the two of them to stuff the envelopes together? 3. Lorraine can tile a room in 20 hours. Juan can tile the same room in 30 hours, and Oliver can tile it in 40 hours. How long would it take

them to tile the room if they worked together? 4. A faucet can fill a barrel in 2 hours. However, there is a hole in the bottom of the barrel that can empty it in 6 hours. Without knowing of the hole, Leo tries to fill the barrel. How long will it take? 5. Tyrone, Jerry, and Dorothy are painting a fence surrounding a field. Tyrone could paint the fence in 4 days by himself. Jerry could paint the same fence, but it would take him 6 days if he did it by himself. Dorothy could paint the fence in 3½ days. How long should it take the three of them, working together, to paint the fence? 6. Michael can decorate the cafeteria for a dance in 4 hours. Emily can decorate the cafeteria in 3 hours. Eylora can decorate the cafeteria in 1½ hours. If the three of them have 1 hour, working together, to decorate for the party, will it be done in time? 7. When turned on, a faucet can fill a tub in 3 hours. The tub has 2 drains — the first can empty the tub in 8 hours and the second can empty the tub in 7 hours. If the faucet is turned on while both drains are open, how long will it take to fill the tub? 8. Manuel and Anita have extensive gardens. Manuel can mow all the lawns with his lawn mower in 5 hours. If Anita helps him with her lawn mower, they can mow all the lawns in 2 hours. How long would it take Anita to mow all the lawns alone with her lawn mower? 9. Albert and Po work in a factory that makes leotards. Albert cuts out the material, and Po stitches the side seams. Albert can cut the parts for 1 box of leotards in 9 hours, and Po can stitch the side seams for 1 box of leotards in 27 hours (including any breaks). If Albert and Po start work at the same time, how long will it be before there is 1 full box of leotards cut and waiting for their side seams to be stitched? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up These work rate problems often include a lot of information, and it’s easy to get the values mixed up. It’s a good idea to

check that your solutions make sense in the original word problem. Section 2.7 Section 2.7 Section 2.7 — Work-Related Tasks Section 2.7 Section 2.7 127127127127127 TTTTTopicopicopicopicopic 2.8.12.8.1 2.8.12.8.1 2.8.1 Section 2.8 tions tions alue Equa alue Equa Absolute VVVVValue Equa Absolute Absolute tions alue Equations tions alue Equa Absolute Absolute Absolute VVVVValue Equa tions tions alue Equa alue Equa Absolute Absolute alue Equations tions alue Equa Absolute tions Absolute California Standards: 3.0:3.0:3.0:3.0:3.0: Students solv Students solveeeee Students solv Students solv Students solv tions tions equa equa tions and inequalities equations equa tions equa inininininvvvvvolving a alues..... alues alues bsolute v bsolute v olving a olving a bsolute values olving absolute v alues bsolute v olving a What it means for you: You’ll solve equations involving absolute values. Key words: absolute value Check it out: In other words, an absolute value is never negative. You already saw absolute values on the number line in Topic 1.2.3. In this Topic you’ll see that absolute values can turn up in equations too — and you’ll learn how to get rid of them to solve the equations. ositivvvvveeeee ositiositi ositi alue is AlAlAlAlAlwwwwwaaaaays Pys Pys Pys Pys Positi alue is Absolute VVVVValue is alue is Absolute TTTTThe he he he he Absolute Absolute alue is Absolute The absolute value of x is defined as: x if x is positive or zero, and –x if x is negative. The absolute value of any real number, x, is written as |x|. Another way to think about it is that the absolute value is the distance of a number from zero on the number line. For example, the distance between 0 and 9 is the same as the distance between 0 and –9. This can be written as |9| = |–9| = 9. –

9 9 –10 –9 –8 –7 –6 –5 –4 –3 –2 – 10 –3 3 The distance between 0 and 3 is the same as the distance between 0 and –3. This can be written as |3| = |–3| = 3. Guided Practice Find the distance that each letter is from zero: B D –5 –4 –3 –2 –. A 3. C Simplify: 5. |–9| 7. –|–2| 2. B 4. D 6. |–20| 8. –|–7| 128128128128128 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 tions Havvvvve e e e e TTTTTwwwwwo Solutions o Solutions o Solutions tions Ha tions Ha alue Equa Absolute VVVVValue Equa alue Equa Absolute Absolute o Solutions alue Equations Ha o Solutions tions Ha alue Equa Absolute Absolute Watch out when you’re solving an equation such as |2x| = 10. There are actually two numbers whose distance from zero (their absolute value) is 10: –10 and 10. So, the equation |2x| = 10 can be rewritten as two separate equations: 2x = 10 or 2x = –10 Here’s the same information about absolute values written in math-speak: Let c ≥≥≥≥≥ 0. If |x| = c, then x = c or x = –c. Example Example Example Example Example 11111 Solve |3x| = 12. Solution Solution Solution Solution Solution There are two separate equations to solve: 3x = 12 or 3x = –12 Solve each equation for x. x = 3 12 3 3 x = 4 12 3 x = − 3 3 x = –4 Check your answers by substituting back into the original equation. |3(4)| = 12 |12| = 12 12 = 12 ✓ |3(–4)| = 12 |–12| = 12 12 = 12 ✓ So x = 4 and x = –4 are the solutions of the equation. Independent Practice Solve: 1. |4x| = 16 4. |3x| = 9 2. |–2x| = 8 3. |–8x| = 24 5. |4.8x| = 144 6. |

0.02x| = 9 7. |–1.04x| = 0.2392 8. |2x + 1x| = 171 9. |10x – 5x| = 100 + 5 Write these as absolute value equations and find the solutions: 10. The product of four and a number is a distance of 20 from 0. 11. A number has a distance of 8 from 0. 12. Twice a number has a distance of 0.6 from 0. 13. The product of 1.6 and a number has a distance of 0.304 from 0. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The main thing to remember with absolute values in an equation is that they result in two possible solutions. In the next Topic you’ll work through more complicated equations involving absolute values. Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 129129129129129 TTTTTopicopicopicopicopic 2.8.22.8.2 2.8.22.8.2 2.8.2 California Standards: 3.0:3.0:3.0:3.0:3.0: Students solv Students solveeeee Students solv Students solv Students solv tions tions equa equa tions and inequalities equations equa tions equa inininininvvvvvolving a alues..... alues alues bsolute v bsolute v olving a olving a bsolute values olving absolute v alues bsolute v olving a What it means for you: You’ll solve equations involving absolute values. Key words: absolute value Absolute VVVVValuealuealuealuealue Absolute Absolute e on e on MorMorMorMorMore on e on Absolute Absolute e on MorMorMorMorMore on Absolute VVVVValuealuealuealuealue Absolute Absolute e on e on e on Absolute Absolute e on This Topic is quite similar to the last one — but this time the absolute values have more than one term. That means there’s a little more solving to do once you’ve removed the absolute value signs. e than One TTTTTererererermmmmm e than One e than One lude Mor lude

Mor y Inc y Inc alues Ma Absolute VVVVValues Ma alues Ma Absolute Absolute lude More than One y Include Mor alues May Inc e than One lude Mor y Inc alues Ma Absolute Absolute If there’s more than just a single term in the absolute value signs, you need to keep those terms together until the absolute value signs have been removed. Example Example Example Example Example 11111 Solve |2x – 5| = 7. Solution Solution Solution Solution Solution Rewrite this as two separate equations: 2x – 5 = 7 or 2x – 5 = –7 Solve both equations for x: 2x = 12 x = 6 2x = –2 x = –1 Check your answers by substituting back into the original equation. |2(6) – 5| = 7 |12 – 5| = 7 |7| = 7 7 = 7 ✓ |2(–1) – 5| = 7 |–2 – 5| = 7 |–7| = 7 7 = 7 ✓ So x = 6 and x = –1 are the correct solutions of the equation. Guided Practice Find all possible solutions to these absolute value equations: 1. |12 – 4x| = 18 3. |5x – 3x – 1| = 10 5. − 3 g = 15 2. |2x – 8| = 4 4. |3j + 1| = 10 6. − 8 x = 1.6 Find all possible solutions to these equations when a = –4: 7. |2x + a| = 10 9. |3x + 8| = a 8. |ax + 8| = 144 10. |ax – a| = –a 130130130130130 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 tion Firststststst tion Fir tion Fir e the Equa e to Rearearearearearrrrrrangangangangange the Equa e the Equa e to R ou Might Havvvvve to R e to R ou Might Ha YYYYYou Might Ha ou Might Ha e the Equation Fir tion Fir e the Equa e to R ou Might Ha If the absolute value is not alone on one side of the equals sign, the equation must be rearranged to get the absolute value on its own before the equation can be split into two parts. To solve an equation

of the form d|ax + b| + c = k, rewrite the equation in the form |ax + b| = k − c d, then solve for x. Example Example Example Example Example 22222 Solve 2|3x – 1| + 4 = 12. Solution Solution Solution Solution Solution Rearrange the equation to get the absolute value on its own: 2|3x – 1| + 4 = 12 2|3x – 13x – 1| = 4 Now split this into the two parts: 3x – 1 = 4 3x = 5 x = 5 3 Check your answers: 2|3 ⎛ ⎜⎜⎜ ⎝ 5 3 ⎞ ⎟⎟⎟ – 1| + 4 = 12 ⎠ 2|5 – 1| + 4 = 12 2|4| + 4 = 12 8 + 4 = 12 ✓ 3x – 1 = –4 3x = –3 x = –1 2|3(–1) – 1| + 4 = 12 2|–4| + 4 = 12 8 + 4 = 12 ✓ So x = 5 3 and x = –1 are the correct solutions of the equation. Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 131131131131131 Guided Practice Find all possible solutions to these equations involving absolute values: 11. |2y – 7| + 5 = 20 12. 11 + |–7 – 2k| = 21 13. |x – 9| – 5 = 1 14. –11 + |5 – 4y| = 2 15. 4|3y – 4| = 8 16. –2|6 – 5y| = –8 17. –3|12 – 7x| = –6 18. 7 – 2|8 – 4x| = –9 3 = 8 2 5 19. 21 20. 4 3 x − 5 22. 10 + = 7 5 2 x − 4 = 11 Absolute VVVVValues on Both Sides alues on Both Sides alues on Both Sides Absolute ou Might Havvvvve e e e e Absolute Absolute ou Might Ha YYYYYou Might Ha ou Might Ha alues on Both Sides alues on Both Sides Absolute ou Might Ha When it comes to solving equations with absolute values on both sides, you end up with four

equations rather than two: Example Example Example Example Example 33333 Solve |3x – 2| = |4 – x|. Solution Solution Solution Solution Solution |3x – 2| = |4 – x| ±(3x – 2) = ±(4 – x) There are four possible solutions: (1) 3x – 2 = 4 – x (2) 3x – 2 = –(4 – x) (3) –(3x – 2) = 4 – x (4) –(3x – 2) = –(4 – x) But — this set of four equations only contains two different equations: if you take equation (4) and divide both sides by –1, you get equation (1). if you take equation (3) and divide both sides by –1, you get equation (2). So if there are absolute values on both sides of an equation, then you can treat one of them as though it is not an absolute value — so writing |3x – 2| = |4 – x| is equivalent to writing |3x – 2| = 4 – x or 3x – 2 = |4 – x|. More generally: |ax + b| = |cx + d| is equivalent to |ax + b| = cx + d or ax + b = |cx + d|. 132132132132132 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 Example Example Example Example Example 44444 Solve |5x – 2| = |10 – x|. Solution Solution Solution Solution Solution Rewrite the equation with only one absolute value: |5x – 2| = 10 – x Write out two separate equations to solve. 5x – 2 = +(10 – x) 5x – 2 = –(10 – x) Solve the equations for x. 5x – 2 = 10 – x 5x + x – 2 = 10 – x + x 6x – 2 = 10 6x – 2 + 2 = 10 + 2 6x = 12 x = 2 5x – x – 2 = –10 + x – x 4x – 2 = –10 4x – 2 + 2 = –10 + 2 4x = –8 x = –2 Check your answers by substituting them into the original equation. |5x – 2| = |10 – x| |5(2) – 2

| = |10 – (2)| |10 – 2| = |8| |8| = |8| 8 = 8 ✓ |5x – 2| = |10 – x| |5(–2) – 2| = |10 – (–2)| |–10 – 2| = |10 + 2| |–12| = |12| 12 = 12 ✓ So x = 2 and x = –2 are the correct solutions of the equation. Guided Practice Find all possible solutions to these absolute value equations: 23. |2x – 8| = |3x – 12| 24. |5x – 7| = |3x + 15| 25. |9x + 18| = |4x – 2| 26. |6x – 13| = |15 – 8x| 27. |4x – 36| = |2x – 4| 28. |8x + 4| = |12x – 2| 29. The distance of (–2x – 8) from 0 is 3. What are the possible values of x? 30. If (2x + 4) and (3x + 8) are the same distance from 0, what are the possible values of x? 31. If (x + 8) is the same distance from 0 as (4x – 8), what are the possible values of x? If c = 10, find all possible solutions to these equations: 32. |2x – c| = |c – x| 33. |cx – 5| = |x – c| 34. |4x – c| = |2x + c| 35. |2cx + 4| = |x – 3c| Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 133133133133133 Independent Practice Solve: 1. |2x| = 84 3. |x + 8| = 24 2. |3z| = –9 4. |–x + 4| = 3 5. |2x + 8| = |4 – 3x| 6. |7x – 4| = |3x + 9| 7. |0.1x – 0.3| = |0.3x + 4.1| 8. –2|x – 20| = –8 9. 1 3 x 1 + = 8 1 12 + x 1 4 11.

1 2 x + 1 = 5 13. − 8 x. 0 1 ( 10 = 4 ) − x 3 10. − x 4 = 12 12 14 14 In Exercises 15–18 you will need to form an absolute value equation and solve it to find the unknown. 15. If (x + 4) is 3x from 0, what are the possible values of x? 16. If (4x – 5) is (2x + 1) from 0, what are the possible values of x? 17. If (3w + 2) and (w – 4) are the same distance away from 0, what are the possible values of w? 18. If (4x – 5 + x) and (7 + 5x + 2) are the same distance from 0, what are the possible values of x? 19. Given that |3x – 5| = |2x + 6|, find the two possible values of b2 – 2bx + x2 if b = –3. If a = 2, b = 4, and c = 6 then solve each absolute value equation for x: 20. |ax – b| = |x + c| 21. |ax + b| + c = a – bx 22. 1 a cx + = − ab ax c ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Equations with absolute values on both sides look really difficult at first. Make sure you understand how Example 3 shows that you still only get two distinct equations when there are two absolute values. 134134134134134 Section 2.8 Section 2.8 Section 2.8 — Absolute Value Equations Section 2.8 Section 2.8 Chapter 2 Investigation ildlife Pe Pe Pe Pe Pararararark k k k k TTTTTrrrrrainsainsainsainsains ildlif ildlif WWWWWildlif ildlif WWWWWildlif ildlife Pe Pe Pe Pe Pararararark k k k k TTTTTrrrrrainsainsainsainsains ildlif ildlif ildlif This Investigation is all about writing and solving equations about rates. A large wildlife park is circular in shape and has a diameter of 7 miles. Around the outskirts of the park is a circular train track. Two automatic trains travel around the park, both in a clockwise direction. The

trains are designed to both average 20 mph so that they never meet. However, one train has developed a fault and now travels at 18 mph. The trains set off from stations on opposite sides of the park at 9 a.m. At what time will the faster train catch up with the slower train? (Assume that the trains instantly reach their average speeds and that you can ignore the lengths of the trains.) North Station South Station Things to think about: • What is the distance around the track? Circumference = p × diameter. Use p = 22 7. When the trains meet, the faster train will have traveled further than the slower train. How much further? If the slower train has traveled x miles, how far has the faster train traveled? When they get to the meeting point, they will have both been traveling for the same amount of time. How long will each train have been traveling for in terms of x? Speed = distance time Extension 1) Where will the trains be when one catches up with the other? How many times will each train pass its starting point before they meet? 2) When the faster train has caught up with the slower train, it changes direction. Both trains are now traveling in opposite directions. After how many minutes will the trains meet again? Open-ended extension The park manager wants the trains to run each day while the park is open without one train catching up with the other. Unfortunately, the speeds of the trains cannot be changed for technical reasons. The opening times of the park are shown on the right. OPENING TIMES Mon - Fri: 9 a.m. - 5 p.m. Saturday: 9 a.m. - 8.30 p.m. Sunday: 10 a.m. - 7 p.m. Write a report to the manager recommending how he can achieve this. You may wish to include diagrams of the track in your report. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Parts of this Investigation are tough because the trains are traveling in a circle — but the main thing with any real-life problem is to write down all the math carefully before you start solving. estigaaaaationtiontiontiontion — Wildlife Park Trains estigestig estig pter 2 Invvvvvestig pter 2 In ChaChaChaChaChapter 2 In pter 2 In pter 2 In 135135135

135135 Chapter 3 Single Variable Linear Inequalities Section 3.1 Inequalities............................................ 137 Section 3.2 Applications of Inequalities................... 147 Section 3.3 Compound Inequalities......................... 155 Section 3.4 Absolute Value Inequalities................... 158 Investigation Mailing Packages.................................. 162 136136136136136 TTTTTopicopicopicopicopic 3.1.13.1.1 3.1.13.1.1 3.1.1 Section 3.1 Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities Inequalities California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... What it means for you: You’ll go over inequality notation from grade 7 and you’ll show inequalities on the number line. Key words: inequality endpoint interval Inequalities work like equations, but they tell you whether one expression is bigger or smaller than the expression on the other side. In grade 7 you solved linear inequalities — so the stuff in the next couple of Topics should be fairly familiar. Inequality Symbols Inequality Symbols Inequality Symbols Inequality Symbols Inequality Symbols Expressions such as 3x > 8, x < –5, x ≥ 10, and x £ 10 are inequalities. An inequality is a mathematical sentence that states that two expressions are not equal. You read the inequality symbols like this: < “is less than” > “is greater than” £££££ “is less than or equal to” ≥≥≥≥≥ “is greater than or equal to” For example, you read m < c as “m is less than c” and you read k ≥ 5 as “k is greater than or equal to 5.

” w Inequalities on the Number Line w Inequalities on the Number Line ou Can Sho YYYYYou Can Sho ou Can Sho w Inequalities on the Number Line ou Can Show Inequalities on the Number Line w Inequalities on the Number Line ou Can Sho The inequality x > 4 represents the interval (part of the number line) where the numbers are greater than 4. Similarly k ≥≥≥≥≥ 5 represents all real numbers greater than or equal to 5 on the number line. The numbers 4 and 5 in these examples represent the endpoints of the intervals of the number line under consideration. However, k ≥≥≥≥≥ 5 includes the endpoint 5, while x > 4 excludes the endpoint 4. Example Example Example Example Example 11111 Show the inequality x > 4 on the number line. Solution Solution Solution Solution Solution You can show x > 4 on the number line like this: Check it out: The symbol • and the closing parenthesis indicate that the list of numbers continues indefinitely (which is also shown by the arrow on the graph). –6 –4 –2 0 2 4 6 Note that the circle at the endpoint is open — this shows that the endpoint 4 is not part of the interval of the number line defined by the inequality x > 4. This is called an open interval, denoted as (4, •••••). Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 137137137137137 Check it out: The “[” bracket indicates that the endpoint is part of the interval, which is also shown by the closed circle in the graph. Check it out: Note also that the interval is closed at one end, but open at the other — which is why it’s called a half-open interval. Example Example Example Example Example 22222 Show the inequality k ≥ 5 on the number line. Solution Solution Solution Solution Solution –6 –4 –2 0 2 654 Note that the closed circle at 5 indicates that the endpoint 5 is included in the interval. This is called a half-open interval, denoted as [5, •••••). Guided Practice Write each of the following inequalities in interval notation, and show each graphically. 1. k > 6 4. j > –5 7. 3 ≥ y 2. m < –2 5. n £ 3.5 8. j ≥ 10 3.

x ≥ 1 6. l £ 3 9. 10 £ x 10. State the inequality represented on the number line opposite. 11. State the inequality represented on the number line opposite using interval notation. Independent Practice –4 –2 –4 –2 0 0 2 2 4 4 In Exercises 1–3 write each inequality in interval notation. 1. r > 8 2. t £ –9 3. 3 £ x In Exercises 4–6 write each interval as an inequality in x. 6. (–•, 0] 5. (–6, •) 4. (–•, 2) In Exercises 7–12 show each inequality or interval graphically. 7. k > –7 10. (–11, •) 9. 1.5 £ y 12. (–•, 5] 8. k £ 2 11. [0, •) 13. Anthony is shopping for a birthday gift for his cousin Robert. He has $25 in his wallet. Write an inequality that shows how many dollars he can spend on the gift. 14. Teresa is only allowed to swim outside if the temperature outside is at least 85 °F. Write an inequality that shows the temperatures in degrees Fahrenheit at which Teresa is allowed to swim. 15. In order to achieve an ‘A’ in math, Ivy needs to score more than 95% on her next test. Write an inequality that shows the test score Ivy needs to achieve in order to earn her ‘A’ in math. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up The most difficult thing is remembering that a “[“ bracket shows the endpoint is included in the interval, and a “(“ parenthesis means the endpoint isn’t included. 138138138138138 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 TTTTTopicopicopicopicopic 3.1.23.1.2 3.1.23.1.2 3.1.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef ess

ions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems,,,,, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve inequalities that contain + and – signs. Key words: inequality endpoint interval action action dition and Subtr dition and Subtr AdAdAdAdAddition and Subtr action dition and Subtraction action dition and Subtr action action dition and Subtr dition and Subtr AdAdAdAdAddition and Subtr action dition and Subtraction action dition and Subtr Inequalities Inequalities ties of ties of oper oper PrPrPrPrProper Inequalities ties of Inequalities operties of Inequalities ties of oper PrPrPrPrProper Inequalities Inequalities ties of ties of oper oper ties of Inequalities operties of Inequalities ties of oper Inequalities To solve inequalities such as x + 1 > 4, 3x + 2 > 7, or x – 7 £ 12, you need to apply properties of inequalities — they’re just like the properties of equality you applied to equations. Inequalities Inequalities ty of ty of oper oper dition Pr AdAdAdAdAddition Pr dition Pr In

equalities ty of Inequalities operty of dition Proper Inequalities ty of oper dition Pr Addition Property of Inequalities Given real numbers a, b, and c, if a > b, then a + c > b + c. In other words, adding the same number to both sides of an inequality gives an equivalent inequality. Example Example Example Example Example 11111 Solve and graph the solution of x – 2 > 5 on a number line. Write the solution in interval notation. Solution Solution Solution Solution Solution Solve Using ad Using ad dition pr dition pr oper oper ty of ty of inequalities inequalities Using addition pr dition proper operty of ty of inequalities inequalities Using ad Using ad dition pr oper ty of inequalities Check it out: The circle is open at the endpoint 7, which shows that 7 isn’t part of the solution interval. Graph: –7 0 7 Solution in interval notation: (7, •••••) Guided Practice Solve and graph each inequality. Write each solution set in interval notation. 1. l – 1 ≥ 3 2. r – 3 < –5 3. x – 4 £ –1 5. –2m < 2 – 3m 7. –2j + 3 ≥ –j 4. m – 7 > –10 6. –k £ –2k + 1 8. –7(j + 1) < –6j Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 139139139139139 Inequalities Inequalities ty of ty of oper oper action Pr action Pr Subtr Subtr Inequalities ty of Inequalities operty of action Proper Subtraction Pr Inequalities ty of oper action Pr Subtr Subtr Subtraction Property of Inequalities Given real numbers a, b, and c, if a > b, then a – c > b – c. That is, subtracting the same number from both sides of an inequality gives an equivalent inequality. Example Example Example Example Example 22222 Solve and then graph the solution of 3x £ 6 + 2x on a number line. Write the solution in interval notation and then state the maximum integer value of x that satisfies the inequality. Solution Solution Solution Solution Solution Solve: 3x £ 6 + 2x 3x – 2x £ 6 + 2x – 2x Subtr action pr action pr Subtr Subtr ty of ty of oper oper inequalities inequalities Sub

traction pr operty of action proper inequalities ty of inequalities Subtr oper action pr inequalities ty of x £ 6 Graph: –1 0 1 2 3 4 5 6 Solution in interval notation: (–•••••, 6] So the maximum integer value of x is 6. opertiestiestiestiesties oper oper ou to Use Both Pr lems Need YYYYYou to Use Both Pr ou to Use Both Pr lems Need Some Proboboboboblems Need lems Need Some Pr Some Pr ou to Use Both Proper oper ou to Use Both Pr lems Need Some Pr Some Pr Example Example Example Example Example 33333 Solve and graph the solution set of 5x – 2 £ 4x – 3 on a number line. State the maximum integer value of x that satisfies the inequality, and write the solution set in interval notation. Solution Solution Solution Solution Solution Solve: 5x – 2 £ 4x – 3 5x – 2 + 2 £ 4x – 3 + 2 AdAdAdAdAddition pr inequalities inequalities ty of ty of oper oper dition pr dition pr inequalities ty of inequalities operty of dition proper inequalities ty of oper dition pr 5x £ 4x – 1 5x – 4x £ 4x – 4x – 1 Subtr inequalities inequalities ty of ty of oper oper action pr action pr Subtr Subtr inequalities ty of inequalities operty of action proper Subtraction pr inequalities ty of oper action pr Subtr opertytytytyty oper oper and commutautautautautatititititivvvvve pre pre pre pre proper and comm and comm oper and comm and comm x £££££ –1 Graph: –4 –3 –2 –1 0 1 2 Solution in interval notation: (–•••••, –1] So the maximum integer value of x is –1. Check it out: In this example, the half-open interval closes with a bracket to show that the endpoint 6 is part of the solution interval. Check it out: The closed circle of the graph shows that the endpoint is included in the solution interval. 140140140140140 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 Guided Practice Solve and graph each inequality, and write each solution set in interval notation. Find also the maximum or minimum integer value that satisfies each inequality. 9. x + 2 < 8 11. j +

2 ≥ –3 13. 3k + 2 ≥ 2k 15. 2t + 1 £ t – 10 10. t + 5 £ 6 12. y + 3 > –9 14. 4x – 3 < 3x 16. 6(x – 1) ≥ 5(x + 2) Independent Practice Solve the following inequalities. Justify each step and write each solution set in interval notation. 1. j – 3 > –1.5 3. x + 3 < –11 5. 5k – 1.25 < 4k – 4 2. k – 2 ≥ –5 4. 2v – 2.5 > v + 1 6. –7j – 5 < –6j Solve and graph each inequality in Exercises 7–18. Write each solution set in interval notation. 7. 6x – 1 ≥ 5x + 3 9. 4(x – 3) – 3x < –11 11. 7t + 6(1 – t) £ –4 13. 2(t + 5) £ 3t – 11 15. 5(–2 + 3x) ≥ –2(1 – 7x) 17. 13 + 3(x – 5) £ 2(–3 + x) 8. 3(k – 2) – 2k ≥ 9 10. 4n – 3(n + 1) ≥ 1 12. 5t – 4(t – 1) > –3 14. –3x + 2 + 7x < –3 + 3x – 4 16. –4(1 – 4x) ≥ –7.5(–1 – 2x) 18. 5x – (3x – 4) > –(2 – x) 19. Find the maximum integer value of x if 6x + 4 £ 5x – 8. 20. Find the maximum integer value of x if –3(3 – 2x) < 5(x – 5). 21. Find the least integer value of x if –2(1 – x) > x – 1. 22. Find the least integer value of x if 4(x – 3) £ 5x + 2 23. Stephen needs to buy a new uniform for soccer. He already has $25, but the uniform costs $55. He participates in car washes to help pay for the uniform. Write an inequality to represent the amount of money, x, that Stephen needs to earn from the car washes in order to be able to afford the new

uniform. Use this inequality to find the minimum amount of money he needs to earn. 24. An art gallery sells Peter’s paintings for $x, and keeps $100 commission. This means Peter is paid $(x – 100) for each painting. If Peter wants to make at least $750 for a particular painting, write an inequality to represent the amount, x, that the gallery needs to sell that painting for. Use this inequality to find the minimum price of the painting. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Adding and subtracting with inequalities is a lot like dealing with normal equations — so there’s nothing in this Topic that should cause you too much trouble. Next up is multiplying and dividing. Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 141141141141141 TTTTTopicopicopicopicopic 3.1.33.1.3 3.1.33.1.3 3.1.3 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems,,,,, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le

and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve inequalities that contain × and ÷ signs. Key words: inequality reverse 142142142142142 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 vision vision tion and Di tion and Di Multiplica Multiplica vision tion and Division Multiplication and Di vision tion and Di Multiplica Multiplica vision vision tion and Di tion and Di Multiplica Multiplica tion and Division vision Multiplication and Di tion and Di vision Multiplica Multiplica Inequalities Inequalities ties of ties of oper oper PrPrPrPrProper Inequalities ties of Inequalities operties of Inequalities ties of oper PrPrPrPrProper Inequalities Inequalities ties of ties of oper oper ties of Inequalities operties of Inequalities ties of oper Inequalities After a Topic on addition and subtraction with inequalities, you know what to expect: multiplication and division of inequalities. Inequalities Inequalities ty of ty of oper oper tion Pr tion Pr Multiplica Multiplica Inequalities ty of Inequalities operty of tion Proper Multiplication Pr Inequalities ty of oper tion Pr Multiplica Multiplica Multiplication Property of Inequalities Given real numbers a, b, and c, if a > b and c > 0 then ac > bc. That is, multiplying both sides of an inequality by a positive number gives an equivalent inequality. For example: Start with an inequality: Multiplying by 2 gives: 2 × 4 < 2 × 10 4 < 10...is true. 8 < 20...which is also true. Example Example Example Example Example 11111 Solve 1 5 x > 2. Solution Solution Solution Solution Solution Multiplica Multiplica Multiplica tion pr tion pr oper oper ty of ty of inequalities inequalities Multiplication pr tion proper operty of ty of inequalities inequalities Multiplica tion pr oper ty of inequalities x > 10 Guided Practice Solve each inequality in Exercises 1–12. 1. 1 2 x £ 3 4. 6 7. j 9 10

. 1 > x 3 12≤ ≤ y 20 − 1 5 2. 5. 1 4 x 4 p > 12 5> − 8. 1 3 1 < a 9 11. 2 5 p < 4 3. 6 15 9. ≥ k 12 1 4 12. − <6 3 4 a Inequalities Inequalities ty of ty of oper oper vision Pr DiDiDiDiDivision Pr vision Pr Inequalities ty of Inequalities operty of vision Proper Inequalities ty of oper vision Pr Division Property of Inequalities Given real numbers a, b, and c, if a > b and c > 0 then a c b >. c In other words, dividing both sides of an inequality by a positive number gives an equivalent inequality. For example: Start with an inequality: Dividing by 2 gives: 4 < 10 4 ÷ 2 < 10 ÷ 2 2 < 5...is true....which is also true. Example Example Example Example Example 22222 Solve 3x < 11. Solution Solution Solution Solution Solution 3x < 11 x 3 3 < x < 11 3 11 3 DiDiDiDiDivision pr vision pr vision pr oper oper ty of ty of inequalities inequalities vision proper operty of ty of inequalities inequalities vision pr oper ty of inequalities Guided Practice Solve each inequality in Exercises 13–26. 13. 10x < 5 15. 32 ≥ 8n 17. 40 < 8y 19. 72 ≥ 9j 21. 20k < 5 23. 4 > 28y 25. 6(6a) ≥ 4 14. 4y ≥ 20 16. 7z £ 28 18. 56 > 7j 20. 6k £ 48 22. 7 < 77x 24. 40a < 5 26. 9 £ 3b(20 + 7) Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 143143143143143 e Numbersssss ty with Negggggaaaaatititititivvvvve Number e Number e Number ty with Ne ty with Ne oper oper tion Pr tion Pr Multiplica Multiplica operty with Ne tion Proper Multiplication Pr e Number ty with Ne oper tion Pr Multiplica Multiplica This is really important — so make sure you read this carefully. You’ve just seen that multiplying both sides of an inequality like 4 > 3 by, say, 3, gives a true new inequality, 12 > 9

. However, if the original inequality is multiplied by a negative number like –3, the resulting inequality is –12 > –9, which is false. To make the resulting inequality true, you have to reverse the inequality sign. That gives –12 < –9, which is true. Multiplication Property of Inequalities — Negative Numbers Given real numbers a, b, and c, if a > b and c < 0 then ac < bc. In other words, if you multiply both sides of an inequality by a negative number, you have to reverse the inequality symbol — otherwise the statement will be false. For example: Check it out: Watch out — the difference between this rule and the normal multiplication property is the “c < 0 then ac < bc” part. Start with an inequality: Multiplying by –2 and reversing the inequality sign gives:...is true. –3 < 8 –2 × –3 > –2 × 8 6 > –16...which is also true. Example Example Example Example Example 33333 Solve – 1 2 x > 2. Solution Solution Solution Solution Solution x > 2 – 1 2 –2 × – 1 2 x < –4 x < –2 × 2 Multiplica Multiplica Multiplica tion pr tion pr oper oper ty of ty of inequalities inequalities Multiplication pr tion proper operty of ty of inequalities inequalities Multiplica tion pr oper ty of inequalities Guided Practice Solve each inequality in Exercises 27–37. 27. 30. 33. ≤1 x 7 − 10 − 4 > −d 3 − ≥j 8 7 28. 31. 34. − <1 c 9 − 1 4 − < −k 11 a 5 29. 11 < < − 12 32. − ≤ 4 1 35 36. − 1 9 1 ⋅ > 4 x( 4 − 13 3 ) 37. 1 + ⋅ 5 10 1 ≥ − g 6 4 144144144144144 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 e Numbersssss ty with Negggggaaaaatititititivvvvve Number e Number e Number ty with Ne ty with Ne oper oper vision Pr DiDiDiDiDivision Pr vision Pr operty with Ne vision Proper e Number ty with Ne oper vision Pr Division Property of Inequalities — Negative Numbers Given real numbers a, b, and c, if a > b

and c < 0 then a c b <. c In other words, if you divide both sides of an inequality by a negative number, you have to reverse the inequality symbol — otherwise the statement will be false. Start with an inequality: Dividing by –3 and reversing the inequality sign gives:...is true. –3...which is also true. Example Example Example Example Example 44444 Solve –13y < 39. Solution Solution Solution Solution Solution –13y < 39 − 39 − − 13 y > –3 13 13 > y DiDiDiDiDivision pr vision pr vision pr oper oper ty of ty of inequalities inequalities vision proper operty of ty of inequalities inequalities vision pr oper ty of inequalities Guided Practice Solve each inequality in Exercises 38–47. 39. 36 < –9x 38. –5x £ 5 41. –x > –1 40. –6j > 48 43. –77c < –11 42. 45 £ –9x 45. 49 > –7a 44. –72y ≥ –8 − ≥ −1 46. − ≥4 y 47. 6c 1 4 10 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 145145145145145 Independent Practice Solve each inequality in Exercises 1–6. 1. 32 < 4h − >1 g 7 5. 2 £ 3. 6. 7 2. 9a £ –45 4. –8c £ 48 w 8 <5 5 7 (1 – 2y) < 5, and state the largest possible integer − h 6 5 7. Solve 3 2 value of y. y – 2 3 Solve each inequality in Exercises 8–21. 8. 2(x – 3) – 3(2 – x) > 8 10. 3(x – 1) < 7 – 2x 12. 3y – (5y + 4) > 7y + 2(y – 5) 13. 5(2x – 3) – 3(x – 7) ≥ 4(3x + 2) + 2x – 9 14. 5 – 4(x + 2) £ 7 + 5(2x – 1) 9. –4(3 – 2x) > 5x + 9 11. 5(y + 3) – 7y £ 3(2y + 3) – 5y 15. 7 – 2(m – 4) £

2m + 11 − + x 2 3 1 5 3 − ≥ 16. 17. x 4 11 12 5 9 18. 0.5(x – 1) – 0.75(1 – x) < 0.65(2x – 1) 19. 7 – 3(x – 7) £ 4(x + 5) + 1 20. 0.35(x – 2) – 0.45(x + 1) ≥ 8 + 0.15(x – 10) 21. 3(2x + 6) – 5(x + 8) £ 2x – 22 22. Laura has $5.30 to spend on her lunch. She wants to buy a chicken salad costing $4.20 and decides to spend the rest on fruit. Each piece of fruit costs 45¢. Write an inequality to represent this situation, and then solve it to find how many pieces of fruit Laura can buy. 23. Audrey is selling magazine subscriptions to raise money for the school library. The library will get $2.50 for every magazine subscription she sells. Audrey wants to raise at least $250 for the library. Write and solve an inequality to represent the number of magazine subscriptions, x, Audrey needs to sell to reach her goal. 24. The total cost of food and supplies for a cat is x dollars per year, and medical expenses can be 1.5 times the cost of food and supplies per year. If Maddie can spend no more than $500 a year on the cat, what is the most that she can spend on food and supplies? Write and solve an inequality to represent the situation. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Multiplication and division with negative numbers can sometimes be difficult. If you multiply by a negative number and forget to reverse the direction of the inequality sign, then your solution will be wrong — so watch out. 146146146146146 Section 3.1 Section 3.1 Section 3.1 — Inequalities Section 3.1 Section 3.1 TTTTTopicopicopicopicopic 3.2.13.2.1 3.2.13.2.1 3.2.1 Section 3.2 Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequalities Multistep Inequ

alities Multistep Inequalities Multistep Inequalities Multistep Inequalities California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems,,,,, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll solve inequalities involving several steps. Key words: inequality isolate Check it out: This example uses the addition, subtraction and division properties of inequalities, as well as the associative property. Inequality problems often involve using more than one of the properties of inequalities that you saw in Topics 3.1.2 and 3.1.3. The multistep inequalities in this Topic are a little harder than the ones you saw in Section 3.1 — but you still solve them using the same methods. p Inequalities Combine Lots of TTTTTececececechniques hniques hniques p Inequalities Combine Lots of p Inequalities Combine Lots of Multiste Multiste

hniques Multistep Inequalities Combine Lots of hniques p Inequalities Combine Lots of Multiste Multiste To simplify and therefore solve an inequality in one variable such as x, you need to isolate the terms in x on one side and isolate the numbers on the other. It’s often easiest to keep the x-terms on the side of the inequality sign where they have a positive value. Example Example Example Example Example 11111 Solve 4x – 7 > 2x. Solution Solution Solution Solution Solution 4x – 7 > 2x 4x – 7 + 7 > 2x + 7 4x > 2x + 7 Then get rid of the 2x on the right 4x – 2x > 2x + 7 – 2x Then get rid of the 2 on the right on the right Then get rid of the 2 Then get rid of the 2 on the right on the right Then get rid of the 2 2x > 7 Elimina Elimina Elimina te the –7 fr te the –7 fr om the left fir om the left fir om the left firststststst Eliminate the –7 fr te the –7 from the left fir Elimina te the –7 fr om the left fir Guided Practice 1. 5x – 2 £ 3 3. –3a – 3 < –9 5. 7 to get et et et et x on its o DiDiDiDiDivide both sides b on its o on its o on its ownwnwnwnwn vide both sides b vide both sides b y 2 to g y 2 to g vide both sides by 2 to g on its o vide both sides b y 2 to g 2. 4x – 1 > 2 4. 5x – 7 £ 8 6. 8. x 2 j 3 + 9 £ –2 – 8 ≥ 7 9. 8g – 10 £ 9g + 4 11. 4c – 9 ≥ 5c + 16 10. 4a + 5 £ 6a + 9 12. k – 4 £ 2k + 20 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 147147147147147 Check it out: Instead of “x,” the variable might be “y,” “z,” or any other letter. ted Inequalities into Smaller Stepspspspsps ted Inequalities into Smaller Ste ted Inequalities into Smaller Ste e

ak Complica BrBrBrBrBreak Complica eak Complica eak Complicated Inequalities into Smaller Ste ted Inequalities into Smaller Ste eak Complica Here’s a useful checklist for tackling more complicated inequality questions by breaking them down into easier steps: Solving Inequalities 1. Multiply out any parentheses. 2. Simplify each side of the inequality. 3. Remove number terms from one side. 4. Remove x-terms from the other side. 5. Multiply or divide to get an x-coefficient of 1. Example Example Example Example Example 22222 Solve the inequality: 7(x – 2) – 3(x – 4) > 2(x – 5) Solution Solution Solution Solution Solution 7x – 14 – 3x + 12 > 2x – 10 4x – 2 > 2x – 10 4x – 2 + 2 > 2x – 10 + 2 4x > 2x – 8 4x – 2x > 2x – 8 – 2x 2x > –8 − 8 2 x > –4 x > 2 2 Multipl Multipl y out the par y out the par entheses entheses Multiply out the par y out the parentheses entheses Multipl Multipl y out the par entheses Simplify Simplify Simplify Simplify Simplify Elimina Elimina te the –2 fr te the –2 fr om the left om the left Eliminate the –2 fr te the –2 from the left om the left Elimina Elimina te the –2 fr om the left Get rid of the 2x on the right on the right on the right Get rid of the 2 Get rid of the 2 on the right on the right Get rid of the 2 Get rid of the 2 vide to get et et et et x on its o on its o on its o DiDiDiDiDivide to g on its ownwnwnwnwn vide to g vide to g on its o vide to g Guided Practice 13. 4x – 2 £ 3(x + 5) 15. 4x + 1 > 2(x + 2) 17. 8(x + 3) £ 7(x + 3) 19. 21 > 3x £ 3(x – 2) 14. 6x – 8 > 5(x + 2) 16. 5x – 4 < 3(x + 6) 18. 2(x + 1) ≥ 4(x – 2) 20. 22 <

3x ≥ 10(x + 1) 23. 2(x – 1) ≥ 4(x – 2) – 8 24. 12(b + 1) – 10b > 7(b + 3) + 6 148148148148148 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 Independent Practice In Exercises 1–7, solve each inequality. 1. 9x – 7 £ 11 3. 6x – 12 > 5x + 8 2. 8c – 10 ≥ 7c + 6 4. 11(x + 8) ≥ 12x – 12 5. 9(p + 5) > 10p – 5 7. a 6 – 3 ≥ 5 6. a 2 < a + 4 In Exercises 8–28, solve each inequality and write the solution set in interval notation. 8. 6x – 2 £ 4(x + 5) 9. 3(x + 1) < 5x + 5 10. 5x + 1 > 3(x + 3) 11. 8(x – 1) ≥ 4x – 4 12. 6(j – 2) > 7(j + 4) 13. 3(x + 2) ≥ 5(x – 2) ( 7 4 ( ( 7 14. 15. 16. ( 8 17 £ 4a > 2x < 2(x + 4) £ 3(x – 2) 18. 3(k – 1) + 2(k + 1) < 4 19. 7(d + 3) + 2(d – 4) > –5 20. 6(x – 4) – 5(x + 1) £ 9 21. 4(t + 0.25) – 8(t – 7) ≥ –3 22. 4(x – 1) ≥ 8(x – 2) – 6x 23. –5(t – 2) + 4 > –(t + 2) 24. 4(t + 0.5) £ 0.5(4t – 12) + 6 25. 5(a – 1) – 2a < 4(a + 4) 26. 4(a + 3) + 4 ≥ –(a – 1) 27. 6(t + 1) – 10 > 7(t + 3) + 4t 28. 5(a – 5) – 3 > 4(a + 8) + 7a ound Up ound

Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up If you ever get stuck when you’re solving inequalities with more than one step, refer to the checklist on the previous page. Just take it one step at a time, as if you were dealing with an equation. Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 149149149149149 TTTTTopicopicopicopicopic 3.2.23.2.2 3.2.23.2.2 3.2.2 California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems ultiste mmmmmultiste p prp pr ultiste ultistep pr lems ultiste incincincincincluding w lems,,,,, d prd proboboboboblems lems lems d prd pr luding wororororord pr luding w luding w luding w lems olving olving inininininvvvvvolving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f

justifica justifica What it means for you: You’ll solve real-life problems involving inequalities. Key words: inequality isolate Check it out: These rules are pretty much the same as when you’re dealing with real-life equation problems. Inequalities Inequalities tions of tions of pplica pplica AAAAApplica Inequalities tions of Inequalities pplications of Inequalities tions of pplica AAAAApplica Inequalities Inequalities tions of tions of pplica pplica tions of Inequalities pplications of Inequalities tions of pplica Inequalities Some real-life problems include phrases like “at least” or “at most,” or deal with maximums or minimums. If you come across these phrases, chances are you’ll need to model the situation as an inequality. Proboboboboblems lems lems Pr eal-Life”e”e”e”e” Pr Pr eal-Lif Inequalities are “Re “Re “Re “Re “Real-Lif eal-Lif Inequalities ar Inequalities ar tions of tions of pplica AAAAApplica pplica lems tions of Inequalities ar pplications of lems Pr eal-Lif Inequalities ar tions of pplica In the same way that applications of equations are real-life problems, applications of inequalities are real-life inequalities problems. Real-life problems involving inequalities could be about pretty much anything — from finding the area of a field to figuring out how many CDs you can buy with a certain amount of money. What they all have in common is that they’ll all be word problems — and you’ll always have to set up and solve an inequality. Solving Real-Life Inequality Problems 1. First decide how you will label the variables... 2....then write the task out as an inequality... 3....make sure you include all the information given... 4....then solve the inequality. Example Example Example Example Example 11111 Find the three smallest consecutive even integers whose sum is more than 60. Solution Solution Solution Solution Solution First you need to label the variables: Let x = first (smallest) even integer x + 2 = next (second)

even integer x + 4 = next (third) even integer 150150150150150 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 Check it out: In this example, you know you need a “>” and not a “≥” because the problem says “more than,” not “more than or equal to.” Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Then you need to write it out as an inequality: In English: The sum of three consecutive even integers is more than 60. In math: x + (x + 2) + (x + 4) > 60 Then simplify > 60 3x + 6 > 60 3x + 6 – 6 > 60 – 6 3x > 54 54 3 x > 18 x > 3 3 Answer in math: x > 18 Answer in English: The smallest even integer is more than 18. So the three smallest consecutive even integers whose sum is greater than 60 are 20, 22, and 24. Guided Practice 1. Find the largest three consecutive odd integers whose sum is at most 147. 2. Find the three smallest consecutive odd integers whose sum is more than 45. 3. Find the smallest three consecutive odd integers such that the sum of the first two integers is greater than the sum of the third integer and 11. 4. Find the smallest three consecutive even integers whose sum is greater than 198. 5. The difference between a number and twice that number is at least 7. Find the smallest possible integer that satisfies this criterion. 6. Three times a number is added to 11, and the result is less than 7 plus twice the number. Find the highest possible integer the number could be. 7. José scored 75 and 85 on his first and second algebra quizzes. If he wants an average of at least 83 after his third quiz, what is the least score that José must get on the third quiz? 8. Lorraine’s test scores for the semester so far are 60%, 70%, 75%, 80%, and 85%. If the cutoff score for a letter grade of B is 78%, what is the least score Lorraine must get on the final test to earn a B? Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 151151151151151 Example Example Example Example Example 22222 The

perimeter of a rectangular field, F, is given by the formula p = 2l + 2w. As shown in the diagram, the length l = (10x – 6) m and the width w = (5x – 3) m. Find the possible values of x for which the given rectangular field would have a perimeter of at least 1182 meters. (5 – 3) m x F x (10 – 6) m Solution Solution Solution Solution Solution The variables are already labeled, so you just need to write it out as an inequality: Substituting the variables into the expression for the perimeter gives: p = 2(10x – 6) + 2(5x – 3) You’re also told that the perimeter is at least 1182 meters — in math, you write that as p ≥ 1182. So the inequality is: 2(10x – 6) + 2(5x – 3) ≥ 1182 20x – 12 + 10x – 6 ≥ 1182 30x – 18 ≥ 1182 30x – 18 + 18 ≥ 1182 + 18 30x ≥ 1200 x ≥ 1200 30 30 30 x ≥≥≥≥≥ 40 Answer in math: x ≥ 40 Answer in English: x would have to be greater than or equal to 40. Guided Practice 9. Lily wants to build a fence along the perimeter of her rectangular garden. She cannot afford to buy more than 14 meters of fencing. The length of the garden, l, is (2x – 3) m and the width, w, is (3x – 10) m. Write an inequality to represent the situation and solve it. What would the dimensions of the garden fence have to be to keep the fencing no longer than 14 meters. 10. A car travels (17x – 5) miles in (x – 7) hours. The car travels at a constant speed not exceeding the speed limit of 55 miles per hour. If speed = distance ÷ time, write an inequality to represent this situation and find the minimum possible number of miles traveled. 11. The formula for calculating the speed of an accelerating car is v = u + at, where v is the final speed, u is the original speed, a is the acceleration, and t is the time taken. Car A starts at 5 m/s and accelerates at 4 m/s2. At the same time, car B starts at 10 m/s and accelerates at 2 m/s2. Write and

solve an inequality to find out how long it will be before Car A is traveling faster than Car B. 152152152152152 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 Example Example Example Example Example 33333 A long-distance telephone call from Los Angeles, California, to Harare, Zimbabwe, costs $9.50 for the first three minutes, plus $0.80 for each additional minute (or fraction of a minute). Colleen has $18.30 to spend on a call. What is the maximum number of additional minutes she can spend on the phone? Solution Solution Solution Solution Solution First you need to label the variables: Let x = the additional number of minutes after the first 3 minutes. Then you need to write it out as an inequality: In English: The total cost is $9.50 plus $0.80 per additional minute. The total cost must be less than or equal to $18.30. In math: 9.50 + 0.80x = Total cost for the call Total cost for the call £ 18.30 \ 9.50 + 0.80x £ 18.30 Then simplify: 950 + 80x £ 1830 950 – 950 + 80x £ 1830 – 950 80x £ 880 x ≤ 880 80 80 80 x £££££ 11 Answer in math: x £ 11 Answer in English: The number of additional minutes must be no more than 11. So Colleen can spend up to 11 additional minutes on the phone. Guided Practice 12. Marisa is buying a new car. Car A costs $20,000, and has an average annual fuel cost of $1000. Car B costs $22,500, and has an average annual fuel cost of $500. After how many years will Car A have cost more than Car B? Assume that all other maintenance costs are equal for both cars. 13. A cell phone company offers its customers either Plan A or Plan B. Plan A costs $90 per month with unlimited air time. Plan B costs $60 per month, plus 50¢ for each minute of cell phone time. How many minutes can a customer who chooses Plan B use the cell phone before the cost of the calls exceeds the amount it would have cost under Plan A? Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 153153153153153 Independent Practice 1. Bank

T’s checking account has monthly charges of an $8 service fee plus 6¢ per check written. Bank S’s checking account has monthly charges of a $10 service fee plus 4¢ per check written. A company has 150 employees, and pays them monthly by check. The company’s financial adviser suggests that Bank S would be cheaper to use. Set up and solve an inequality that supports this recommendation. 2. Jack is doing a sponsored swim to raise money for charity. His mom sponsors him $10, plus $1 for every length of the pool he completes. His uncle sponsors him just $1.50 for every length he completes. How many lengths will Jack have to complete for his uncle to pay more than his mom? 3. On average Sendi uses 350 minutes of air time per month. Company M offers a cell phone plan of $70 per month plus 85¢ for each minute of air time. Company V offers a cell phone plan of $130 per month plus 65¢ for each minute of air time. Sendi chooses Company V. Use an inequality to show that this plan is cheaper for her. 4. A group of friends want to drive to a beach resort and spend 5 days there. A car rental firm offers them two rental plans; $15 a day plus 30¢ per mile traveled, or $20 a day plus 10¢ per mile. Which rental plan would be better if the beach resort is 150 miles from home, and why? 5. A bank charges a $10 monthly service fee plus 5¢ handling fee per check processed through its Gold checking account. The bank also offers a Platinum checking account and charges a $15 monthly service fee plus 3¢ handling fee per check drawn from this account. What is the highest number of checks per month for which the Gold account is cheaper than the Platinum account? 6. A group of executives is traveling to a meeting, so they decide to hire a car and travel together. The car rental agency rents luxury cars at $65 per day plus 65¢ per mile traveled, or $55 per day plus 85¢ per mile traveled. What is the maximum number of miles that they can drive before the $55 per day plan becomes more expensive than the $65 per day plan? ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up This Topic is very similar to real-life applications of equations, which is covered in Sections

2.4–2.7. Always remember to give your solution as a sentence that answers the original problem. 154154154154154 Section 3.2 Section 3.2 Section 3.2 — Applications of Inequalities Section 3.2 Section 3.2 TTTTTopicopicopicopicopic 3.3.13.3.1 3.3.13.3.1 3.3.1 Section 3.3 Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities Compound Inequalities California Standards: 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify 4.0: Students simplify eeeeexprxprxprxprxpressions bef essions befororororore solving e solving e solving essions bef essions bef e solving e solving essions bef linear linear linear equations and linear linear inequalities in one variaariaariaariaariabbbbblelelelele,,,,, inequalities in one v inequalities in one v inequalities in one v inequalities in one v such as 3(2x – 5) + 4(x – 2) = 12..... Students solveeeee 5.0:5.0:5.0:5.0:5.0: Students solv Students solv Students solv Students solv lems,,,,, lems lems p prp proboboboboblems mmmmmultiste p prp pr ultiste ultiste ultistep pr lems ultiste including word problems, olving inininininvvvvvolving olving olving linear equations olving linear inequalities in linear inequalities in and linear inequalities in linear inequalities in linear inequalities in le and prooooovidevidevidevidevide one variaariaariaariaariabbbbble and pr le and pr le and pr one v one v one v le and pr one v h steppppp..... h ste h ste or eac or eac tion f tion f justifica justifica or each ste tion for eac justification f h ste or eac tion f justifica justifica What it means for you: You’ll learn about problems involving two inequalities, and how to solve them. Key words: conjunction disjunction inequality Check it

out: The solution set of the conjunction is the numbers for both both both inequalities are which both both true. Sometimes a math problem gives you two different restrictions on a solution, using inequality signs. A compound inequality is two inequalities together — for example, 2x + 1 < 5 and 2x + 1 > –1. lude the WWWWWororororord “d “d “d “d “And”And”And”And”And” lude the lude the lems Inc Conjunction Proboboboboblems Inc lems Inc Conjunction Pr Conjunction Pr lems Include the lude the lems Inc Conjunction Pr Conjunction Pr The word “and” means the compound inequality below is a “conjunction.” You can rewrite a conjunction as a single mathematical statement, usually involving two inequality signs, like this: 2x + 1 < 5 and 2x + 1 > –1 can be rewritten as –1 < 2x + 1 < 5. Guided Practice In Exercises 1–9, express each conjunction as a single mathematical statement. 1. 3y – 1 > 5 and 3y – 1 < 11 2. 4a + 7 > –10 and 4a + 7 < –2 3. –8c + 2 £ 16 and –8c + 2 ≥ –3 4. 7x – 2 < 14 and 7x – 2 > 4 5. 10a – 7 < 2 and 10a – 7 > –5 6. 9t + 4 £ 4 and 9t + 4 ≥ 3 7. –4g – 5 < –5 and –4g – 5 > –10 8. 7c – 9 < 7 and 7c – 9 > –4 9. 8y + 9 £ 2 and 8y + 9 > –6 Solving Conjunctions Solving Conjunctions Solving Conjunctions Solving Conjunctions Solving Conjunctions The solution to a conjunction must satisfy both inequalities — both inequalities must be true. Example Example Example Example Example 11111 Solve and graph the inequality –1 < 2x + 1 < 5. Solution Solution Solution Solution Solution The aim is to get x by itself. –1 – 1 < 2x + 1 – 1 < 5 – 1 Subtr act 1 act 1 Subtr Subtr act 1 Subtract 1 act 1 Subtr –2 < 2x < 4 –1 < x < 2

y 2 to get et et et et x in the mid y 2 to g y 2 to g in the middledledledledle vide b vide b in the mid in the mid DiDiDiDiDivide b vide by 2 to g y 2 to g vide b in the mid So the solution is any number greater than –1 but less than 2. This is graphed as: –1 0 1 2 Section 3.3 Section 3.3 Section 3.3 — Compound Inequalities Section 3.3 Section 3.3 155155155155155 Guided Practice In Exercises 10–13, solve and graph each inequality. 10. 5 < 3y – 1 < 11 11. –10 < 4a + 6 < –2 12. 7 £ 7x – 7 £ 14 13. –5 £ 9t + 4 £ 22 In Exercises 14–19, solve each inequality. 14. –5 < a 10 16. –11 < –4g + 5 < –3 < 3 18. –15 < 8y + 9 £ 9 15. –11 < c − 9 7 < 5 17. –9 < 6y – 9 < 3 a + − 6 19. 7 < £ 9 6 Check it out: In this example, the lines head in opposite directions — they have no points in common. Any number in the solution set only satisfies one of the inequalities. lude the WWWWWororororord “Or” d “Or” d “Or” lude the lude the lems Inc Disjunction Proboboboboblems Inc lems Inc Disjunction Pr Disjunction Pr d “Or” lems Include the d “Or” lude the lems Inc Disjunction Pr Disjunction Pr Here’s an example of a disjunction: 3x – 4 < –4 or 3x –4 > 4 The solution to a disjunction is all the numbers that satisfy either one inequality or the other. Example Example Example Example Example 22222 Solve and graph the solution set of 3x – 4 < –4 or 3x – 4 > 4. Solution Solution Solution Solution Solution 3x – 4 + 4 < –4 + 4 3x < 0 x < 0 –1 0 or or or 1 Guided Practice 3x – 4 + 4 > 4 + 4 AdAdAdAdAdd 4d 4d 4d 4d 4 3

x > 8 x > 8 3 2 8 3 DiDiDiDiDivide b vide b vide b vide by 3y 3y 3y 3y 3 vide b In Exercises 20–23, solve the inequality and graph each solution set. 20. 7a – 7 < –7 or 7a – 7 > 21 21. 5x – 4 £ 6 or 5x – 4 ≥ 26 22. c − 9 5 < 3 or c − 9 5 ≥ 9 23. t − 7 3 £ –9 or t − 7 3 > 6 In Exercises 24–27, solve each disjunction. 24. 8c – 4 > 92 or 8c – 4 < –12 25. –9g – 7 £ 2 or –9g – 7 > 20 26. 6c + 5 > 8 or 6c + 5 £ 5 27. j − 13 6 £ –10 or j − 13 6 ≥ 5 156156156156156 Section 3.3 Section 3.3 Section 3.3 — Compound Inequalities Section 3.3 Section 3.3 Independent Practice Solve each conjunction or disjunction in Exercises 1–19. 1. –7 < 2x + 3 < 9 3. 9 < 4x + 5 < 17 5. 3 £ 3(2x – 5) £ 9 2. 8 < 3x – 4 < 14 4. –1 £ x – 3 £ 5 6. –6 £ 9 – 5x £ 19 7. 5 £ 7 – 2(x – 3) £ 21 8. –. –3 £ 11. – 10. 3 £ 4x – 9 and 4x – 9 £ 15 2 ) and 2 ) ( 3 x − 4 < 10 12. 2y + 2 < 4y – 4 or 4y – 4 > 5y + 2 13. 5y + 7 < –13 or 7 – 3y < –5 14. 11 + 2y < 4y – 3 or 4y – 3 > 6y + 7 15. –2x £ –3x + 4 and –3x + 4 £ 4x + 18 16. 3 8 x + 3 < –4 or 3 – 3 7 x < –6 17 18. –15 £ 4x – 6 £ –10 19. 4x – 9 < 27 and 10x – 16 > 2x + 8 20. The sum of three consecutive even integers is between 82 and 85. Find the numbers. 21. The sum of

three consecutive odd integers is between 155 and 160. Find the consecutive odd integers. The formula C = degrees Celsius. Use this fact to answer Exercises 22–23. (F – 32) is used to convert degrees Fahrenheit to 5 9 22. The temperature inside a greenhouse falls to a minimum of 65 °F at night and rises to a maximum of 120 °F during the day. Find the corresponding temperature range in degrees Celsius. 23. The usual temperature range of liquid water is 0 degrees Celsius (freezing point) to 100 degrees Celsius (boiling point). Find the corresponding temperature range of water in degrees Fahrenheit. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up This Topic can be a little hard to understand at first. You can write the solution to a conjunction in one statement, but disjunction solutions usually have to be written in two parts because they cover two different parts of the number line. Section 3.3 Section 3.3 Section 3.3 — Compound Inequalities Section 3.3 Section 3.3 157157157157157 TTTTTopicopicopicopicopic 3.4.13.4.1 3.4.13.4.1 3.4.1 Section 3.4 alue Inequalities alue Inequalities Absolute VVVVValue Inequalities Absolute Absolute alue Inequalities alue Inequalities Absolute Absolute Absolute VVVVValue Inequalities alue Inequalities alue Inequalities Absolute Absolute alue Inequalities Absolute alue Inequalities Absolute California Standards: 3.0:3.0:3.0:3.0:3.0: Students solv Students solveeeee Students solv Students solv Students solv inequalities inequalities equations and inequalities inequalities inequalities alues..... alues alues bsolute v bsolute v olving a inininininvvvvvolving a olving a bsolute values olving absolute v alues bsolute v olving a What it means for you: You’ll solve absolute value inequalities of the form Ωmx ± cΩ < v or Ωmx ± cΩ > v. Key words: absolute value conjunction disjunction inequality endpoint interval Check it out: The solution interval is the open interval (–4, 18). On the graph, the endpoints –4 and

18 aren’t included. You last saw absolute value equations in Section 2.8 — now you’re going to see inequalities involving absolute values. As with normal inequalities, you can have conjunctions and disjunctions with absolute value inequalities. e Compound Inequalities e Compound Inequalities alue Inequalities ar Absolute VVVVValue Inequalities ar alue Inequalities ar Absolute Absolute e Compound Inequalities alue Inequalities are Compound Inequalities e Compound Inequalities alue Inequalities ar Absolute Absolute An absolute value inequality can have a form such as |x| < m. |x| < m means that x is restricted to points on the number line less than m units from 0 — in either the positive or the negative direction, as shown on this number line: |x| < m is equivalent to –m < x < m. –m unitsm 0 m units m Guided Practice In Exercises 1–4, write the equivalent compound inequality and graph the inequality on a number line. 1. |x| < 3 3. |g| < 2 2. |a| < 7 4. |x – 1| < 10 nequalities CCCCCan an an an an BBBBBe e e e e CCCCConjunctions onjunctions onjunctions nequalities alue IIIIInequalities nequalities alue Absolute VVVVValue alue Absolute Absolute onjunctions onjunctions nequalities alue Absolute Absolute You can write an absolute value inequality of the form |mx + c| < v as the conjunction mx + c < v and mx + c > –v (or you could write it mx + c < v and –(mx + c) < v). So, to solve an inequality like this, you can solve the compound inequality –v < mx + c < v. Example Example Example Example Example 11111 Solve |x – 7| < 11. Write the solution in interval notation and graph its solution interval on a number line. Solution Solution Solution Solution Solution x – 7 < 11 and x – 7 > –11 –11 < x – 7 < 11 –11+ 7 < x – 7 + 7 < 11 + 7 –4 < x < 18 Solution interval: (–4, 18) rite the inequality as a conjunction rite the inequality as a conjunction WWWWWrite the inequality as a conjunction rite the inequality as a

conjunction rite the inequality as a conjunction e the conjunction to get et et et et x b b b b by itself y itself y itself e the conjunction to g e the conjunction to g SolvSolvSolvSolvSolve the conjunction to g y itself y itself e the conjunction to g Graph: –4 0 18 158158158158158 Section 3.4 Section 3.4 Section 3.4 — Absolute Value Inequalities Section 3.4 Section 3.4 ndpoints are e e e e IIIIIncncncncncluded luded luded ndpoints ar Sign Means the he he he he EEEEEndpoints ar ndpoints ar Sign Means t TTTTThe he he he he £££££ Sign Means t Sign Means t luded luded ndpoints ar Sign Means t Example Example Example Example Example 22222 Solve |4x – 9| £ 11. Write the solution in interval notation and graph its solution interval on a number line. Solution Solution Solution Solution Solution –11 £ 4x – 9 £ 11 –11 + 9 £ 4x – 9 + 9 £ 11 + 9 –2 £ 4x £ 20 rite the inequality as a conjunction rite the inequality as a conjunction WWWWWrite the inequality as a conjunction rite the inequality as a conjunction rite the inequality as a conjunction e the conjunction to get et et et et x b b b b by itself y itself y itself e the conjunction to g e the conjunction to g SolvSolvSolvSolvSolve the conjunction to g y itself y itself e the conjunction to g Check it out: In this example, the solution interval is closed — so the endpoints are included in the graph. – 1 2 £££££ x £££££ 5 Solution interval: [– 1 2, 5] Graph: 0 –0.5 5 Guided Practice Solve each conjunction and write each solution set in interval notation. 5. |t – 10| < 1 6. |a + 2| £ 4 7. |3a| £ 21 8. |15j| < 5 9. r 8 £ 2 10. h 12 < 1 11. |x + 8| < 10 12. |x – 11| < 4 14. |3x – 7| < 8 15. |5y + 7| £ 22 13. |2x – 5| < 11 y + 2 16. 4 5 £ 13 1 4 17. y − 3 20. |

3(2 – x)| £ 5 £ 5 1 5 18. 2 x − 5 21. |5x – 11| £ 19 £ 4 19. 3 2 x − 2 £ 4 Check it out: The graphs head in opposite directions — they have no points in common. junctions junctions nequalities CCCCCan an an an an BBBBBe e e e e DisDisDisDisDisjunctions nequalities nequalities alue IIIIInequalities alue alue Absolute VVVVValue Absolute Absolute junctions junctions nequalities alue Absolute Absolute An absolute value inequality can have a form such as |x| > m. |x| > m means that x is restricted to points on the number line more than m units from 0 — in either the positive or the negative direction, as shown on this number line: unitsm |x| > m is equivalent to x < –m or x > m. –m 0 m units m Section 3.4 Section 3.4 Section 3.4 — Absolute Value Inequalities Section 3.4 Section 3.4 159159159159159 Check it out: In this example, the solution set is in two parts. The endpoints –2 and 5 are not included. Don’t forget: The “» ” sign is the “union” sign — here it means that the solution set is (–•, –2) andandandandand (5, •). See Section 1.1 for a reminder about unions. Guided Practice In Exercises 22–25, write the equivalent compound inequality, and graph the inequality on a number line. 22. |x| > 14 24. |a| > 10 23. |t| > 8 25. |t + 8| > 23 junctions junctions mal Dis mal Dis s into Nor Absolute VVVVValuealuealuealuealues into Nor s into Nor Absolute ConConConConConvvvvvererererert t t t t Absolute Absolute junctions mal Disjunctions s into Normal Dis junctions mal Dis s into Nor Absolute An absolute value inequality of the form |mx + c| > v is the same as the disjunction mx + c < –v or mx + c > v. To solve this type of absolute value inequality, you solve the disjunction. Example Example Example Example Example 33333 Solve the compound inequality |2x – 3