Table of Contents Algebra Lesson 18: Constant, Variables, and Algebraic Expressions …… 113 Lesson 19: Verbal Phrases and Mathematical Phrases ………… 118 Lesson 20: Polynomials …………………………………………….. 123 Lesson 21: Laws of Exponents …………………………………….. 126 Lesson 22: Addition and Subtraction of Polynomials ……………. 130 Lesson 23: Multiplying Polynomials ……………………………….. 135 Lesson 24: Dividing Polynomials …………………………………... 141 Lesson 25: Special Products ……………………………………….. 146 Lesson 26: Solving Linear Equations and Inequalities in One Variable Using Guess and Check ……………………. 153 Lesson 27: Solving Linear Equations and Inequalities Algebraically ……………………………………………. 160 Lesson 28: Solving Linear Inequalities Algebraically …………….. 169 Lesson 29: Solving Absolute Value Equations and Inequalities ... 176
What about the second activity? Did you pick out the correct tiles? 5x + 3x Get five (+x tiles) and three more (+x) tiles. How many do you have in all? There are eight (+x) altogether. Therefore, 5x + 3x = 8x. (3x - 4) - 6x Get three (+x) tiles and four (-1) tiles to represent (3x - 4). Add six (-x) tiles. [Recall that subtraction also means adding the negative of the quantity.] Now, recall further that a pair of one (+x) and one (-x) is zero. What tiles do you have left? That’s right, if you have with you three (-x) and four (-1), then you are correct. That means the sum is (-3x -4). (2x2 – 5x + 2) + (3x2 + 2x)
3. (30x3 – 50x2 + 20x – 80) – (17x3 + 26x + 19) 30x3 – 50x2 + 20x – 80 + -17x3 - 26x – 19 III. Exercises A. Perform the indicated operation, first using the tiles when applicable, then using paper and pen. 1. 3x + 10x 6. 10xy – 8xy 2. 12y – 18y 7. 20x2y2 + 30x2y2 3. 14x3 + (-16x3) 8. -9x2y + 9x2y 4. -5x3 -4x3 9. 10x2y3 – 10x3y2 5. 2x – 3y 10. 5x – 3x – 8x + 6x B. Answer the following questions. Show your solution. What is the sum of 3x2 – 11x + 12 and 18x2 + 20x – 100? What is 12x3 – 5x2 + 3x + 4 less than 15x3 + 10x + 4x2 – 10? What is the perimeter of the triangle shown at the right? If you have (100x3 – 5x + 3) pesos in your wallet and you spent (80x3 – 2x2 + 9) pesos in buying foods, how much money is left in your pocket? What must be added to 3x + 10 to get a result of 5x – 3? Summary In this lesson, you learned about tiles and how to use them to represent algebraic expressions. You learned how to add and subtract terms and polynomials using these tiles. You were also able to come up with the rules in adding and
If you multiply (2x + 3) and (x – 7) by F-O-I-L method, a. the product of the first terms is __________. b. the product of the outer terms is __________. c. the product of the inner terms is __________. d. the product of the last terms is __________. e. Do you see any similar terms? What are they? f. What is the result when you combine those similar terms? g. The final answer is _____________. Another Way of Multiplying Polynomials 1. Consider this example. 2. Now, consider this. 7 8 2x + 3 X 5 9 x – 7 7 0 2 14x + 21 _3 9 0 2x2 + 3x 4 6 0 2 2x2 + 17x + 21 Consider the example below. 3a – 5b 4a + 7 21a – 35b 12a2 – 20ab 12a2 – 20ab + 21a – 35b D. To multiply a polynomial with more than one term by a polynomial with three or more terms, simply distribute the first term of the first polynomial to each term of the other polynomial. Repeat the procedure up to the last term and simplify the results by combining similar terms. Examples: 1) (x + 3) (x2 – 2x + 3) = x(x2 – 2x + 3) – 3(x2 – 2x + 3) = x3 – 2x2 + 3x – 3x2 + 6x – 9 = x3 – 5x2 + 9x – 9 2) (x2 + 3x – 4) (4x3 + 5x – 1) = x2(4x3 + 5x – 1) + 3x(4x3 + 5x – 1) - 4(4x3 + 5x – 1)
= 4x5 + 5x3 – x2 + 12x4 + 15x2 – 3x – 16x3 – 20x + 4 = 4x5 + 12x4 – 11x3 + 14x2 – 23x + 4 3) (2x – 3) (3x + 2)(x2 – 2x – 1) = (6x2 – 5x – 6)(x2 – 2x – 1) = 6x4 – 17x3 – 22x2 + 17x + 6 *Do the distribution one by one. III. Exercises A. Simplify each of the following by combining like terms. 1) 6x + 7x 2) 3x – 8x 3) 3x – 4x – 6x + 2x 4) x2 + 3x – 8x + 3x2 5) x2 – 5x + 3x – 15 B. Call a student or ask for volunteers to recite the basic laws of exponent but focus more on the “product of a power” or ”multiplying with the same base”. Give follow up exercises through flashcards. 1) x12 ÷ x5 2) a-10 • a12 3) x2 • x3 4) 22 • 23 5) x100 • x C. Answer the following. 1. Give the product of each of the following. a) (12x2y3z)(-13ax3z4) b) 2x2(3x2 – 5x – 6) c) (x – 2)(x2 – x + 5) 2. What is the area of the square whose side measures (2x – 5) cm? (Hint: Area of the square = s2) 3. Find the volume of the rectangular prism whose length, width and height are (x + 3) meter, (x – 3) meter and (2x + 5) meter. (Hint: Volume of rectangular prism = l x w x h) 4. If I bought (3x + 5) pencils which cost (5x – 1) pesos each, how much will I pay for it? Summary
Find someone who Can give the value of x so that x + 3 = 5 is a true equation. Can determine the smallest integer value for x that can hold x > 1.5 true. Can solve by guess and check for the solution of 9x – 1=8. Can give the value of 3x – 1 if x = 3. Can give the numerical value of 3(22 – 32). Knows which is greater between x3 and 3x when x = 2. Can translate the phrase ‘a number x increased by 3 is 2’ to algebraic expression. Can determine which of these {0,1, 2,…, 8, 9} is/are solution/s of 3x < 9. Can write an inequality for which all positive numbers are NOT solutions. Knows the largest integer value of x that can satisfy the inequality 2x – 1 < 3? Knows what Arabic word is known to be the origin of the word Algebra. Can write an equation that is true when x = 4. Can write a simple inequality that will is satisfied by the elements in the set {–1, 0, 1.1, , 3, 4, …}. Can name the set of numbers satisfying the inequality x < 0. Can explain what an open sentence is. Can give the positive integer values of x that can satisfy x + 3 < 6. Summary In this lesson, you learned how to evaluate linear equations at a specific value of x. You also learned to determine whether particular values of x are solutions to linear equations and inequalities in one variable.
What number/expression must be placed in the box to make each statement correct? 1. x – 20 < –12 Given x – 20 + < –12 + API x < 8 2. –7x ≥ 49 Given (–7x) ( ) ≥ (49)( ) MPI x ≤ –7 3. – 8 > 3 Given – 8 + > 3 + API > 11 > (3) MPI x > 12 4. 13x + 4 < –5 + 10x Given 13x + 4 + < –5 + 10x + API 3x + 4 < –5 3x + 4 + < –5 + API 3x < –9 ( )3x < (–9)( ) MPI x < –3
POINTS TO REMEMBER: The last statement in each item in the preceding set of exercises is the solution set of the given inequality. For example, in no. 4, the solution to 13x + 4 < –5 + 10x consists of all numbers less than –3 (or x < –3). This solution represents all numbers that make the inequality true. The solution can be written using set notation as {x | x < –3}. This is read as the “set of all numbers x such that x is less than –3”. To solve linear inequalities in one variable, isolate the variable that you are solving for on one side of the inequality by applying the properties of inequality. V. Questions/Points to Ponder Observe how the properties of inequality may be used to find the solution set of linear inequalities: 1. b + 14 > 17 b + 14 – 14 > 17 – 14 b > 3 Solution Set: {b | b > 3} ‘2. 4t – 17 < 51 4t – 17 + 17 < 51 + 17 4t < 68 t < 17 Solution Set: {t | t < 17} 2r – 32 > 4r + 12 2r – 32 – 4r > 4r + 12 – 4r –2r – 32 > 12 –2r – 32 + 32 > 12 + 32 –2r > 44 r ≤ –22 Solution Set: {r | r < –22} VI. Exercises Find the solution set of the following inequalities. 1. b – 19 ≤ 15 6. 3w + 10 > 5w + 24 2. 9k ≤ –27 7. 12x – 40 ≥ 11x – 50 3. –2p > 32 8. 7y + 8 < 17 + 4y 4. 3r – 5 > 4 9. h – 9 < 2(h – 5) 5. 2(1 + 5x) < 22 10. 10u + 3 – 5u > –18 – 2u VII. Questions/Points to Ponder
Step 3: If the absolute value of an expression is equal to a positive number, say a, then the expression inside the absolute value can either be a or –a (Again, think, why?). Equate the expression inside the absolute value sign to a and to –a, and solve both equations. Example 1: Solve |3a – 4| – 9 = 15. Step 1: Let the expression on one side of the equation consist only of a single absolute value expression. |3a – 4| – 11 = 15 |3a – 4| = 26 Step 2: Is the number on the other side of the equation negative? No, it’s a positive number, 26, so proceed to step 3 Step 3: To satisfy the equation, the expression inside the absolute value can either be +26 or –26. These correspond to two equations. 3a – 4 = 26 3a – 4 = –26 Step 4: Solve both equations. 3a – 4 = 26 3a = 30 a = 10 3a – 4 = –26 3a = –22 a = We can check that these two solutions make the original equation true. If a = 10, then |3a – 4| – 9 = |3(10) – 4| – 9 = 26 – 9 = 15. Also, if a = –22/3, then |3a – 4| – 9 = |3(–22/3) – 4| – 9 = |–26| – 9 = 15. Example 2: Solve |5x + 4| + 12 = 4. Step 1: Let the expression on one side of the equation consist only of a single absolute value expression. |5x + 4| + 12 = 4 |5x + 4| = –8 Step 2: Is the number on the other side of the equation negative? Yes, it’s a negative number, –8. There is no solution because |5x + 4| can never be negative, no matter what we substitute for x.
Example 3: Solve |c – 7| = |2c – 2|. Step 1: Let the expression on one side of the equation consist only of a single absolute value expression. Done, because the expression on the left already consists only of a single absolute value expression. Step 2: Is the number on the other side of the equation negative? No, because |2c – 2| is surely not negative (the absolute value of a number can never be negative). Proceed to Step 3. Step 3: To satisfy the equation, the expression inside the first absolute value, c – 7, can either be +(2c – 2) or –(2c – 2). These correspond to two equations. [Notice the similarity to Step 3 of Example 1.] c – 7 = +(2c – 2) c – 7 = –(2c – 2) Step 4: Solve both equations. c – 7 = +(2c – 2) c – 7 = 2c – 2 –c – 7 = –2 –c = 5 c = –5 c – 7 = –(2c – 2) c – 7 = –2c + 2 3c – 7= 2 3c = 9 c = 3 Again, we can check that these two values for c satisfy the original equation. Example 4: Solve |b + 2| = |b – 3| Step 1: Let the expression on one side of the equation consist only of a single absolute value expression. Done, because the expression on the left already consists only of a single absolute value expression. Step 2: Is the number on the other side of the equation negative? No, because |b – 3| is surely not negative (the absolute value of a number can never be negative). Proceed to Step 3. Step 3: To satisfy the equation, the expression inside the first absolute value, b + 2, can either be equal to +(b – 3) or –(b – 3). These correspond to two equations. [Notice the similarity to Step 3 of Example 1.] b + 2 = +(b – 3) b + 2 = –(b – 3)
Step 4: Solve both equations. b + 2 = +(b – 3) b + 2 = b - 3 2 = –3 This is false. There is no solution from this equation. b + 2 = –(b – 3) b + 2 = –b + 3 2b + 2 = 3 2b = 1 b = Since the original equation is satisfied even if only of the two equations in Step 3 were satisfied, then this problem has a solution: b = . This value of b will make the original equation true. Example 5: Solve |x – 4| = |4 – x|. Step 1: Let the expression on one side of the equation consist only of a single absolute value expression. Done, because the expression on the left already consists only of a single absolute value expression. Step 2: Is the number on the other side of the equation negative? No, because |4 – x| is surely not negative (the absolute value of a number can never be negative). Proceed to Step 3. Step 3: To satisfy the equation, the expression inside the first absolute value, x – 4, can either be equal to +(4 – x) or –(4 – x). These correspond to two equations. [Notice the similarity to Step 3 of Example 1.] x – 4 = +(4 – x) x – 4 = –(4 – x) Step 4: Solve both equations. x – 4 = +4 – x 2x – 4 = 4 2x = 8 x = 4 x – 4 = –(4 – x) x – 4 = –4 + x –3 = –3 This is true no matter what value x is. All real numbers are solutions to this equation. Since the original equation is satisfied even if only of the two equations in Step 3 were satisfied, then the solution to the absolute value equation is the set of all real numbers.
4. Solve for c: |c + 12| + 3 > 17 A. {c | c > –2 or c < 2} B. {c | c > –26 or c < 2} C. {c | c < –2 or c > 2 } D. {c | c < –26 or c > 2} 5. Solve the absolute value inequality: |1 – 2w| < 5 A. {c | 3 < c < –2} B. {c | –3 < c < 2} C. {c | 3 > c > –2} D. {c | –3 > c > 2} VII. Questions/Points to Ponder Solve the following problems involving absolute value. You need to cut a board to a length of 13 inches. If you can tolerate no more than a 2% relative error, what would be the boundaries of acceptable lengths when you measure the cut board? (Hint: Let x = actual length, and set up an inequality involving absolute value.) A manufacturer has a 0.6 oz tolerance for a bottle of salad dressing advertised as 16 oz. Write and solve an absolute value inequality that describes the acceptable volumes for “16 oz” bottles. (Hint: Let x = actual amount in a bottle, and set up an inequality involving absolute value.) Summary In this lesson you learned how to solve absolute value equations and absolute value inequalities. If a is a positive number, then the solution to the absolute value equation |x| = a is x = a or x = –a. There are two types of absolute value inequalities, each corresponding to a different procedure. If |x| < k, then –k< x < k. If |x| > k, then x < –k or x > k. These principles work for any positive number k.