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| > 7. Write the solution set in interval notation and graph the solution set. Solution Solution Solution Solution Solution 2x β 3 < β7 or 2x < β4 x < β2 2x β 3 > 7 WWWWWrite the inequality as a disjunction rite the inequality as a disjunction rite the inequality as a disjunction rite the inequality as a disjunction rite the inequality as a disjunction 2x > 10 e the disjunction to get et et et et x b b b b by itself x > 5 Solv Solv Solv y itself y itself e the disjunction to g e the disjunction to g Solve the disjunction to g y itself Solv y itself e the disjunction to g Solution set: (ββ’β’β’β’β’, β2) »»»»» (5, β’β’β’β’β’) Graph: β2 0 5 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 β₯β₯β₯β₯β₯ Sign Means t Sign Means t luded luded ndpoints ar Sign Means t Remember that the solution interval is closed if there is a βgreater than or equal toβ sign. Example Example Example Example Example 44444 Solve and graph |5x + 11| β₯ 6. Write the solution set in interval notation. Solution Solution Solution Solution Solution 5x + 11 Β£ β6 or 5x Β£ β17 x £££££ β 17 5 5x + 11 β₯ 6 5x β₯ β5 x β₯β₯β₯β₯β₯ β1 e to get et et et et x b b b b by itself SolvSolvSolvSolvSolve to g y itself y itself e to g e to g y itself y itself e to g Check it out: Here, the solution intervals are half-open. The endpoints β3.4 and β1 are included. Solution set: (ββ’β’β’β’β’, β3.4] »»»»» [β1, β’β’β’β’β’) Graph: β3.4 β1 0 160160160160160 Section 3.4 Section 3.4 Section 3.4 β Absolute Value Inequalities Section 3.4 Section 3.4 Guided Practice Solve each disjunction and write each |
solution set in interval notation. 26. |a β 8| > 1 27. |t + 2| > 8 28. |7a| β₯ 14 29. |4j| β₯ 16 30. c 3 > 6 31. c 12 β₯ 1 3 32. |x + 9| > 7 33. |x β 11| > 12 34. |3x β 7| > 13 35. |5y + 11| > 21 36. |4m + 9| β₯ 11 37. |7b β 8| β₯ 13 38. 3 4 x β 3 β₯ 5 39. 2 1 x β 7 β₯ 3 40. |3(x β 2) + 7| β₯ 8 41. |4(3 + x)| β₯ 13 42. |2(3x β 7) + 15| β₯ 11 Independent Practice 1. Write |a| > 5 as a compound inequality. 2. Write |w| Β£ c as a compound inequality. In Exercises 3β4 write the equivalent compound inequality in interval notation. 3. |a| < 4 4. |b| β₯ 2 In Exercises 5β12, solve each inequality. 5. g 2 β₯ 3 7. |w β 17| Β£ 8 9. |5t + 10| β₯ 15 11. 8 t β 7 3 β₯ 2 6. |c + 2| > 24 8. |4a β 3| Β£ 9 x + 2 10. 3 1 < 1 12. 17 2 c + 5 > 4 13. Solve and graph the solution set of 7|2(x β 7) + 5| β₯ 14. 14. Given m > 0, graph the solution set of |x| Β£ m on a number line. 15. Given m > 0, solve and graph the solution set of |x| β₯ m. 16. Solve |3m β 5| > m + 7. 17. Solve |5m + 3| > 2m β 1. 18. A floor tile must measure 20 cm along its length, to within 2 mm. Write and solve an absolute value inequality to find the maximum and minimum possible lengths for the tile. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Like in Section 3.3, youβve got to watch out for the difference between conjunctions and disjunctions. Also, look back at Section 1 |
.1 if youβve forgotten what the βΒ»β sign means. Section 3.4 Section 3.4 Section 3.4 β Absolute Value Inequalities Section 3.4 Section 3.4 161161161161161 Chapter 3 Investigation Mailing Pacacacacackakakakakagggggeseseseses Mailing P Mailing P Mailing P Mailing P Mailing Pacacacacackakakakakagggggeseseseses Mailing P Mailing P Mailing P Mailing P Inequalities turn up all the time in everyday life. In this investigation youβll see that using inequalities makes it much easier to model some real-life problems. A shipping company will deliver packages of any weight, as long as the conditions on the right are satisfied. Part 1: Design a set of six boxes of different shapes and sizes that meet these requirements. Part 2: What is the maximum volume a box can have while still satisfying the shipping companyβs requirements? Packages must satisfy the following: Length + Width + Height Β£ 120 cm Length Β£ 80 cm Width Β£ 80 cm Height Β£ 80 cm Volume = length Γ width Γ height Things to think about: Try to include a wide range of different sizes and shapes in your set of boxes. Aim to make your set as useful as possible for mailing different types of item. Extension For these extensions, only boxes in the shapes of rectangular prisms should be considered. 1) What is the maximum surface area a box can have while still satisfying the shipping companyβs requirements? What are the dimensions of this box? 2) You need to ship a picture that has a height of 83 cm. How wide could the picture be so that it fits in a rectangular prism shaped box that satisfies the shipping companyβs requirements? Write an inequality to represent the possible width of the picture. width 83 c m Open-ended Extension 1) You work for an organization that produces pocket dictionaries. The dimensions of the dictionaries are shown on the right. What size box will hold the greatest number of dictionaries, while still satisfying the shipping companyβs requirements? The dictionaries donβt all have to be placed in the box the same way. Experiment with placing some dictionaries vertically and some horizontally. 10 cm 6cm c m 3 2) Your company is concerned about the environment and wants to use the smallest area of cardboard possible to package the books in. Design the box that |
you think will be most efficient. (Assume the box does not need any tabs to stick it together.) ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up In general, problems that use the words βmaximum,β βminimum,β βlimits,β or βtoleranceβ often mean that you need to use inequalities to model the situations. 162162162162162 estigaaaaationtiontiontiontion β Mailing Packages estigestig estig pter 3 Invvvvvestig pter 3 In ChaChaChaChaChapter 3 In pter 3 In pter 3 In Chapter 4 Linear Equations and Their Graphs Section 4.1 The Coordinate Plane........................... 164 Section 4.2 Lines...................................................... 175 Section 4.3 Slope..................................................... 189 Section 4.4 More Lines............................................ 197 Section 4.5 Inequalities............................................ 212 Investigation Tree Growth........................................... 227 163163163163163 TTTTTopicopicopicopicopic 4.1.14.1.1 4.1.14.1.1 4.1.1 Section 4.1 te Plane te Plane he Coordinadinadinadinadinate Plane he Coor he Coor TTTTThe Coor te Plane te Plane he Coor he Coordinadinadinadinadinate Plane TTTTThe Coor te Plane te Plane he Coor he Coor te Plane he Coor te Plane California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students g Students g ph a linear ph a linear Students g tion tion equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: Youβll refresh your knowledge of plotting coordinates. Key words: coordinate horizontal vertical point of intersection origin Check it out: The x-coordinate is sometimes called the abscissa, and the y-coordinate is sometimes called the ordinate. Check |
it out: The coordinates of the origin are (0, 0). Plotting coordinates isnβt anything new to you β youβve had lots of practice in earlier grades. This Topic starts right at the beginning though, to remind you of the earlier work. e Used to Locate Pte Pte Pte Pte Points on a Plane oints on a Plane oints on a Plane e Used to Loca e Used to Loca tes ar CoorCoorCoorCoorCoordinadinadinadinadinates ar tes ar oints on a Plane tes are Used to Loca oints on a Plane e Used to Loca tes ar A plane is a flat surface, kind of like a blackboard, a tabletop, or a sheet of paper. However, a plane extends indefinitely in all directions β it goes on and on forever. Planes are made up of an infinite number of points. To locate one of these points, two perpendicular number lines are drawn in the plane. The horizontal number line is called the x-axis. y-axis The vertical number line is called the y-axis. The point of intersection of the two number lines is called the origin. origin x-axis You can locate each point in the plane using an ordered pair of numbers (x, y), where x represents the horizontal distance and y represents the vertical distance of the point from the origin. The numbers in an ordered pair are often called coordinates. Example Example Example Example Example 11111 Plot the coordinates (3, 4) on a graph. Solution Solution Solution Solution Solution (3, 4) x-coordinate y-coordinate 3 units across (3, 4) 4 units up 164164164164164 Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 Guided Practice In Exercises 1β4, plot the coordinates on a coordinate plane. 1. (4, 0) 3. (8, 2) 2. (β2, 5) 4. (5, β2) Use the graph opposite to answer Exercises 5β6. 5. What are the coordinates of point A? 6. What are the coordinates of point B? B β4 β2 A 2 4 4 2 0 β2 β4 han One Pointointointointoint han One P y Need to Plot More e e e e TTTTThan One P han One P y Need to Plot |
Mor y Need to Plot Mor ou Ma YYYYYou Ma ou Ma ou May Need to Plot Mor han One P y Need to Plot Mor ou Ma You can often join up several plotted points to create the outline of a shape. Example Example Example Example Example 22222 Draw a coordinate plane and plot and label the points M (3, 4), A (β3, 4), C (β3, β4), and V (3, β4). Connect each pair of consecutive points and find the perimeter of the resulting quadrilateral. Solution Solution Solution Solution Solution A negative x-coordinate means the point is left of the y-axis. A -5-6 -4 -3 -2 C 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 - negative y-coordinate means the point is below the x-axis. The length of each square on the coordinate grid represents one unit of measure. So, the quadrilateral is 8 units long and 6 units wide. Perimeter (distance around the edge) = 2(6) + 2(8) = 12 + 16 = 28 units Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 165165165165165 Guided Practice Work out Exercises 7β9 by plotting and labeling the points on copies of the coordinate plane used in Example 2. 7. A (0, 5), B (β5, 2), C (β5, β4), D (5, β4), and E (5, 2). Name the figure formed when the points are connected in order. 8. A (3, 3), B (β3, 3), C (β3, β3), and D (3, β3). Connect the points in order and then name and find the area of the figure formed. 9. A (2, 5), B (2, β2), C (5, β2), D (5, β4), E (β5, β4), F (β5, β2), G (β2, β2), and H (β2, 5). Connect the points in order and find the perimeter of the figure formed. Independent Practice 1. What is the difference between the x-axis and the y-axis? 2. Explain in words how to graph the coordinates (5, 3). Use the graph opposite to answer Exerc |
ises 3β7. 3. What are the coordinates of point A? 4. What are the coordinates of point B? 5. What are the coordinates of point C? 6. What are the coordinates of point D? 7. What are the coordinates of point E? D β4 β2 4 2 0 β2 β4 C A 2 B 4 E In Exercises 8β11, plot the coordinates on a coordinate plane. 8. The origin 9. (2, 1) 10. (β4, 1) 11. (3, β2) 12. Plot and label the following points: A (β1, β2), B (2, 4), C(5, β2), D (β2, 2), and E (6, 2). Connect the points in order, then connect E to A, and name the figure formed. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That Topic should have felt quite familiar. Remember that coordinate pairs always list the x-coordinate first, then the y-coordinate β and watch out for any negative numbers. In the next Topic youβll look at each part of the coordinate plane in more detail. Section 4.1 166166166166166 Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 TTTTTopicopicopicopicopic 4.1.24.1.2 4.1.24.1.2 4.1.2 California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students g Students g ph a linear ph a linear Students g tion tion equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: Youβll look in more detail at the four main regions of the coordinate plane, and the axes. Key words: coordinate quadrant axes the Plane the Plane ants of ants of Quadr Quadr the Plane ants of the Plane Quadrants of the Plane ants of Quad |
r Quadr the Plane the Plane ants of ants of Quadr Quadr ants of the Plane Quadrants of the Plane ants of Quadr the Plane Quadr There are four main regions of the coordinate plane β theyβre divided up by the x- and y-axes. This Topic is about spotting where on the coordinate plane points lie. our Quadrantsantsantsantsants our Quadr he Plane Consists of F F F F Four Quadr our Quadr he Plane Consists of TTTTThe Plane Consists of he Plane Consists of our Quadr he Plane Consists of The coordinate axes divide the plane into four regions called quadrants. The quadrants are numbered counterclockwise using Roman numerals. The signs of the coordinates differ from quadrant to quadrant, as shown in the diagram below. If the x-coordinate is negative and the y-coordinate is positive, the point is in quadrant II. If the x- and y-coordinates are both positive, the point is in quadrant I. If the x- and y-coordinates are both negative, the point is in quadrant III. If the x-coordinate is positive and the y-coordinate is negative, the point is in quadrant IV. So, you can easily tell which quadrant a particular point is in by simply looking at the signs of its coordinates. Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 167167167167167 Example Example Example Example Example 11111 State the quadrant where each point is located. Justify your answer. a) (β2, 3) c) (a, βa) where a > 0 b) (111, 2111) d) (βm, βm) where m > 0 Solution Solution Solution Solution Solution a) (β2, 3) is in quadrant II, since the x-coordinate is negative and the y-coordinate is positive. b) (111, 2111) is in quadrant I, since the x- and y-coordinates are both positive. c) (a, βa) is in quadrant IV, since the x-coordinate is positive and the y-coordinate is negative (the problem states that a > 0). d) (βm, βm) is in quadrant III, since the x and y-coordinates are both negative (the problem states that m |
> 0). Guided Practice In Exercises 1β6, name the quadrant or axis where each point is located. Justify your answers. 1. (1, 0) 4. (0, β5) In Exercises 7β12, a > 0, k > 0, m < 0, v < 0, and p Ε R. Name the quadrant or axis where each point is located. Justify your answers. 7. (a, k) 10. (p, 0) 3. (β4, β1) 6. (βp, p) 8. (m, k) 11. (0, p) 2. (2, 3) 5. (β5, 8) 9. (a, v) 12. (v, v) In Exercises 13β18, a < β2. Name the quadrant where each point is located. Justify your answers. 13. (3, a) 16. (2a, 2) 15. (β3, βa) 18. (4, a β 3) 14. (a, β1) 17. (a β 2, 3) 168168168168168 Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 - or y-Ax-Ax-Ax-Ax-Axeseseseses e on the x- or - or - or e on the e on the oints ar oints ar Some P Some P oints are on the Some Points ar - or e on the oints ar Some P Some P Points in the plane are located in one of the four quadrants or on the axes. All points whose coordinates are in the form (x, 0) are on the x-axis β for example, (3, 0), (1, 0), (β2, 0), (β3.5, 0). All points whose coordinates are in the form (0, y) are on the y-axis β for example, (0, 2), (0, 4), (0, β3), (0, β2). (0, 4) (0, 2) (β3.5, 0) (β2, 0) (1, 0) (3, 0) (0, β2) (0, β3) Example Example Example Example Example 22222 State which axis, if either, these |
points lie on. Justify your answer. a) (0, 3) b) (12, 2) c) (45, 0) Solution Solution Solution Solution Solution a) (0, 3) is on the y-axis, since x = 0. b) (12, 2) isnβt on an axis, since neither x nor y is 0. c) (45, 0) is on the x-axis, since y = 0. Guided Practice In Exercises 19β24, let t > 0. State which axis, if either, each point lies on. 19. (0, 1) 22. (t, 0) 20. (0, β3) 23. (β4, 0) 21. (t, 1) 24. (t + 1, 0) Independent Practice In Exercises 1β12, let a > 0 and b < 0. Name the quadrant or axis where each point is located, and justify your answers. 1. (5, a) 4. (0, β3) 7. (b, 1) 10. (2b, a) 2. (a, 4) 5. (β2, 0) 8. (b, 3) 11. (βb, a) 3. (a + 2, 4) 6. (0, 0) 9. (2a, b) 12. (βa, β2b) ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Another straightforward Topic, really. Working out which quadrant a coordinate pair appears in is all about checking the signs of each of the numbers β and a coordinate pair can only be on one of the axes if one of the numbers is 0. In the next Topic youβll look at lines plotted on the coordinate plane. Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 169169169169169 TTTTTopicopicopicopicopic 4.1.34.1.3 4.1.34.1.3 4.1.3 Lines Lines Lines Lines Lines Lines Lines Lines Lines Lines California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students |
g Students g ph a linear ph a linear Students g tion tion equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: Youβll learn a formal definition of a line, and youβll refresh your knowledge of line equations. Key words: coordinate quadrant axes Thatβs enough of learning about the coordinate plane itself β now itβs time to plot lines. If you join two points in a coordinate plane youβll form part of a line. Lines Havvvvve No Endpoints e No Endpoints e No Endpoints Lines Ha Lines Ha e No Endpoints e No Endpoints Lines Ha Lines Ha A straight line extends indefinitely in opposite directions. Lines have: no endpoints (they have no beginning or end) infinite length no thickness To identify a straight line you just need two points on it. So, to draw a straight line in a plane, simply plot two points and connect them using a straightedge. Example Example Example Example Example 11111 Draw the straight line defined by the points (β2, 3) and (5, β1). Solution Solution Solution Solution Solution -5-6 -4 -3 -2 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 -6 1 2 3 4 5 6 The arrowheads show that the line continues indefinitely (without ending). Guided Practice Complete the sentence in Exercise 1. 1. A straight line extends ___________ in opposite directions. In Exercises 2β8, draw the straight line defined by the pairs of coordinates. 2. (3, 1) and (β4, 2) 3. (0, 0) and (β3, β3) 4. (β2, 1) and (3, 1) 5. (3, β2) and (β1, β2) 6. (1, 3) and (1, β4) 7. (5, 1) and (β3, 2) 8. (3, 3) and (β1, β1) 170170170170170 Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4. |
1 y an Equationtiontiontiontion y an Equa y an Equa A Line Can Be Described b A Line Can Be Described b A Line Can Be Described by an Equa y an Equa A Line Can Be Described b A Line Can Be Described b Check it out: At the point (3, 5), x = 3. You can check that (3, 5) is of the form (x, 2x β 1) by substituting 3 for x. In other words: (x, 2x β 1) ο¬ (3, 2(3) β 1) ο¬ (3, 5). All the points that lie on the line shown have coordinates (x, y), where y = 2x β 1. This means that the coordinates all have the form (x, 2x β 1) β for example, (β1, β3), (1, 1), (2, 3), (3, 5). y = 2x β 1 is called the equation of the line. (3, 5) (2, 3) (1, 11 -2 -3 -4 -5 -6 -6 -5 -4 -3 -2 -1 (β1, β3) y x= 2 β 1 Example Example Example Example Example 22222 The points on the line y = 7x + 3 are defined by (x, 7x + 3). Find the coordinates of the points where x Ε {0, 1, 2}. Solution Solution Solution Solution Solution Just substitute each value of x into (x, 7x + 3) to find the coordinates. x = 0 means (x, 7x + 3) = (0, 7(0) + 3) = (0, 3) x = 1 means (x, 7x + 3) = (1, 7(1) + 3) = (1, 10) x = 2 means (x, 7x + 3) = (2, 7(2) + 3) = (2, 17) So the coordinates of the points are (0, 3), (1, 10), and (2, 17). Guided Practice Draw out a coordinate grid spanning β6 to 6 on the x-axis and β6 to 6 on the y-axis. Draw and label the following lines on the grid. 9. y = 3 12. x = 2 10. y = β4 13. x = β5 11. y |
= 0 14. x = 0 Draw the graphs for Exercises 15β16 on coordinate grids spanning β6 to 6 on the x-axis and β6 to 6 on the y-axis. 15. Draw the graph of the set of all points (x, y) such that x = y. 16. Draw the graph of the set of points (x, y) such that x = 5 and y Ε R. Describe the line you have drawn. 17. If x Ε {β1, 1, 3}, find the set M of points defined by (x, β2x + 1). Exercises 18 and 19 are about the set of ordered pairs (x, 5x β 6). 18. Name the five members of the set if x is a natural number less than 6. 19. Which members of this set are also members of the set of ordered pairs (x, x2)? Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 171171171171171 Independent Practice In Exercises 1β2, x is an integer greater than β4 and less than 0. 1. Find the set of points defined by (x, 3x β 2). 2. Which member of this set is also a member of the set of ordered pairs (x, x β 4)? Work out Exercises 3β4 by plotting and labeling the points on a grid spanning β6 to 6 on the x-axis and β3 to 9 on the y-axis. 3. If x Ε {β3, β2, β1, 0, 1, 2, 3}, plot the set of points defined by (x, x2). 4. If x Ε {β5, β4, β3, β2, β1, 0, 1, 2, 3}, plot the set of points defined by (x, |x + 1|). 5. What shape would you expect the graph to be for the set of points defined by (x, |x|)? Work out Exercises 6β11 by plotting and labeling the points on a grid spanning β6 to 6 on the x-axis and β6 to 6 on the y-axis. 6. Plot two lines l1 and l2 whose points are defined by l1 = (x, 2x + 1) and l2 = (x, 2x + 3). 7. Describe |
the two lines defined in Exercise 6 and the relationship between them. 8. Plot the two lines whose points are defined by (x, β2x + 1) and (x, β2x β 2). 9. Describe the two lines defined in Exercise 8 and the relationship between them. 10. Draw two lines whose points are defined by (x, 2x + 1) and (x, β0.5x + 1). 11. Give the coordinates of the point where the two lines defined in Exercise 10 intersect. Work out Exercises 12β14 by plotting and labelling the points on a grid spanning β6 to 6 on the x-axis and β9 to 3 on the y-axis. 12. If x Ε {β5, β4, β3, β2, β1, 0, 1, 2, 3}, plot all points defined by (x, β|x + 1|). 13. Compare the graph of the set of points defined by (x, |x + 1|) (in Exercise 4) with the graph of the set of points defined by (x, β|x + 1|) above. Describe the relationship between the two graphs. 14. What would you expect the relationship to be between the graph of the set of points defined by (x, |x|) and the graph of the set of points defined by (x, β|x|)? ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You know what a line is from everyday life β but notice that the Math definition is a little more precise. In Math, a line is infinitely long and doesnβt actually have any thickness. Itβs hard to imagine, but itβs OK to just carry on without worrying too much about how strange that seems. 172172172172172 Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 TTTTTopicopicopicopicopic 4.1.44.1.4 4.1.44.1.4 4.1.4 California Standards: Students grrrrraaaaaph a linear 6.0:6.0:6.0:6.0:6.0: Students g ph a linear ph a linear Students g Students g ph a linear ph a linear Students g tion tion |
equa equa tion and compute the xequation equa tion equa and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). What it means for you: Youβll check whether lines are horizontal or vertical. Key words: horizontal vertical constant tical Lines tical Lines ontal and VVVVVererererertical Lines ontal and ontal and HorizHorizHorizHorizHorizontal and tical Lines tical Lines ontal and ontal and VVVVVererererertical Lines HorizHorizHorizHorizHorizontal and tical Lines tical Lines ontal and ontal and tical Lines ontal and tical Lines You can tell a lot about a line just by looking at the points it goes through. One of the simplest things to spot without plotting the graph is whether the line is horizontal or vertical. e the Same x-Coor -Coordinadinadinadinadinatetetetete -Coor -Coor e the Same tical Line Havvvvve the Same e the Same tical Line Ha oints on a VVVVVererererertical Line Ha tical Line Ha oints on a PPPPPoints on a oints on a -Coor e the Same tical Line Ha oints on a The x-coordinate tells you how far to the left or right of the y-axis a point is. Points with the same x-coordinate are all the same horizontal distance from the y-axis. So, if a set of points all have the same x-coordinate, that set will fall on a vertical line. The equation of a vertical line is x = c, where c is a constant (fixed) number. For example, x = 3, x = β1. Example Example Example Example Example 11111 Draw and label the lines x = β3 and x = 2. Solution Solution Solution Solution Solution (β3, 4) -3 -4 (β3, β2) -2 4 3 2 1 -1 0 -1 -2 -3 -4 (2, 0) 1 2 3 4 (2, β4) The values of the y-coordinates are different for each point on the line, but the values |
of x are the same (β3 for the line x = β3, 2 for the line x = 2). PPPPPoints on a Horiz oints on a Horiz oints on a Horiz ontal Line ha ontal Line ha ontal Line havvvvve the Same e the Same e the Same -Coor -Coor -Coordinadinadinadinadinatetetetete oints on a Horizontal Line ha e the Same y-Coor oints on a Horiz ontal Line ha e the Same -Coor The y-coordinate tells you how far above or below the x-axis a point is. Points with the same y-coordinate are all the same vertical distance from the x-axis. So, if a set of points all have the same y-coordinate, that set will fall on a horizontal line. Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 173173173173173 The equation of a horizontal line is y = c, where c is a constant (fixed) number. For example, y = 6, y = β3. Example Example Example Example Example 22222 Draw and label the lines y = 4 and y = β1. Solution Solution Solution Solution Solution y = 4 (β3, 4) y = β1 -3 -2 -4 (β3, β1) (2, 4) 1 2 3 4 (4, β1) 4 3 2 1 -1 0 -1 -2 -3 -4 Guided Practice The values of the x-coordinates are different for each point on the line, but the values of y are the same (β1 for the line y = β1, 4 for the line y = 4). In Exercises 1β10, draw and label each set of lines on a coordinate plane. 1. x = 3 and x = β1 3. x = 2 and x = β3 5. y = 4 and y = 1 7. y = 3 and x = 2 9. y = β7 and x = β5 2. x = 4 and x = 1 4. y = 3 and y = β1 6. y = 2 and y = β3 8. y = 6 and x = β4 10. x = 6 and y = 4 Independent Practice In Exercises 1β6, draw and label each set of lines on a |
coordinate plane. 1. x = 8 and x = 0 3. x = 4 and x = β6 5. y = β4 and y = β6 2. x = β3 and x = 1 4. y = 0 and y = 3 6. y = 2 and y = β3 In Exercises 7β12, write the equation for each line on the graph. y 4 3 2 1 β6 β5 β4 β3 β2 β1 0 β1 β2 β3 β4 β5 β6 β7 β8 7. 654321 x 8. 9. 10. y 20 18 16 14 12 10 8 6 4 2 β12β10β8β6 β4 β2 0 β2 β4 8642 10 x 12 11. 8642 10 x 12 y 8 6 4 2 β12β10β8β6 β4 β2 0 β2 β4 β6 β8 β10 β12 β14 β16 12. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up One thing that can look confusing at first is that the line x = 0 is actually the y-axis, while the line y = 0 is the x-axis. And a line of the form x = c (for any constant c) is a vertical line (that is, it is parallel to the y-axis) β while a line y = c is a horizontal line (that is, itβs parallel to the x-axis). 174174174174174 Section 4.1 Section 4.1 Section 4.1 β The Coordinate Plane Section 4.1 Section 4.1 Topic 4.2.1 Section 4.2 Points on a Line Points on a Line California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: Youβll learn how to show mathematically that points lie on a line. Key words: linear equation variable solution set |
verify You already dealt with lines in Topics 4.1.3 and 4.1.4. In this Topic youβll see a formal definition relating ordered pairs to a line β and youβll also learn how to show that points lie on a particular line. Graphs of Linear Equations are Straight Lines An equation is linear if the variables have an exponent of one and there are no variables multiplied together. For example, linear: 3x + y = 4, 2x = 6, y = 5 β x nonlinear: xy = 12, x2 + 3y = 1, 8y3 = 20 Linear equations in two variables, x and y, can be written in the form Ax + By = C. The solution set to the equation Ax + By = C consists of all ordered pairs (x, y) that satisfy the equation. All the points in this solution set lie on a straight line. This straight line is the graph of the equation. If the ordered pair (x, y) satisfies the equation Ax + By = C, then the point (x, y) lies on the graph of the equation. The ordered pair (β2, 5) satisfies the equation 2x + 2y = 6. So the point (β2, 5) lies on the line 2x + 2y = 6. y-axis The point (2, 1) lies on the line 2x + 2y = 6. So the ordered pair (2, 1) satisfies the equation 2x + 2y = 6. (2, 1) 1 2 3 4 5 6 x-axis (β2, 5) β4 β3 β2 6 5 4 3 2 1 0 β1 β1 β2 β3 Verifying That Points Lie on a Line To determine whether a point (x, y) lies on the line of a given equation, you need to find out whether the ordered pair (x, y) satisfies the equation. If it does, the point is on the line. You do this by substituting x and y into the equation. Section 4.2 β Lines 175 Check it out: This is the method for showing that (2, β3) is a solution of x β 3y = 11. Example 1 a) Show that the point (2, β3) lies on the graph of x β 3y = 11. b) Determine whether the point (β1, 1) lies on the graph of 2x + 3y = |
4. Solution a) 2 β 3(β3) = 11 2 + 9 = 11 11 = 11 Substitute 2 for x and β3 for y A true statement So the point (2, β3) lies on the graph of x β 3y = 11, since (2, β3) satisfies the equation x β 3y = 11. b) If (β1, 1) lies on the line, 2(β1) + 3(1) = 4. But 2(β1) + 3(1) = β2 + 3 = 1 Since 1 Ο 4, (β1, 1) does not lie on the graph of 2x + 3y = 4. Guided Practice Determine whether or not each point lies on the line of the given equation. 1. (β1, 2); 2x β y = β4 3. (β3, β1); β5x + 3y = 11 5. (β2, β2); y = 3x + 4 7. (β2, β1); 8x β 15y = 3 2. (3, β4); β2x β 3y = 6 4. (β7, β3); 2y β 3x = 15 6. (β5, β3); βy + 2x = β7 8. (1, 4); 4y β 12x = 3 β βββ; β3x β 10y = 2 β 10.,β β βββ β 2 3 2 5 β βββ β 1 3,β 1 4 β βββ; 6x β 16y = 7 β 9. Independent Practice In Exercises 1β4, determine whether or not each point lies on the graph of 5x β 4y = 20. 1. (0, 4) 3. (2, β3) 2. (4, 0) 4. (8, 5) In Exercises 5β8, determine whether or not each point lies on the graph of 6x + 3y = 15. 5. (2, 1) 8. (3, β1) 7. (β1, 6) 6. (0, 5) In Exercises 9β12, determine whether or not each |
point lies on the graph of 6x β 6y = 24. 9. (4, 0) 11. (100, 96) 10. (1, β3) 12. (β400, β404) 13. Explain in words why (2, 31) is a point on the line x = 2 but not a point on the line y = 2. 14. Determine whether the point (3, 4) lies on the lines 4x + 6y = 36 and 8x β 7y = 30. Round Up Round Up You can always substitute x and y into the equation to prove whether a coordinate pair lies on a line. Thatβs because if the coordinate pair lies on the line then itβs actually a solution of the equation. 176 Section 4.2 β Lines Topic 4.2.2 California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: Youβll learn how to graph a straight line by joining two points. Key words linear equation Graphing Ax + By = C Graphing Ax + By = C Every point on a line is a solution to the equation of the line. If you know any two solutions (any two coordinate pairs), then you can join the points with a straight line. Graphing the Line Ax + By = C Using Two Points The graph of the equation Ax + By = C consists of all points (x, y) whose coordinates satisfy Ax + By = C. To graph the line, you just need to plot two points on it and join them together with a straight line. Rearrange the equation so it is in the form y = Px + Q. Choose two values of x and substitute them into your equation to find the corresponding values of y. Plot the two points and draw a straight line through them. Plot a third point to check that the line is correct β the point should lie on the line. Example 1 Plot and label the graph of the equation x β y = β3. Solution Rearrange this first to get |
y = x + 3. Choose two values of x, then draw a table to help you find the y-values. x 2β 4 y (x, y) y = x 3+ 3+2β= 1= )1,2β( y = x 3+ 3+4= 7= )7,4( When you plot the graph, the line should be straight. y-axis β3 (4, 7) y = x β (1, 4) 1 2 3 4 5 6 x-axis 1 β1 β2 β3 (β2, 1) β6 β5 β4 β3 β2 Checkx, y) = (1, 4) (1, 4) lies on the line β which means the line is correct. Section 4.2 β Lines 177 Example 2 Graph and label the line of the equation y = β2x β 4. Solution x 2β 2 y y 2β= x 4β 4β)2β(2β= 0= y 2β= x 4β 4β)2(2β= 8β= (x, y) )0,2β( )8β,2( (β2, 0) β3 β5 β4 β2 β6 y-axis 2 1 1 2 3 4 5 6 (0, β4) y = β 2 x β 4 (2, β8) 0 β1 β1 β2 β3 β4 β5 β6 β7 β8 β9 β10 x-axis Check: x = 0 ο¬ y = β2x β 4 = β2(0) β 4 = 0 β 4 = β4 ο¬ (x, y) = (0, β4) (0, β4) lies on the line β which means that the line is correct. Guided Practice Graph the line through the two points in each of Exercises 1β2. 1. (β1, β3) and (3, 5) 2. (β3, 4) and (4, β3) Graph and label the lines of the equations in Exercises 3β6. 3. βx β 2y = 4 5. 5y β 3x = 15 4. 2x β 3y = 6 6. 7y β 2x = β14 Independent Practice In Exercises 1β4, graph the line through each set of two points. 1. (β1, β2 |
) and (2, 4) 3. (0, 0) and (2, 6) 2. (β1, β1) and (1, 3) 4. (0, β2) and (1, 1) Graph and label the lines of the equations in Exercises 5β16. 5. x + y = 8 7. 2x + y = β3 9. β3x + y = β6 11. 2x β y = β14 13. 8x + 4y = 24 15. 3x β 9y = β27 6. y β x = 10 8. 5x + y = β12 10. β10x + y = 21 12. 6x + 2y = 18 14. 12x β 4y = 8 16. 2x β 8y = 16 Round Up Round Up Itβs easy to make a mistake when working out y-values, so choose x-values that will make the algebra easy (for example, 0 and 1). And itβs always a good idea to check your line by plotting a third point. 178 Section 4.2 β Lines Topic 4.2.3 California Standards: 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: Youβll rearrange line equations to find unknown values. Key words: linear equation substitute Check it out: Because (2k, 3) lies on the line you can replace x and y with 2k and 3, and the equation will still hold. Using Line Equations Using Line Equations This Topic carries straight on from Topic 4.2.2. If thereβs an unknown in a pair of coordinates then you need to substitute the x and y values into the equation, and rearrange to find the unknown value. Points on a Line Satisfy the Equation of the Line In Lesson 4.2.1 you saw that if the ordered pair (x, y) satisfies the equation Ax + By = C, then the point (x, y) lies on the corresponding graph. Similarly, if a point (x, y) lies on the line Ax + By = C, then its coordinates satisfy the equation. So if you substitute the coordinates for x and y into the equation, it will make a true statement. Here are some examples showing how you can use this property |
to calculate unknown values: Example 1 (2k, 3) is a point on the line x β 3y = 7. Find k. Justify your answer. Solution x β 3y = 7 2k β 3(3) = 7 2k β 9 = 7 2k β 9 + 9 = 7 + 9 2k = 16 k = 2 2 k = 8 16 2 Substituting 2k for x and 3 for y Addition Property of Equality Division Property of Equality So the point on the line x β 3y = 7 with y-coordinate 3 is (16, 3). Section 4.2 β Lines 179 Example 2 β βββ β 2, m 4 3 β βββ is a point on the line 2x β 9y = β10. Find the value of m. β Solution 2x β 9y = β10 β βββ = β10 β 2(4) β 9 β βββ β m 2 3 8 β 6m = β10 β6m = β18 β β m 18 6 β β 6 6 m = 3 = So the point on the line 2x β 9y = β10 with x-coordinate 4 is (4, 2). Guided Practice β βββ β 2 3 1. 4p, β βββ lies on the graph of 6x β 5y = β32. Find the value of p. β β βββ 2. The point β β 4 5 Find the value of h. 2, β βββ h lies on the graph of βx β 3y = 3. β 3. The point (2a, β3a) lies on the graph of βx β 2y = β12. Find the value of a. 4. Find the coordinates of the point in Exercise 5. β βββ β 1 4 5. k, β 3 4 β βββ is a point on the line x β 3y = 5. Find the value of k. οΏ½ |
οΏ½ k 6. Find the coordinates of the point in Exercise 7. 180 Section 4.2 β Lines Example 3 The point (1, 3) lies on the line bx + y = 6. Find the value of b. Solution Here you have to use the coordinates to identify the equation. The question is different, but the method is the same. bx + y = 6 b(1 Guided Practice 7. The point (β1, 2) lies on the line βbx + 2y = β4. Find the value of b. 8. The point (β3, β5) lies on the graph of 2x β 3ky = 24. Find the value of k. 9. The point (8, 7) lies on the line bx + y = 11. Find b. 10. The point (6, 3) lies on the line bx β y = 9. Find b. 11. The point (14, 3) lies on the line x + by = β10. Find b. 12. The point (23, β4) lies on the line x β ky = β21. Find k. 13. The point (4, 4) lies on the line 4x β 2ky = 0. Find k. 14. The point (5, β1) lies on the line 9x + 3ky = 30. Find k. 15. The point (β2, β2) lies on the line 7kx β 3y = 13. Find k. 16. The point (β1, β4) lies on the line 7kx β 2ky = 0.25. Find k. Independent Practice In Exercises 1β5, the given point lies on the line 3x β 4y = 24. Find the value of each k and find the coordinates of each point. 1. (2k, 0) 4. (4k, k) 2. (4k, β6) 5. (β4k, β6k) 3. (4k, β18) In Exercises 6β10, the point (2, β4) lies on the given line. Find the value of b in each case. 6. bx + 3y = β6 9. 7x + by = β18 7. βbx + 5y = 10 10. 9bx β 2by = 4 8. 6x β by = 32 11. The point |
(1, 1) lies on the line bx β 2by = 4. Find b. 12. The point (β3, 6) lies on the line 4x + 6ky = 24. Find k. 13. The point (4k, 2k) lies on the line 2x β 6y = 12. Find k and the coordinates of the point. Round Up Round Up This Topicβs really just an application of the method you learned in Topic 4.2.1. Once youβve substituted the x and y values into the equation then you just solve it as normal. Section 4.2 β Lines 181 The x- and y-Intercepts The x- and y-Intercepts The intercepts of a graph are the points where the graph crosses the axes. This Topic is all about how to calculate them. The x-Intercept is Where the Graph Crosses the x-Axis The x-axis on a graph is the horizontal line through the origin. Every point on it has a y-coordinate of 0. That means that all points on the x-axis are of the form (x, 0). The x-intercept of the graph of Ax + By = C is the point at which the graph of Ax + By = C crosses the x-axis. The x-intercept here is (β1, 0). y-axis 3 2 1 1 2 3 4 x-axis β4 β3 β2 0 β1 β1 β2 β3 Computing the x-Intercept Using βy = 0β Since you know that the x-intercept has a y-coordinate of 0, you can find the x-coordinate by letting y = 0 in the equation of the line. Example 1 Find the x-intercept of the line 3x β 4y = 18. Solution Let y = 0, then solve for x: 3x β 4y = 18 3x β 4(0) = 18 3x β 0 = 18 3x = 18 x = 6 So (6, 0) is the x-intercept of 3x β 4y = 18. Topic 4.2.4 California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2 |
x + 6y < 4). What it means for you: Youβll learn about x- and y-intercepts and how to compute them from the equation of a line. Key words: intercept linear equation 182 Section 4.2 β Lines Check it out: Always write the x-intercept as a point, not just as the value of x where the graph crosses the x-axis. For example, (6, 0), not 6. Example 2 Find the x-intercept of the line 2x + y = 6. Solution Let y = 0, then solve for x: 2x + y = 6 2x + 0 = 6 2x = 6 x = 3 So (3, 0) is the x-intercept of 2x + y = 6. Guided Practice In Exercises 1β8, find the x-intercept. 1. x + y = 5 2. 3x + y = 18 3. 5x β 2y = β10 4. 3x β 8y = β21 5. 4x β 9y = 16 6.15x β 8y = 5 7. 6x β 10y = β8 8. 14x β 6y = 0 The y-Intercept is Where the Graph Crosses the y-Axis The y-axis on a graph is the vertical line through the origin. Every point on it has an x-coordinate of 0. That means that all points on the y-axis are of the form (0, y). The y-intercept of the graph of Ax + By = C is the point at which the graph of Ax + By = C crosses the y-axis. The y-intercept here is (0, 3). y-axis 6 5 4 3 2 1 β4 β3 β2 β1 0 1 2 3 x-axis Section 4.2 β Lines 183 Computing the y-Intercept Using βx = 0β Since the y-intercept has an x-coordinate of 0, find the y-coordinate by letting x = 0 in the equation of the line. Example 3 Find the y-intercept of the line β2x β 3y = β9. Solution Let x = 0, then solve for y: β2x β 3y = β9 β2(0) β 3y = β9 0 β 3y = β9 β3y = β9 y = 3 So (0, 3 |
) is the y-intercept of β2x β 3y = β9. Example 4 Find the y-intercept of the line 3x + 4y = 24. Solution Let x = 0, then solve for y: 3x + 4y = 24 3(0) + 4y = 24 0 + 4y = 24 4y = 24 y = 6 So (0, 6) is the y-intercept of 3x + 4y = 24. Guided Practice In Exercises 9β16, find the y-intercept. 9. 4x β 6y = 24 10. 5x + 8y = 24 11. 8x + 11y = β22 12. 9x + 4y = 48 13. 6x β 7y = β28 14. 10x β 12y = 6 15. 3x + 15y = β3 16. 14x β 5y = 0 Check it out: Always write the y-intercept as a point, not just as the value of y where the graph crosses the y-axis. For example, (0, 3), not 3. 184 Section 4.2 β Lines Independent Practice 1. Define the x-intercept. 2. Define the y-intercept. Find the x- and y-intercepts of the following lines: 3. x + y = 9 5. βx β 2y = 4 7. 3x β 4y = 24 9. β5x β 4y = 20 4. x β y = 7 6. x β 3y = 9 8. β2x + 3y = 12 10. β0.2x + 0.3y = 1 11. 0.25x β 0.2y = 2 1 12 13. ( 3 5 ) is the x-intercept of the line β10x β 3y = 12. 0g, Find the value of g. )is the y-intercept of the line 2x β 15y = β3. ( 14. 0 1, k 5 Find the value of k. 15. The point (β3, b) lies on the line 2y β x = 8. Find the value of b. 16. Find the x-intercept of the line in Exercise 15. 17. Another line has x-intercept (4, 0) and equation 2y + kx = 20. Find the value of k. In Exercises 18-22 |
, use the graph below to help you reach your answer6 β5 β4 β3 β2 β1 0 β1 1 2 3 4 5 x 6 β2 β3 β4 β5 β6 18. Find the x- and y-intercepts of line n. 19. Find the x-intercept of line p. 20. Find the y-intercept of line r. 21. Explain why line p does not have a y-intercept. 22. Explain why line r does not have an x-intercept. Round Up Round Up Make sure you get the method the right way around β to find the x-intercept, put y = 0 and solve for x, and to find the y-intercept, put x = 0 and solve for y. In the next Topic youβll see that the intercepts are really useful when youβre graphing lines from the line equation. Section 4.2 β Lines 185 Topic 4.2.5 California Standards: 6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). 7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: Youβll graph lines by first calculating the x- and y-intercepts. Key words: linear equation intercept 186 Section 4.2 β Lines Graphing Lines Graphing Lines In Topic 4.2.2 you learned how to graph a straight line by plotting two points. If youβre not given points on the line, itβs easiest to use the x- and y-intercepts. Graphing Lines by Computing the Intercepts The method below for plotting a straight-line graph is the same as in Topic 4.2.2. To graph the line, you plot two points β except this time you use the x-intercept and the y-intercept, then draw a straight line through them. Graphing a Line Find the x-intercept β let y = 0, then solve the equation for x. Find the y-intercept β let x = 0, then solve the equation for |
y. Draw a set of axes and plot the two intercepts. Draw a straight line through the points. Check your line by plotting a third point. Example 1 Draw the graph of 5x + 3y = 15 by computing the intercepts. Solution x-intercept: 5x + 3(0) = 15 5x + 0 = 15 5x = 15 x = 3 y-intercept: 5(0) + 3y = 15 0 + 3y = 15 3y = 15 y = 5 Therefore (3, 0) is the x-intercept and (0, 5) is the y-intercept. Check: x = 1 5x + 3y = 15 3 β2 0 β1 β1 β2 β3 y-axis 1 2 3 4 5 6 x-axis β 15 5 3 ) lies on the line β which means the line is correct. 10 Finding the intercepts is the quickest way of finding two points. When you substitute 0 for y to solve for x, the y-term disappears, and vice versa β making the equations easier to solve. Example 2 Draw the graph of y = βx + 3 by computing the intercepts. Solution x-intercept: 0 = βx + 3 x = 3 y-intercept: y = β(0) + 3 y = 3 Therefore (3, 0) is the x-intercept and (0, 3) is the y-intercept. β3 β2 y-axis 6 5 4 3 2 1 Check: x = 1 y = βx + 3 ο¬ y = β1 + 3 = 2 (1, 2) lies on the line β which means the line is correct. 1 2 3 4 5 6 x-axis 0 β1 β1 β2 β3 Guided Practice Draw the graphs of the following equations by computing the intercepts. 1. 5x + 2y = 10 3. 6x β y = 3 5. 2x + y = 3 7. x + y = 10 9. 2x β y = 4 11. 3x + y = 9 13. y = 2x + 4 2. β3x β 3y = 12 4. β4x + 5y = 20 6. βx β 8y = 2 8. x β y = 4 10. x + 5y = 10 12. x β 4y = 8 14. y = 5x β 10 15. y = 3x + 9 17. |
y = β 2 5 x β 2 16. y = 18 Section 4.2 β Lines 187 Independent Practice Draw graphs of the lines using the x- and y-intercepts in Exercises 1β6. 1. x-intercept: (β3, 0) 2. x-intercept: (1, 0) 3. x-intercept: (4, 0) 4. x-intercept: (β6, 0) 5. x-intercept: (β1, 0) 6. x-intercept: (2, 0) y-intercept: (0, 2) y-intercept: (0, 6) y-intercept: (0, β3) y-intercept: (0, β4) y-intercept: (0, 7) y-intercept: (0, β5) Draw the graphs of the equations in Exercises 7β18 by computing the intercepts. 7. x + y = 6 9. x β y = β5 11. 3x + y = 6 13. 2x β y = β4 15. 4x + 3y = β12 17. 6x β 3y = 24 8. x + y = β4 10. x β y = 7 12. 2x + y = 8 14. 3x β y = β3 16. 5x β 2y = 10 18. 10x β 12y = 60 19. Show that the graphs of x + y = 6 and β6x β 6y = β36 are the same. 20. Explain why the graph of 5x + 8y = 0 cannot be drawn using the intercepts. Round Up Round Up This Topic follows on neatly from Topic 4.2.2, where you graphed lines by plotting two points and joining them with a straight line. You can use any two points β the main reason for using the intercepts is that theyβre usually easier to calculate. 188 Section 4.2 β Lines TTTTTopicopicopicopicopic 4.3.14.3.1 4.3.14.3.1 4.3.1 Section 4.3 Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line Slope of a Line California Standards: 7.0:7 |
.0:7.0:7.0:7.0: Students verify that a point lies on a line, given an equation of the line. le to derivvvvveeeee le to deri Students are ae ae ae ae abbbbble to deri le to deri Students ar Students ar le to deri Students ar Students ar tions tions linear equa linear equa tions by using the linear equations tions linear equa linear equa point-slope formula. What it means for you: Youβll find the slope of a line given any two points on the line. Key words: slope steepness horizontal vertical rise over run Check it out: Dx is pronounced βdelta xβ and just means βchange in x.β By now youβve had plenty of practice in plotting lines. Any line can be described by its slope β which is what this Topic is about. The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The Slope of a Line is Its Steepness The slope (or gradient) of a line is a measure of its steepness. The slope of a straight line is the ratio of the vertical change to the horizontal change between any two points lying on the line. The vertical change is usually written Dy, and itβs often called the rise. In the same way, the horizontal change is usually written Dx, and itβs often called the run. D = rise y D = run x O Slope = vertical change horizontal change = rise run = Ξ Ξ y x, provided DDDDDx ΟΟΟΟΟ 0 If you know the coordinates of any two points on a line, you can find the slope. The slope, m, of a line passing through points P1 (x1, y1) and P2 (x2, y2) is given by this formula, provided x2 β x1 ΟΟΟΟΟ 0 There is an important difference between positive and negative slopes β a positive slope means the line goes βuphillβ ( ), whereas a line with a negative slope goes βdownhillβ ( ). Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 189189 |
189189189 a Line a Line ula to Find the Slope of Use the Fororororormmmmmula to Find the Slope of ula to Find the Slope of Use the F Use the F a Line ula to Find the Slope of a Line a Line ula to Find the Slope of Use the F Use the F Example Example Example Example Example 11111 Find the slope of the line that passes through the points (2, 1) and (7, 4) and draw the graph. Check it out: In this example, (x1, y1) = (2, 1) and (x2, y2) = (7, 4). Solution Solution Solution Solution Solution So the slope is 3 5. You know that the line passes through (2, 1) and (7, 4), so just join those two points up to draw the graph. y (7, 4) (2, 1) In the graph above, notice how the line has a positive slope, meaning it goes βuphillβ from left to right. In fact, since the slope is 3 5, the line goes 3 units up for every 5 units across. Guided Practice In Exercises 1β4, find the slope of the line on the graph opposite. (4) 5. Find the slope of the line that passes through the points (1, 5) and (3, 2), and draw the graph. 6. Find the slope of the line that passes through the points (3, 1) and (2, 4), and draw the graph. β6 β5 β4 β3 β2 β1 (3) (2) (11 β2 β3 β4 β5 β6 190190190190190 Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 e Coordinadinadinadinadinatestestestestes e Coor e Nee Negggggaaaaatititititivvvvve Coor e Coor e Nee Ne eful If TTTTTherherherherhere are are are are are Ne eful If eful If Be Car Be Car Be Careful If e Coor eful If Be Car Be Car Example Example Example Example Example 22222 Find the slope of the line that passes through the points (3, 4) and (6, β2). Solution Solution Solution Solution Solution 2 So the slope is β2 |
. Check it out: It doesnβt matter which point you call (x1, y1) and which you call (x2, y2) β choose whichever makes the math easier. This time the line has a negative slope, meaning it goes βdownhillβ from left to right. Here the slope is β2, which means that the line goes 2 units down for every 1 unit across. w a Graaaaaphphphphph w a Gr e to Draaaaaw a Gr w a Gr e to Dr t Hat Havvvvve to Dr e to Dr t Hat Ha ou Donβββββt Ha ou Don YYYYYou Don ou Don w a Gr e to Dr ou Don Example Example Example Example Example 33333 Find the slope of the lines through: a) (2, 5) and (β4, 2) b) (1, β6) and (3, 3) Solution Solution Solution Solution Solution a) m = b Guided Practice Find the slope m of the line through each pair of points below. 7. (β1, 2) and (3, 2) 9. (5, β7) and (β3, β7) 11. (β1, β3) and (1, β4) 13. (β2, β2) and (β3, β17) 15. (0, β1) and (1, 0) 8. (0, β5) and (β6, 1) 10. (4, β1) and (β3, 5) 12. (5, 7) and (β11, β12) 14. (18, 2) and (β32, 7) 16. (0, 0) and (β14, β1) Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 191191191191191 lems Invvvvvolvolvolvolvolve e e e e VVVVVariaariaariaariaariabbbbbleslesleslesles lems In Some Proboboboboblems In lems In Some Pr Some Pr lems In Some Pr Some Pr Example Example Example Example Example 44444 If the slope of the line that passes through the points (4, β1) and (6, 2k) is 3, find the value of k. Solution Solution Solution Solution |
Solution Even though one pair of coordinates contains a variable, k, you still use the slope formula in exactly the same way as before. m = y x 2 2 β β, which means that But the slope is 3, so 1 2 k + 2 = 3 ο¬ 2k + 1 = 6 ο¬ 2k = 5 ο¬ k = 5 2 Guided Practice Find the slope m of the lines through the points below. 17. (7, β2c) and (10, βc) 19. (2, 2k) and (β5, β5k) 21. (3d, 7d) and (5d, 9d) 23. (9c, 12v) and (12c, 15v) 25. (10, 14d) and (d, β7) 18. (b, 1) and (3b, β3) 20. (3q, 1) and (2q, 7) 22. (4a, 5k) and (2a, 7k) 24. (p, q) and (q, p) 26. (2t, β3s) and (18s, 14t) In Exercises 27β32 youβre given two points on a line and the lineβs slope, m. Find the value of the unknown constant in each Exercise. 27. (β2, 3) and (3k, β4), m = 28. (4, β5t) and (7, β8t), m = 2 5 2 7 29. (4b, β6) and (7b, β10), m = 3 4 30. (β8, β6) and (12, 4), m = β 31. (7k, β3) and (k, β1), m = 4 2 5 v 32. (1, β17) and (40, 41), m = β 174 78 t 192192192192192 Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 Independent Practice 1. Find the slope m of the lines shown below. (iii) (v) -5 -4 y -4 -5 (i) (ii) (iv) In Exercises 2β5, find the slope of the line that passes through the given points, and draw the graph. 2. (β2, 1) |
and (0, 2) 4. (β5, 2) and (β1, 3) 3. (4, 4) and (1, 0) 5. (3, β3) and (7, 3) In Exercises 6β10, find the slope of the line through each of the points. 6. (β3, 5) and (2, 1) 8. (2, 3) and (4, 3) 10. (2s, 2t) and (s, 3t) 7. (0, 4) and (β4, 0) 9. (6d, 2) and (4d, β1) In Exercises 11β15, youβre given two points on a line and the lineβs slope, m. Find the value of the unknown constant in each Exercise. 11. (3t, 7) and (5t, 9), m = β 12. (3k, 1) and (2k, 7), m = 1 2 1 3 13. (0, 14d) and (10, β6d), m = β1 14. (2t, β3) and (β3t, 5), m = 5 4 15. (0, 8d) and (β1, 4d), m = β 1 3 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Slope is a measure of how steep a line is β itβs how many units up or down you go for each unit across. If you go up or down a lot of units for each unit across, the line will be steep and the slope will be large (either large and positive if it goes up from left to right, or large and negative if it goes down from left to right). Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 193193193193193 oint-Slope Fororororormmmmmulaulaulaulaula oint-Slope F oint-Slope F PPPPPoint-Slope F oint-Slope F PPPPPoint-Slope F oint-Slope Fororororormmmmmulaulaulaulaula oint-Slope F oint-Slope F oint-Slope |
F The point-slope formula is a really useful way of calculating the equation of a straight line. aight Line aight Line a Str a Str tion of tion of ula to Find the Equa Use the Fororororormmmmmula to Find the Equa ula to Find the Equa Use the F Use the F aight Line a Straight Line tion of a Str ula to Find the Equation of aight Line a Str tion of ula to Find the Equa Use the F Use the F If you know the slope of the line and a point on the line, you can use the point-slope formula to find the equation of the line. The point-slope formula for finding the equation of a line is: y β y1 = m(x β x1) where m is the slope and (x1, y1) is a point on the line. You substitute the x-coordinate of a point on the line for x1 and the y-coordinate of the same point for y1. Watch out though β x and y are variables and they stay as letters in the equation of the line. Example Example Example Example Example 11111 Find the equation of the line through (β4, 6) that has a slope of β3. Solution Solution Solution Solution Solution (x1, y1) = (β4, 6) and m = β3 y β y1 = m(x β x1) ο¬ y β 6 = β3[x β (β4)] ο¬ y β 6 = β3(x + 4) ο¬ y β 6 = β3x β 12 ο¬ y + 3x = β6 Guided Practice Write the equation of the line that passes through the given point and has the given slope. 1. Point (β2, β3), slope = β1 2. Point (3, β5), slope = 2 3. Point (β7, β2), slope = β5 4. Point (4, β3), slope = 2 3 5. Point (2, 6), slope = β 3 4 6. Point (β2, β3), slope = 5 8 7. Point (β5, β3), slope = β 6 7 β βββ 8. Point β β 2 3, β βββ β 1 4, slope = |
2 5 TTTTTopicopicopicopicopic 4.3.24.3.2 4.3.24.3.2 4.3.2 California Standards: 7.0:7.0:7.0:7.0:7.0: Students verify that a point lies on a line, given an equation of the line. Students are ae ae ae ae abbbbble to deri le to derivvvvveeeee le to deri le to deri Students ar Students ar Students ar le to deri Students ar y using y using tions b tions b linear equa linear equa y using tions by using linear equations b y using tions b linear equa linear equa the point-slope fororororormmmmmula.ula.ula.ula.ula. the point-slope f the point-slope f the point-slope f the point-slope f What it means for you: Youβll learn about the pointslope formula and use it to find the equation of a line. Key words: slope point-slope formula Check it out: The point-slope formula is just the one from Topic 4.3.1 rearranged. (x, y) is any point on the line, so (x1, y1) and (x, y) are the two points on the line. 194194194194194 Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 Check it out You can use either point as (x1, y1) here β youβll still get the same equation. Find the Slope Firststststst Find the Slope Fir e Givvvvven,en,en,en,en, Find the Slope Fir Find the Slope Fir e Gi e Gi oints ar IfIfIfIfIf TTTTTwwwwwo Po Po Po Po Points ar oints ar oints are Gi Find the Slope Fir e Gi oints ar If you know the coordinates of two points on a straight line, you can still find the equation using the point-slope formula β but you have to find the slope first. Example Example Example Example Example 22222 Write the equation of the straight line that contains the points (3, β2) and (β1, 5). Solution Solution Solution Solution Solution Step 1: Find the slope using the given points |
Slope Step 2: Write the equation y β y1 = m(x β x1) (x β 3) y β (β2 4y + 8 = β7(x β 3) ο¬ 4y + 8 = β7x + 21 ο¬ 4y + 7x = 13 (x β 3) Guided Practice Write the equation of the line that passes through the given pair of points. 9. (β1, 0) and (3, β4) 11. (β5, 7) and (3, 9) 13. (8, 7) and (β7, β5) 15. (3, 1) and (5, 4) 17. (6, 2) and (4, 1) 19. (1, 0) and (2, 0) 21. (4, β1) and (β1, β3) 23. (4, 5) and (2, 6) 25. (3, β5) and (0, 8) 10. (β1, 1) and (β3, β3) 12. (6, β8) and (β2, β10) 14. (β10, 11) and (5, β12) 16. (2, β5) and (3, β1) 18. (β3, 5) and (4, 3) 20. (β2, β2) and (7, 5) 22. (7, 2) and (3, 3) 24. (β2, β3) and (β3, β2) 26. (1, 1) and (4, β6) Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 195195195195195 Independent Practice In Exercises 1β6, write the equation of the line that passes through the given point and has the given slope. 1. Point (1, 5), slope = β3 2. Point (2, 0), slope = 1 2 3. Point (3, 1), slope = 1 4 5. Point (β8, 6), slope = β 4 5 4. Point (β3, 4), slope = β 6. Point (β3, β4), slope = 2 3 3 8 In Exercises 7β16, write the equation of the line that passes through the given pair of points. 7. (0, |
3) and (4, β1) 9. (3, 8) and (4, 4) 11. (β6, 9) and (β4, β6) 13. (β4, β8) and (β5, 4) 15. (10, 5) and (4, 6) 8. (β1, 6) and (7, 5) 10. (4, β7) and (β3, 5) 12. (4, β9) and (β3, β9) 14. (β8, 3) and (8, 4) 16. (0, 0) and (β4, β6) 17. The points (5, 6) and (8, 7) lie on a line. Find the equation of this line. 18. The line in Exercise 17 forms one side of a triangle that has one vertex at the point (5, β4). If the slope of one of the edges of the triangle is β3, find the equation of this edge. 19. The point (8, k) lies on the third edge of the triangle in Exercise 18. Given that the triangle is isosceles, find, by graphing, the value of k. 20. Joshua is an architect who must build a wheelchair-accessible office building. To make a ramp that is easy to maneuver in a wheelchair, Joshua designs a ramp that is 27 inches high and 540 inches long. What is the slope of the ramp? 21. If the student population at a high school changes from 1372 in 1996 to 1768 in 2006, what is the average rate of change of the student population? (Hint: Use the pairs of coordinates (1996, 1372) and (2006, 1768) to reach your answer.) ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up You should look over the point-slope formula until you can write it down from memory. Itβs a really useful formula and makes finding the equation of a line much easier β but only if you remember it. 196196196196196 Section 4.3 Section 4.3 Section 4.3 β Slope Section 4.3 Section 4.3 TTTTTopicopicopicopicopic 4.4.14.4.1 4.4.14.4.1 4.4.1 Section 4. |
4 allel Lines allel Lines PPPPParararararallel Lines allel Lines allel Lines PPPPParararararallel Lines allel Lines allel Lines allel Lines allel Lines California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students under Students under stand Students understand stand Students under allel allelallel par par pts of pts of the conce the conce allel parallel pts of par the concepts of the conce par pts of the conce lines lines lines and perpendicular lines lines lines and how their slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point..... What it means for you: Youβll work out the slopes of parallel lines and youβll test if two lines are parallel. Key words: parallel intersect Now that youβve practiced finding the slope of a line, you can use the method on a special case β parallel lines. allel Lines Nevvvvver Meet er Meet er Meet allel Lines Ne PPPPParararararallel Lines Ne allel Lines Ne er Meet er Meet allel Lines Ne Parallel lines are two or more lines in a plane that never intersect (cross). These lines are all parallel. No matter how long you draw them, theyβll never meet. The symbol || is used to indicate parallel lines β you read this symbol as βis parallel to.β So, if l1 and l2 are lines, then l1 || l2 means βline l1 is parallel to line l2.β allel Lines Havvvvve Identical Slopes e Identical Slopes e Identical Slopes allel Lines Ha PPPPParararararallel Lines Ha allel Lines Ha e Identical Slopes e Identical Slopes allel Lines Ha You can determine whether lines are parallel by looking at their slopes. Two lines are parallel if their slopes are equal. Example Example Example Example Example 11111 Prove that the three lines A, B, and C shown on the graph are parallel. Solution Solution Solution Solution Solution Using the rise over run formula (see Topic 4.3.1), you can see that they all have a slope of Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 197197197197197 Guided Practice 1. Two lines on |
the same plane that never intersect are called _________________ lines. 2. To determine if two lines are parallel you can look at their _________________. 3. Prove that the line f defined by y β 3 = 2 3 (x β 4) is parallel to line g defined by y β 6 = 2 3 (x + 1). t Havvvvve Defined Slopes e Defined Slopes e Defined Slopes t Hat Ha t Ha tical Lines Donβββββt Ha tical Lines Don VVVVVererererertical Lines Don tical Lines Don e Defined Slopes e Defined Slopes tical Lines Don Vertical lines are parallel, but you canβt include them in the definition on page 197 because their slopes are undefined. Points on a vertical line all have the same x-coordinate, so they are of the form (c, y1) and (c, y2). The slope of a vertical line is undefined because is not defined. y Finding Slopes y Finding Slopes allel b Lines are Pe Pe Pe Pe Parararararallel b allel b Lines ar Lines ar est if TTTTTest if est if y Finding Slopes allel by Finding Slopes est if Lines ar y Finding Slopes allel b Lines ar est if To check if a pair of lines are parallel, just find the slope of each line. If the slopes are equal, the lines are parallel. Example Example Example Example Example 22222 Show that the straight line through (2, β3) and (β5, 1) is parallel to the straight line joining (7, β1) and (0, 3). Solution Solution Solution Solution Solution Step 1: Find the slope of each line using the formula m1 = m2 = Step 2: Compare the slopes and draw a conclusion. β = β4 7 4 7, so m1 = m2. So the straight line through (2, β3) and (β5, 1) is parallel to the straight line through (7, β1) and (0, 3). Check it out: The slope of a vertical line is undefined because division by 0 is undefined. Donβt forget: See Topic 4.3.1 for more on calculating the slope of a line. Check it out: Itβs possible that all the points lie on the same line β meaning there is actually only one line. To be |
absolutely accurate, you should check that this isnβt the case before you say the lines are parallel. You can do this by finding the equations of the lines and comparing them. 198198198198198 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 Guided Practice 4. Show that line a, which goes through points (7, 2) and (3, 3), is parallel to line b joining points (β8, β4) and (β4, β5). 5. Show that the line through points (4, 3) and (β1, 3) is parallel to the line though points (β6, β1) and (β8, β1). 6. Determine if line f joining points (1, 4) and (6, 2) is parallel to line g joining points (0, 8) and (10, 4). 7. Determine if the line through points (β5, 2) and (3, 7) is parallel to the line through points (β5, 1) and (β3, 6). 8. Determine if the line through points (β8, 4) and (β8, 3) is parallel to the line through points (6, 3) and (β4, 3). lems are e e e e TTTTTougher ougher ougher lems ar allel Line Proboboboboblems ar lems ar allel Line Pr Some Parararararallel Line Pr allel Line Pr Some P Some P ougher ougher lems ar allel Line Pr Some P Some P Example Example Example Example Example 33333 Find the equation of a line through (β1, 4) that is parallel to the straight line joining (5, 7) and (β6, β8). Solution Solution Solution Solution Solution Step 1: Find the slope m1 of the line through (5, 7) and (β6, β8). m1 = = β β 15 11 = 15 11 Step 2: The slope m2 of the line through (β1, 4) must be equal to since the lines are parallel. 15 11 So, m1 = m2 = 15 11 Step 3: Now use the point-slope formula to find the equation of the line through point (β1, 4) with slope 15 11. y β y1 = m(x β x1) Check |
it out: This is the equation of the line through (β1, 4) that is parallel to the straight line joining (5, 7) and (β6, β8). [x β (β1)] ο¬ y β 4 = 15 11 ο¬ 11y β 44 = 15(x + 1) ο¬ 11y β 44 = 15x + 15 Equation: 11y β 15x = 59 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 199199199199199 Guided Practice 9. Find the equation of the line through (β3, 7) that is parallel to the line joining points (4, 5) and (β2, β8). 10. Find the equation of the line through (6, β4) that is parallel to the line joining points (β1, 6) and (7, 3). 11. Find the equation of the line through (β1, 7) that is parallel to the line joining points (4, β3) and (8, 6). 12. Write the equation of the line through (β3, 5) that is parallel to the line joining points (β1, 2.5) and (0.5, 1). 13. Write the equation of the line through (β2, β1) that is parallel to the line x + 3y = 6. Independent Practice 1. Line l1 has slope about l1 and l2? 1 2 and line l2 has slope 1 2. What can you conclude 2. Line l1 has a slope of β 1 3. If l1 || l2, then what is the slope of l2? 3. Show that all horizontal lines are parallel. 4. Show that the line through the points (5, β3) and (β8, 1) is parallel to the line through (13, β7) and (β13, 1). 5. Determine if the line through the points (5, 4) and (0, 9) is parallel to the line through (β1, 8) and (4, 0). 6. Determine if the line through the points (β2, 5) and (6, 0) is parallel to the line through (8, β1) and (0, 4). 7. Determine if the line through the points (4, β7) and (4, β4) is parallel to |
the line through (β5, 1) and (β5, 5). 8. Determine if the line through the points (β2, 3) and (β2, β2) is parallel to the line through (1, 7) and (β6, 7). 9. Find the equation of the line through (1, β2) that is parallel to the line joining the points (β3, β1) and (8, 7). 10. Find the equation of the line through (β5, 3) that is parallel to the line joining the points (β2, 6) and (8, β1). 11. Write the equation of the line through (0, 6) that is parallel to the line 3x + 2y = 6. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up When you draw lines with different slopes on a set of axes, you might not see where they cross. But remember, you are only looking at a tiny bit of the lines β they go on indefinitely in both directions. If they donβt have identical slopes, theyβll cross sooner or later. 200200200200200 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 TTTTTopicopicopicopicopic 4.4.24.4.2 4.4.24.4.2 4.4.2 California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students under Students under stand Students understand stand Students under the concepts pts pts pts pts of parallel lines the conce the conce the conce the conce and perperperperperpendicular lines pendicular lines pendicular lines pendicular lines and pendicular lines how their slopes are related. le to find le to find Students are ae ae ae ae abbbbble to find Students ar Students ar le to find le to find Students ar Students ar a line a line tion of tion of the equa the equa a line tion of a line the equation of the equa a line tion of the equa pendicular to a givvvvvenenenenen perperperperperpendicular to a gi pendicular to a gi pendicular to a gi pendicular to a gi t passes |
thr line tha line tha ough a ough a t passes thr line that passes thr ough a t passes through a line tha t passes thr line tha ough a gigigigigivvvvven point. en point. en point. en point. en point. What it means for you: Youβll work out the slopes of perpendicular lines and youβll test if two lines are perpendicular. Key words: perpendicular reciprocal Check it out: This definition doesnβt work for horizontal or vertical lines, since the slope of a vertical line canβt be defined in the same way as for other lines. But remember that the lines x = c and y = k (for constants c and k) are perpendicular. pendicular Lines pendicular Lines PPPPPerererererpendicular Lines pendicular Lines pendicular Lines PPPPPerererererpendicular Lines pendicular Lines pendicular Lines pendicular Lines pendicular Lines Math problems about parallel lines often deal with perpendicular lines too. βPerpendicularβ might sound like a difficult term, but itβs actually a really simple idea. t Right AngAngAngAngAngleslesleslesles t Right t Right pendicular Lines Meet a PPPPPerererererpendicular Lines Meet a pendicular Lines Meet a pendicular Lines Meet at Right t Right pendicular Lines Meet a Two lines are perpendicular if they intersect at 90Β° angles, like in the graphs on the right. ocals ocals ecipr e Negggggaaaaatititititivvvvve Re Re Re Re Recipr ecipr e Nee Ne e Ne pendicular Lines ar Slopes of P P P P Perererererpendicular Lines ar pendicular Lines ar Slopes of Slopes of ocals eciprocals pendicular Lines are Ne ocals ecipr pendicular Lines ar Slopes of Slopes of Two lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. To get the reciprocal of a number you divide 1 by it. For example, the reciprocal of x is 1 x and so the negative reciprocal is β 1 x. The reciprocal of x y is 1 x y = and so the negative reciprocal is β y x y x. Example Example Example Example Example 11111 Prove that lines A and B, shown on the graph, are perpendicular to each other. Solution Solution Solution Solution Solution Using the rise over |
run formula: Donβt forget: See Topic 4.3.1 for the βrise over runβ formula. Slope of A = m1 = Slope of B = m2 = 1 2 = 2 4 β4 2 = β2 4 2 2 4 B A 1 2 is the negative reciprocal of β2, so A and B must be perpendicular. Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 201201201201201 Guided Practice 1. Perpendicular lines meet at ________________ angles. 2. Find the negative reciprocal of 3. 3. Find the negative reciprocal of β 4. Find the negative reciprocal of β 1 4 4 5.. 5. Use the graph to prove that A and B are perpendicular. A β6 β4 β2 y 6 4 2 0 β2 β4 β6 2 4 6 X B = β1 = β1 Γ ines: m11111 Γ Γ ines: pendicular LLLLLines: ines: pendicular PPPPPerererererpendicular pendicular = β1 Γ m22222 = β1 = β1 Γ ines: pendicular When you multiply a number by its reciprocal, you always get 1. For example, 1 5 5 Γ = = and So because two perpendicular slopes are negative reciprocals of each other, their product is always β1. Hereβs the same thing written in math-speak: Check it out: ^ is the symbol for βis perpendicular to.β If two lines l1 and l2 have slopes m1 and m2, l1 ^ l2 if and only if m1 Γ m2 = β1. Example Example Example Example Example 22222 P and Q are two straight lines and P ^ Q. P has a slope of β4. What is the slope of Q? Solution Solution Solution Solution Solution mP Γ mQ = β1 ο¬ β 4 Γ mQ = β1 1 4 β =1 β 4 ο¬ mQ =. So the slope of Q is 1 4. 202202202202202 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 Guided Practice 6. Lines l1 and l2 are perpendicular. If the slope of l1 is 1 5, find the slope of l2. 7. Lines A and B are perpendicular. If the slope of |
A is β 5 8, find the slope of B. 8. Lines R and T are perpendicular. If R has slope β 7 11, what is the slope of T? 9. The slope of l1 is β0.8. The slope of l2 is 1.25. Determine whether l1 and l2 are perpendicular. inding SSSSSlopes lopes lopes inding pendicular by y y y y FFFFFinding inding pendicular b ines are e e e e PPPPPerererererpendicular b pendicular b ines ar ShoShoShoShoShow w w w w TTTTThahahahahat t t t t LLLLLines ar ines ar lopes lopes inding pendicular b ines ar Example Example Example Example Example 33333 Determine the equation of the line passing through (3, 1) that is perpendicular to the straight line through (2, β1) and (4, 2). Solution Solution Solution Solution Solution Step 1: Find slope m1 of the line through (2, β1) and (4, 2): Step 2: Find the slope m2 of a line perpendicular to that line: m1 Γ m2 = β1 3 2 Γ m2 = β1 ο¬ m2 = β 2 3 Step 3: Now use the point-slope formula to find the equation of the line through (3, 1) with slope β 2 3. y β y1 = m(x β x1x β 3) ο¬ 3y β 3 = β2(x β 3) ο¬ 3y β 3 = β2x + 6 Equation: 3y + 2x = 9 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 203203203203203 Guided Practice 10. Show that the line through the points (5, β3) and (β8, 1) is perpendicular to the line through (4, 6) and (8, 19). 11. Show that the line through (0, 6) and (5, 1) is perpendicular to the line through (4, 8) and (β1, 3). 12. Show that the line through (4, 3) and (2, 2) is perpendicular to the line through (1, 3) and (3, β1). 13. Determine the equation of the line through |
(3, β4) that is perpendicular to the line through the points (β7, β3) and (β3, 8). 14. Determine the equation of the line through (6, β7) that is perpendicular to the line through the points (8, 2) and (β1, 8). 15. Find the equation of the line through (4, 5) that is perpendicular to the line β3y + 4x = 6. Independent Practice In Exercises 1β8, J and K are perpendicular lines. The slope of J is given. Find the slope of K. 1. mJ = β3 5 2 2. mJ = β14 4. mJ = 3. mJ = 6 7 5. mJ = β 8 3 6. mJ = β 14 15 8. mJ = 0.45 7. mJ = β0.18 9. Show that the line through (2, 7) and (β2, 8) is perpendicular to the line through (β3, β3) and (β2, 1). 10. Show that the line through (β4, 3) and (3, β2) is perpendicular to the line through (β7, β1) and (β2, 6). 11. Determine the equation of the line through (5, 9) that is perpendicular to a line with slope 1 3. 12. Determine the equation of the line through (3, β5) that is perpendicular to the line through the points (β3, 2) and (β6, β4). ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up βPerpendicularβ is just a special math word to describe lines that are at right angles to each other. Remember that the best way to show that two lines are at right angles is to calculate their slopes β if they multiply together to make β1, then the lines are perpendicular. 204204204204204 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 TTTTTopicopicopicopicopic 4.4.34.4.3 4.4.34.4.3 4.4.3 California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students |
under Students under stand Students understand stand Students under allel allelallel par par pts of pts of the conce the conce allel parallel pts of par the concepts of par the conce pts of the conce pendicular pendicular lines and per lines and per pendicular lines and perpendicular lines and per pendicular lines and per lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes ted. Students are ted. ararararare re re re re relaelaelaelaelated. ted. ted. able to find the equation of a line perpendicular to a given line that passes through a given point..... What it means for you: Youβll plot graphs and solve equations using a method called the slope-intercept form of an equation. he Slope-Intercececececeptptptptpt he Slope-Inter he Slope-Inter TTTTThe Slope-Inter he Slope-Inter he Slope-Intercececececeptptptptpt he Slope-Inter he Slope-Inter TTTTThe Slope-Inter he Slope-Inter a Line a Line FFFFForororororm ofm ofm ofm ofm of a Line a Line a Line FFFFForororororm ofm ofm ofm ofm of a Line a Line a Line a Line a Line Now that youβve practiced calculating the slope and intercept of a line, you can use the two things together to make plotting graphs easier. a Line: y = mx + b a Line: he Slope-Intercececececept Fpt Fpt Fpt Fpt Forororororm ofm ofm ofm ofm of a Line: a Line: he Slope-Inter TTTTThe Slope-Inter he Slope-Inter a Line: he Slope-Inter In the slope-intercept form the y is alone on one side of the equation. The slope-intercept form of the equation is: y = mx + b Key words: slope-intercept form intercept Here are a few examples of equations in the slope-intercept form: y = 3x + Donβt forget: In the slope-intercept form of the line equation, m and b are numbers, and x and y are variables. Guided Practice In Exercises 1β6, decide whether each equation is in slope-inter |
cept form or not. 1. y = 3x + 7 3. y β 3 = 2(x β 4) 2. 3x + 4y = 7 4. y β 8 = 3(x β 4) 5. y = 3 2 x + 18 6. y = β4x β 1 t Easy to Plot Graaaaaphsphsphsphsphs t Easy to Plot Gr m Makes es es es es IIIIIt Easy to Plot Gr t Easy to Plot Gr m Mak he Slope-Intercececececept Fpt Fpt Fpt Fpt Forororororm Mak m Mak he Slope-Inter TTTTThe Slope-Inter he Slope-Inter t Easy to Plot Gr m Mak he Slope-Inter The slope-intercept form of an equation is really useful for plotting graphs because m is the slope of the line and b is the y-coordinate of the y-intercept. y = mx + b slope y-coordinate of y-intercept Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 205205205205205 Example Example Example Example Example 11111 Plot the graph of y = 1 2 x + 2. Solution Solution Solution Solution Solution y = mx + b = 1 2 x + 2 Slope = m = 1 2 β for a slope of 1 2, go up 1 unit for every 2 units across. y-coordinate of the y-intercept = b = 2 β so the y-intercept is (0, 2). Example Example Example Example Example 22222 Plot the graph of y = β3x β 4. Solution Solution Solution Solution Solution y = mx + b = β3x β 4 Slope = m = β3 β for a slope of β3, go down 3 units for every unit across. y-coordinate of the y-intercept = b = β4 β so the y-intercept is (0, β4). y 1 x6 -5 -4 -3 -2 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 -6 y x = β3 β 4 -6 -5 -4 -3 -2 6 5 4 3 2 1 -1 0 -1 -2 -3 -4 -5 -6 1 2 3 4 5 6 Guided Practice 7. In the equation y = 8x + 5 |
, find the slope. 8. In the equation y = βx + 10, find the slope. 9. In the equation y = 2x + 5, find the y-intercept. 10. In the equation y = 7b β 3, find the y-intercept. In Exercises 11β18, plot each equation on a graph. 11. y = 2x + 3 12. y = x β 6 13. y = β7x β 8 15. y = β 1 2 x + 6 17. y = 3 4 x 14. y = β 1 3 x β 4 16. y = 1 5 x β 3 18. y = 6 Donβt forget: You can write y = β3x β 4 as y = β3x + (β4), so the y-intercept is β4. 206206206206206 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 lope-IIIIInternternternterntercececececept pt pt pt pt FFFFFororororormmmmm lopeet the SSSSSlopelopeet the to GGGGGet the et the to SolvSolvSolvSolvSolve fe fe fe fe for or or or or y to to lopeet the to OK, so you know the slope-intercept form of an equation makes drawing graphs a lot easier. The trouble is, youβll often be given an equation which isnβt in slope-intercept form. To get the equation into slope-intercept form, solve for y. Example Example Example Example Example 33333 A line has the equation Ax + By = C, where B Ο 0. Solve this equation for y, justifying each step. Solution Solution Solution Solution Solution Ax + By = C Ax β Ax + By = βAx + C By = βAx + C By B Ax = β + en equationtiontiontiontion en equa en equa GiGiGiGiGivvvvven equa en equa equality equality ty of ty of oper oper action pr action pr Subtr Subtr equality ty of equality operty of action proper Subtraction pr equality ty of oper action pr Subtr Subtr equality equality ty of ty of oper oper vision pr vision pr DiDiDiDiDivision pr equality ty of equality operty of vision proper equality ty of oper vision pr The equation is now |
in slope-intercept form. The slope, m = β the y-coordinate of the y-intercept, b = β C B. A B and Example Example Example Example Example 44444 Determine the slope and y-intercept of the line 2x β 3y = 9. Solution Solution Solution Solution Solution Step 1: Solve the given equation for y. 2x β 3y = 9 β3y = β2x + Now youβve got the equation in slope-intercept form, y = mx + b. Step 2: Get the slope and y-intercept from the equation. The slope, m = 2 3. The y-coordinate of the y-intercept, b = β3. So, the y-intercept = (0, β3). Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 207207207207207 Guided Practice In Exercises 19-24, find the slope and y-intercept of the line. 19. 3x + 3y = 9 20. 2y β 6x = 10 21. 2x β 2y = 5 22. β7y + 5x = 14 23. β2y + 3x = 8 24. β5x + 4y = 12 Write down the equations of the following lines in slope-intercept form. 25. The line with slope 4 that passes through the point (0, 2). 26. The line with slope 2 that passes through the point (0, β6). 27. The line with slope β3 that passes through the point (0, 1). 28. The line with slope β 6 7 that passes through the point (0, β3). Independent Practice In Exercises 1β10, find the slope and y-intercept of each equation thatβs given. x β 5 1. y = 1 2 3. y = 3x + 6 5. y = 2(x + 2) 7. 7 β y = 5(x + 4) 9. 3x + 4y = 8 x + 1 2. y = 2 3 4. β6y = 3x + 12 6. y β 4 = 2(x + 1) 8. y β 3 = 2(x β 9) 10. 2x + 3y = 9 In Exercises 11β15, plot the graph of the given equation. 11. y = |
1 3 x + 5 13. y = x + 2 15. y = 2x 12. y = β 1 3 x β 6 14. y = βx + 2 In Exercises 16β20, write the equations of the lines in slope-intercept form. 4 3 1 2 16. A line with slope 17. A line with slope 18. 4x + 2y = 8 20. 3x β 4y = β16 that passes through the point (0, 4) that passes through the point (0, β2) 19. 6x β 3y = 15 In Exercises 21β22, write the equations of the lines in slope-intercept form. 21. The line with slope 0 that passes through the point (2, 6) 22. The line with slope 2 that passes through the point (6, 3) 208208208208208 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 TTTTTopicopicopicopicopic 4.4.44.4.4 4.4.44.4.4 4.4.4 California Standards: 8.0:8.0:8.0:8.0:8.0: Students under stand stand Students under Students under stand Students understand stand Students under allel allelallel par par pts of pts of the conce the conce allel parallel pts of par the concepts of par the conce pts of the conce pendicular pendicular lines and per lines and per pendicular lines and perpendicular lines and per pendicular lines and per lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes lines and how their slopes Students areeeee Students ar Students ar ted. ted. ararararare re re re re relaelaelaelaelated. ted. Students ar Students ar ted. aaaaabbbbble to find the equa tion of tion of le to find the equa le to find the equa tion of le to find the equation of tion of le to find the equa pendicular to a pendicular to a a line per a line per a line perpendicular to a pendicular to a pendicular to a a line per a line per gigigigigivvvvven line tha t passes t passes en line tha en line tha en line that passes t passes en line tha t passes en point. en point. ough a givvvvven point. o |
ugh a gi ough a gi thrthrough a gi thrthr en point. en point. ough a gi thr What it means for you: Youβll learn how to tell whether lines are parallel or perpendicular by looking at the slope-intercept form. Key words: parallel perpendicular reciprocal Donβt forget: See Topic 4.3.1 for more on the slope of a line. Donβt forget: Use the point-slope formula once you have all the information you need. About Slopes About Slopes MorMorMorMorMore e e e e About Slopes About Slopes About Slopes MorMorMorMorMore e e e e About Slopes About Slopes About Slopes About Slopes About Slopes This Topic carries on from the material on parallel and perpendicular lines that you learned earlier in this Section. Lines are Pe Pe Pe Pe Parararararallel allel allelallel Lines ar Lines ar ou if alues of m TTTTTell ell ell ell ell YYYYYou if ou if alues of VVVVValues of alues of allel ou if Lines ar Lines ar ou if alues of Parallel lines all have the same slope, so the slope-intercept forms of their equations all have the same value of m. For example, the lines y = 3x + 2, y = 3x β 1 and y = 3x β 6 are all parallel. Example Example Example Example Example 11111 Find the equation of the line through (4, β4) that is parallel to the line 2x β 3y = 6. Solution Solution Solution Solution Solution Step 1: Write 2x β 3y = 6 in the slope-intercept form β that is, solve the equation for y. 2x β 3y = 6 ο¬ β3y = β2x + 6 ο¬ y = 2 3 x β 2 Step 2: Get the slope from the equation. The slope of the line y = 2 3 x β 2 is 2 3. Since the required line through (4, β4) is parallel to the line y = 2 3 x β 2, its slope is also 2 3. Step 3: Now write the equation of the line through (4, β4) with a slope of 2 3. y β y1 = m(x β x1) ο¬ y β (β4) = 2 3 (x β 4) οΏ½ |
οΏ½ 3(y + 4) = 2(x β 4) ο¬ 3y + 12 = 2x β 8 ο¬ 33333y β 2x = β20 Guided Practice 1. Give an example of a line that is parallel to y = 1 2 x + 1. 2. Is the line y = 4 5 x β 2 parallel to the line y = 4 5 x + 6? Explain. 3. Find the equation of the line through (β4, 3) that is parallel to the line y = 3x + 9. 4. Find the equation of the line through (3, 8) that is parallel to the line 3x + y = 1. Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 209209209209209 Donβt forget: See Topic 4.4.2 for a reminder on perpendicular lines. Lines are Pe Pe Pe Pe Perererererpendicular pendicular pendicular Lines ar Lines ar ou if Also TTTTTell ell ell ell ell YYYYYou if ou if Also alues of m Also Also alues of VVVVValues of alues of pendicular ou if Lines ar pendicular Lines ar ou if Also alues of The slope-intercept forms of equations of perpendicular lines have values of m that are negative reciprocals of each other. For example, the lines y = 3x + 2 and y = β because the negative reciprocal of 3 is β 1 3. 1 3 x β 1 must be perpendicular, Example Example Example Example Example 22222 Find the equation of the line through (2, β4) that is perpendicular to the line β3y β x = 5. Solution Solution Solution Solution Solution Step 1: Write β3y β x = 5 in the slope-intercept form (that is, solve the equation for y). β3y β x = 5 ο¬ β3y = Step 2: Get the slope (m1) of the line y = β 3 slope (m2) of the required line through (2, β4). x β 5 3 and determine the Since m1 = β 1 3, and m2 is the negative reciprocal of β 1 3, m2 must be 3. Step 3: Write the equation of the line through (2, β4) with a slope of 3. Use the point-slope formula here: y β y1 = m( |
x β x1) ο¬ y β (β4) = 3(x β 2) ο¬ y + 4 = 3x β 6 ο¬ y β 3x = β10 Guided Practice 5. Give an example of a line thatβs perpendicular to the line y = 6x. 6. Is the line y = 4x + 2 perpendicular to the line y = β 1 4 x + 4? Explain your answer. 7. Find the equation of the line through (β2, 0) that is perpendicular to the line y = β2x β 4. 8. Find the equation of the line through (β4, 6) that is perpendicular to the line 3x β 4y = 24. 210210210210210 Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 Independent Practice In Exercises 1β8, determine whether the pairs of lines are parallel, perpendicular, or collinear. Check it out: βCollinearβ means βon the same straight line.β If two lines have the same slope, and pass through the same point, they must be collinear. 1. y = 2x + 1 and y = β 1 2 x β 6 2. y = 1 3 x + 5 and y = β3x β 4 3. y = 4x β 8 and y = β 1 4 x + 2 4. y = 6 and y = 3 5. x = 2 and y = β4 6. 5x β 2y = β10 and 10x β 4y = β20 7. 3x + y = 6 and 6x + 2y = β4 8. 2x β y = β4 and 6x β 3y = β12 In Exercises 9β20, find the equations of the lines. 9. The line through (5, 2) thatβs parallel to a line with slope 1 2. 10. The line through (3, β3) thatβs parallel to a line with slope 2 5. 11. The line through (2, β9) thatβs perpendicular to a line with slope β3. 12. The line through (β3, 1) thatβs perpendicular to a line with slope 5 8. 13. The line through (0, 0) thatβs parallel to 3x + y = 18. 14. The |
line through (3, 5) thatβs parallel to 3x β 7y = β21. 15. The line through (4, β3) thatβs parallel to 3x β 4y = 16. 16. The line through (β2, 6) thatβs parallel to 6x β 10y = β20. 17. The line through (0, 6) thatβs perpendicular to 2x + y = 18. 18. The line through (β3, β5) thatβs perpendicular to 3x β 6y = β24. 19. The line through (6, β2) thatβs perpendicular to 3x β 5y = β10. 20. The line through (8, 2) thatβs perpendicular to the line joining the points (β6, 3) and (β2, 6). 21. Use slopes to decide whether the points (β3, β8), (3, β2), and (8, 3) are collinear (on the same line) or noncollinear. 22. Use slopes to decide whether the points (4, 5), (3, β2), and (8, 3) are collinear or noncollinear. 23. Use slopes to decide whether the points B (2, 10), K (β1, 3), and J (5, β3) are vertices of a right triangle. 24. Show that the points M (3, 11), A (β4, 4), and T (3, β3) are vertices of a right triangle. ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Hopefully youβll see now why the slope-intercept form of a line is so useful β you can just glance at the equations to see whether lines are parallel or perpendicular, without having to plot the graphs. Section 4.4 β More Lines Section 4.4 Section 4.4 Section 4.4 Section 4.4 211211211211211 Section 4.5 gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined gions Defined gions Defined by Inequalities by Inequalities by Inequalities by Inequalities |
by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities by Inequalities Just like with equations, you can graph inequalities on the coordinate plane. The only tricky bit is showing whether the solution set is above or below the line. This Topic will show you how. vides the Plane into TTTTThrhrhrhrhree Ree Ree Ree Ree Reeeeegions gions gions vides the Plane into vides the Plane into A Line Di A Line Di gions A Line Divides the Plane into gions vides the Plane into A Line Di A Line Di The graph of a linear equation divides the plane into three regions: The set of points that lie on the line. The set of points that lie above the line. The set of points that lie below the line. The regions above and below the line of a linear equation are each represented by a linear inequality. Example Example Example Example Example 11111 Graph y = βx + 3 and show that it divides the plane into three regions. Solution Solution Solution Solution Solution Set of points bbbbbeloeloeloeloelowwwww the line β all these points satisfy the inequality. β6 β5 β4 β3 β2 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 y-axis Set of points abababababooooovvvvveeeee the line β all these points satisfy the inequality. 1 2 3 4 5 6 x-axis Set of points on on on on on the line β all these points satisfy the equation y = βx + 3 y = βx + 3 y = βx + 3. The line y = βx + 3 y = βx + 3 forms a bbbbborororororder der der der der between the other two regions. The points on the line donβt satisfy either of the inequalities β thatβs why the line is dashed. TTTTTopicopicopicopicopic 4.5.14.5.1 4.5.14.5.1 4.5.1 California Standards: 6.0:6.0:6.0:6.0:6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4)..... TTTTThehehehehey ary ary ary ary are also e also e |
also e also e also h the reeeeegiongiongiongiongion h the r le to sketcetcetcetcetch the r h the r le to sk aaaaabbbbble to sk le to sk h the r le to sk defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality (e(e(e(e(e.g.g.g.g.g.,.,.,.,., the the the the they sky sky sky sky sketcetcetcetcetch the r h the reeeeegiongiongiongiongion h the r h the r h the r < 4). < 4). defined by 2x + 6 + 6 + 6 + 6 + 6y < 4). defined by 2 defined by 2 < 4). < 4). defined by 2 defined by 2 What it means for you: Youβll learn to sketch the region defined by a linear inequality. Key words: inequality plane region Check it out: Youβll find more about the dashed line in Topic 4.5.2. 212212212212212 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 Identifying the Reeeeegiongiongiongiongion Identifying the R Identifying the R Identifying the R Identifying the R To identify the inequality defining a region, choose a point from the region and substitute its coordinates into the equation of the line that borders the region. Since the point doesnβt lie on the line, the equation wonβt be a true statement. To make the statement true, you need to replace the β=β with a β<β or β>β sign. The resulting inequality defines the region containing the point. Example Example Example Example Example 22222 State the inequality that defines the shaded region. y-axis Solution Solution Solution Solution Solution Choose a point in the shaded region, for example, (0, 2). Test this point in the equation of the line: y = 2x + 1 β6 β5 β4 β3 β2 So at (0, 2), you get 2 = 2(0) + 1, that is, 2 = 1, which is a false statement. 0 β1 β1 β2 β3 β4 β5 β6 1 2 3 4 5 6 x-axis Since 2 > |
1, a β>β sign is needed to make it a true statement. So (0, 2) satisfies the inequality y > 2x + 1. Therefore the inequality that defines the shaded region is y > 2x + 1. Guided Practice In Exercises 1β2, state the inequality that defines the shaded region on each of the graphs. 1. β6 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 β5 β4 β2 β3 2 + x 2 = y 26 β5 β4 β3 β2 1 2 3 4 5 6 0 β1 β1 β2 β3 β4 β5 β6 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 213213213213213 y a Linear Inequality y a Linear Inequality gion Defined b hing the Reeeeegion Defined b gion Defined b hing the R SkSkSkSkSketcetcetcetcetching the R hing the R y a Linear Inequality gion Defined by a Linear Inequality y a Linear Inequality gion Defined b hing the R An ordered pair (x, y) is a solution of a linear inequality if its x and y values satisfy the inequality. The graph of a linear inequality is the region consisting of all the solutions of the inequality (the solution set). To sketch the region defined by a linear inequality, you need to plot the graph of the corresponding equation (the border line), then shade the correct region. To decide which is the correct region, just test a point. Example Example Example Example Example 33333 Sketch the region defined by 6x β 3y < 9. Solution Solution Solution Solution Solution First plot the graph of the corresponding equation. This is the border line. Rearrange the equation into the form y = mx + b: 6x β 3y = 9 ο¬ 3y = 6x β 9 ο¬ y = 2x β 3 Table of values for sketching the line: x 0 3 y ( y,x ) y 2= x 3β 3β)0(2= 3β= )3β,0( y 2= x 3β 3β)3(2= 3= )3,3( So the border line goes through the points (0, β3) and (3, 3), as shown in the |
graph below. Now test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. 6x β 3y < 9 0 β 0 < 9 0 < 9 β This is a true statement. Therefore (0, 0) lies in the region 6x β 3y < 9 β so shade the region containing (0, 0). y 6x β 6 β5 β4 β3 β2 0 β1 β1 β2 β3 β4 β5 β6 y-axis 1 2 3 4 5 6 x-axis Check it out: Use whichever method you find easiest for sketching the line β see Section 4.2. For example, you could plot the intercepts if you prefer. Check it out: (0, 0) has been used as a test point to make the algebra easy, but you can use any point β as long as it doesnβt lie on the line. 214214214214214 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 Example Example Example Example Example 44444 Graph the solution set of y + 2x > 4. Solution Solution Solution Solution Solution First plot the graph of the corresponding equation (the border line). Table of values for sketching the line: x 0 3 y ( y,x ) y 2β= x 4+ 4+)0(2β= 4= )4,0( y 2β= x 4+ 4+)3(2β= 2β= )2β,3( So the border line goes through the points (0, 4) and (3, β2), as shown below. Test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. y + 2x > 4 0 + 0 > 4 0 > 4 β This is a false statement. Therefore (0, 0) does not lie in the region y + 2x > 4 β so shade the region that does not contain (0, 0). y-axis y x-axis 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 β6 β5 β4 β3 β2 Guided Practice In each of Exercises 3β8, use a set of axes spanning from β6 to 6 on the x- and y-axes. For each Exercise, shade the region defined by the inequality |
. 3. y > 0.5x + 2 5. y + x > β2 7. β2y + 3x > 6 4. y + 2x < 0 6. 4x + 3y < 12 8. y < βx + 3 In Exercises 9β14, show whether the given point is a solution of β5x + 2y > β8. 9. (0, 0) 12. (2, 1) 11. (β3, 9) 14. (β15, 13) 10. (6, β3) 13. (39, β36) Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 215215215215215 Independent Practice In Exercises 1β4, use the graph opposite to determine if the given point is in the solution set. 1. (0, 0) 2. (1, 2) 3. (3, β2) 4. (β2, β2) y-axis 6 β5 β4 β3 β2 0 β1 β1 β2 β3 β4 β5 β6 x-axis In Exercises 5β8, state the inequality that defines the shaded region on each of the graphs. 2x + y = 4 y-axis 6 β5 β4 β3 β2 0 β1 β1 β2 β3 β4 β5 β6 x-axis 5. 7. y = 1 β6 β5 β4 β3 β2 y-axis 1 β1 β2 β3 β4 β5 β6 2 + 1 x 2 = y β6 β5 β4 β3 β2 y-axis 1 β1 β2 β3 β4 β5 β6 x-axis 6. 8. y-axis 2 0 β1 β1 β2 β3 β4 β5 β6 x-axis x-axis β6 β5 β4 β3 β2 9. Show whether (β2, β1) is a solution of 2x β 5y Β£ 10. 10. Show whether (β3, 5) is a solution of 4x + y < 5. 11. Show whether (2, 4) is a solution of x + 6y < 0. 12. Show whether (0, β3) is a solution of y β₯ β6x + 5. Graph the solution set in Exercises 13β22. 13. |
y < 3 4 x + 6 15. y < β 2 5 x β 2 17. x > 0 19. x + 2y > 8 21. 4x β 6y > 24 14. y < 4 5 x + 4 16. y < 1 18. x β 4y > 8 20. 4x + 3y < β12 22. 5x + 8y < 24 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Well, that was quite a long Topic, with lots of graphs. Inequality graphs arenβt easy, so itβs always a good idea to check whether youβve shaded the correct part by seeing whether a test point satisfies the original inequality. 216216216216216 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 TTTTTopicopicopicopicopic 4.5.24.5.2 4.5.24.5.2 4.5.2 California Standards: 6.0:6.0:6.0:6.0:6.0: Students graph a linear equation and compute the xand y-intercepts (e.g., graph 2x + 6y = 4)..... TTTTThehehehehey ary ary ary ary are also e also e also e also e also h the reeeeegiongiongiongiongion h the r le to sketcetcetcetcetch the r h the r le to sk aaaaabbbbble to sk le to sk h the r le to sk defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality defined by linear inequality (e(e(e(e(e.g.g.g.g.g.,.,.,.,., the the the the they sky sky sky sky sketcetcetcetcetch the r h the reeeeegiongiongiongiongion h the r h the r h the r < 4). < 4). defined by 2x + 6 + 6 + 6 + 6 + 6y < 4). defined by 2 defined by 2 < 4). < 4). defined by 2 defined by 2 What it means for you: Youβll learn how to show the different types of inequality on a graph. Key words: inequality |
strict inequality border region Donβt forget: Remember, Β£ means βless than or equal to,β β₯ means βgreater than or equal to.β gions gions s of R R R R Reeeeegions s of s of BorBorBorBorBorderderderderders of gions gions s of s of R R R R Reeeeegions BorBorBorBorBorderderderderders of gions gions s of s of gions s of gions In Topic 4.5.1 you were dealing with regions defined by strict inequalities β the ones involving a < or > sign. This Topic shows you how to graph inequalities involving Β£ and β₯ signs too. Borderderderderdersssss Bor Bor ypes of e Difffffferererererent ent ent ent ent TTTTTypes of ypes of e Dif gions Can Havvvvve Dif e Dif gions Can Ha RRRRReeeeegions Can Ha gions Can Ha ypes of Bor Bor ypes of e Dif gions Can Ha The region defined by a strict inequality doesnβt include points on the border line, and you draw the border line as a dashed line. For example, the region defined by y > βx + 3 doesnβt include any points on the line y = βx + 3. Regions defined by inequalities involving a £££££ or β₯β₯β₯β₯β₯ sign do include points on the border line. In this case, you draw the border line as a solid line. For example, the region defined by y β₯ βx + 3 includes all the points on the line y = βx + 3. Example Example Example Example Example 11111 Graph 2x β y = β2 and show the three regions of the plane that include all the points on this line. Solution Solution Solution Solution Solution e and e and Set of points abababababooooovvvvve and e and e and on on on on on the line β all these points satisfy the inequality 22222x β β β β β y £££££ β2 β2 β2 β2 β2. Set of points on on on on on the line β all these points satisfy the equation = β2 22222x β β β β β y = β2 = β2 = β2 and both = β2 the inequalities. 22222 |
x β β β β β y £££££ β2 β2 β2 β2 β2 β6 β5 β4 β3 β2 y-axis 22222x β β β β β y β₯β₯β₯β₯β₯ β2 β2 β2 β2 β2 0 β1 β1 β2 β3 β4 β5 β6 Set of points bbbbbeloeloeloeloelowwwww and on and on and on the line β all and on and on these points satisfy the inequality 22222x β β β β β y β₯β₯β₯β₯β₯ β2 β2 β2 β2 β2. x-axis Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 217217217217217 der Line der Line the Bor lusivvvvve ofe ofe ofe ofe of the Bor the Bor lusi lusi gions Inc hing Reeeeegions Inc gions Inc hing R SkSkSkSkSketcetcetcetcetching R hing R der Line the Border Line gions Inclusi der Line the Bor lusi gions Inc hing R The method for sketching the region is the same as the method in Topic 4.5.1, except now thereβs an extra step for showing the border type β using a dashed line, or a solid line. Example Example Example Example Example 22222 Sketch the region of the coordinate plane defined by y Β£ βx β 5. Solution Solution Solution Solution Solution First plot the border line. The border line equation is y = βx β 5. Table of values for sketching the line: x 0 y ( y,x ) y β= x 5β 5β0β= 5β= )5β,0( 3β y β= x 5β 5β3= 2β= )2β,3β( So the border line goes through the points (0, β5) and (β3, β2), as shown in the graph below. Identify the border type: the border line is solid, since the sign is £££££. Test whether the point (0, 0) satisfies the inequality β substitute x = 0 and y = 0 into the inequality. y Β£ βx β 5 ο¬ 0 Β£ β0 β 5 0 Β£ β5 β this is a false |
statement. y-axis Therefore (0, 0) doesnβt lie in the region y Β£ βx β 5 β so shade the region that doesnβt contain (0, 0). 1 2 3 4 5 6 x-axis β6 β5 β4 β3 β2 y Β£ xβ β 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 β7 β8 Guided Practice In Exercises 1β4, show whether the given point is in the solution set of β2x + 3y Β£ β15. 1. (0, 0) 4. (β3, β5) 3. (4, β7) 2. (6, β1) In each of Exercises 5β8 use a set of axes spanning from β6 to 6 on the x- and y-axes, and shade the region defined by the inequality. 5. 4x + 3y Β£ 9 7. β2y β₯ x 6. y Β£ 2x + 3 8. 2x + y β₯ 4 218218218218218 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 phing Reeeeegions gions gions phing R Graaaaaphing R phing R Gr Gr e Examples of MorMorMorMorMore Examples of e Examples of gions e Examples of Gr gions phing R Gr e Examples of Graphing regions isnβt always straightforward, so here are a couple more examples and some more practice exercises. Example Example Example Example Example 33333 Graph the solution set of 4y β 3x β₯ 12. Solution Solution Solution Solution Solution First form the border-line equation: 4y β 3x = 12 4y = 3x + 12 y = 3 4 x + 3 Table of values for sketching the line+ x 3,x ) 3+)0( 3= )3,0( 3+)4( 6= )6,4( So the border line goes through the points (0, 3) and (4, 6). The border line is solid, since the sign is β₯β₯β₯β₯β₯. Test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. 4y β 3x β₯ 12 4(0) β 3(0) β₯ 12 0 β₯ 12 β This is a |
false statement. y-axis Therefore (0, 0) doesnβt lie in the region 4y β 3x β₯ 12 β so shade the region that doesnβt contain (0, 0). 4 β 3 y x β₯ 12 8 7 6 5 4 3 2 1 β6 β5 β4 β3 β2 0 β1 β1 β2 β3 1 2 3 4 5 6 x-axis Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 219219219219219 Check it out: After youβve sketched a region, it can be a good idea to test another point to make sure youβve got it right. Pick a point in your shaded region and make sure the inequality is satisfied there as well. Example Example Example Example Example 44444 Sketch the region of the coordinate plane defined by 2y < 2x + 6. Solution Solution Solution Solution Solution First form the border-line equation: 2y = 2x + 6 y = x + 3 Table of values for sketching the line: x 0 3 y y = x 3+ 3+0= 3= y = x 3+ 3+3= 6= y,x ) ( )3,0( )6,3( So the border line goes through the points (0, 3) and (3, 6). The border line is dashed, since the sign is <. Test whether the point (0, 0) satisfies the inequality. Substitute x = 0 and y = 0 into the inequality. 2y < 2x + 6 2(0) < 2(0) + 6 0 < 6 β This is a true statement. Therefore (0, 0) lies in the region 2y < 2x + 6 β so shade the region containing (0, 0). β6 β5 β4 β3 β2 y-axis -axis 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 Guided Practice In each of Exercises 9β18, use a set of axes spanning from β6 to 6 on the x- and y-axes. For each exercise, shade the region defined by the inequality. 9. y Β£ x 11. x + 2 > 0 13. y + 3 Β£ 0 15. y Β£ 2x β 5 17. y > 2x + 6 10. y β 2 < 0 12. x β 4 |
Β£ 0 14. y β₯ β3x 16. x + 4y < 4 18. 4x β 3y β₯ 8 19. Show whether (5, 4) is in the solution set of 4x β 3y Β£ 8. 20. Show whether (β4, 2) is in the solution set of 2x + y > β6. 220220220220220 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 Independent Practice In Exercises 1β4, show whether the given point is in the solution set of the given line. 1. (β1, 3); 2x + y < 1 3. (0.5, 0.25); x + 2y Β£ 1 2. (β2, β8); x β y β₯ 6 4. (1, β5); x + 2y > β9 In Exercises 5β6, determine if the given point is in the solution set shown by the shaded region of the graph. 5. (4, 6) 6. (β1, 1) y-axis 6 5 4 3 2 1 y-axis 6 β5 β4 β3 β2 x-axis 1 2 3 4 5 6 0 β1 β1 β2 β3 β4 β5 β6 In Exercises 7β8, determine if the given point is in the solution set shown by the shaded region of the graph. 7. (3, β2) 8. (β2, 3) x-axis 1 2 3 4 5 6 y = β2 β6 β5 β4 β3 β2 0 β1 β1 β2 β3 β4 β5 β6 Graph the solution set in Exercises 9β20. 9. x > 4 11. x β 3y Β£ 9 13. x + y β₯ β1 15. x + 2y < 2 17. y β₯ 3x 19. x + y β₯ 3 10. y Β£ β2 12. x + y < 5 14. y β₯ 2x β 7 16. y Β£ 2x + 2 18. y > β2x + 1 20. 2x + y Β£ β3 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up Remember β graphs of inequalities including < and > signs will always have a dashed line, and graphs |
including Β£ and β₯ signs will always have a solid line. Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 221221221221221 TTTTTopicopicopicopicopic 4.5.34.5.3 4.5.34.5.3 4.5.3 California Standards: 9.0:9.0:9.0:9.0:9.0: Students solve a system of two linear equations in two variables algebraically and are able to interpret the Students Students Students answer graphically. Students Students ararararare ae ae ae ae abbbbble to solv e a system e a system le to solv le to solv le to solve a system e a system e a system le to solv ofofofofof tw tw tw tw two linear inequalities in o linear inequalities in o linear inequalities in o linear inequalities in o linear inequalities in twtwtwtwtwo vo vo vo vo variaariaariaariaariabbbbbles and to sk les and to sketcetcetcetcetchhhhh les and to sk les and to sk les and to sk the solution sets..... the solution sets the solution sets the solution sets the solution sets What it means for you: Youβll graph two inequalities to show the solution set that satisfies both inequalities. Key words: system of linear inequalities region point-slope formula gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined gions Defined gions Defined RRRRReeeeegions Defined gions Defined gions Defined o Inequalities o Inequalities bbbbby y y y y TTTTTwwwwwo Inequalities o Inequalities o Inequalities bbbbby y y y y TTTTTwwwwwo Inequalities o Inequalities o Inequalities o Inequalities o Inequalities The idea of graphing two different inequalities on one graph is really not as hard as it sounds. When youβve finished, your graph will show the region where all the points satisfy both inequalities. y More e e e e TTTTThan One Linear Inequality han One Linear Inequality han One Linear Inequality y Mor y Mor gions Defined b R R R R Reeeeegions Defined b gions Defined b han One Linear Inequality g |
ions Defined by Mor han One Linear Inequality y Mor gions Defined b A system of linear inequalities is made up of two or more linear inequalities that contain the same variables. For example, 3x + 2y > 6 and 4x β y < 5 are linear inequalities both containing the variables x and y. An ordered pair (x, y) is a solution of a system of linear inequalities if it is a solution of each of the inequalities in the system. For example, (1, β2) is a solution of the system of inequalities y < βx + 2 and 2y < 2x + 6. The graph of two linear inequalities is the region consisting of all the points satisfying both inequalities. Example Example Example Example Example 11111 Sketch the region satisfying both y < βx + 2 and 2y < 2x + 6. y-axis Line 2y = 2x + 6 1 2 3 4 5 6 x-axis x< β + 2 = region defined by y = region defined by 2 < 2 + 6 x y = region defined by < β + 2 x y and 2 < 2 + 6 x y Line y = βx + 2 Solution Solution Solution Solution Solution 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 β6 β5 β4 β3 β2 Set of points satisfying bbbbbothothothothoth y < βx + 2 andandandandand 2y < 2x + 6. 222222222222222 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 Check it out: The problem in Example 2 is just another way of asking you to sketch the region defined by 5y + 3x β₯ β25 and y β x Β£ β5. gions Defined by y y y y TTTTTwwwwwo Linear Inequalities o Linear Inequalities o Linear Inequalities gions Defined b hing Reeeeegions Defined b gions Defined b hing R SkSkSkSkSketcetcetcetcetching R hing R o Linear Inequalities o Linear Inequalities gions Defined b hing R To sketch the region defined by two linear inequalities, shade the regions defined by each inequality. The region defined by both inequalities is the area where your shading overlaps. Example Example Example Example Example 22222 Graph the solution set satisfying 5y + 3x |
β₯ β25 and y β x Β£ β5. Solution Solution Solution Solution Solution First line: 5y + 3x = β25 ο¬ 5y = β3x β 25,x ) ( y x 0 5 y β= y β= 3 5 3 5 x 5β x 5β 5β)0(β= 3 5 5β)5(β= 3 5 5β= )5β,0( 8β= )8β,5( The border line y = β 3 5 and is a solid line. x β 5 goes through the points (0, β5) and (5, β8), Second line: y β x = β,x ) y = x 5β 5β0= 5β= )5β,0( y = x 5β 5β3= 2β= )2β,3( The border line y = x β 5 goes through the points (0, β5) and (3, β2), and is a solid line. Test whether the point (0, 0) satisfies each inequality: 5y + 3x β₯ β25 5(0) + 3(0) β₯ β25 0 β₯ β25 This is a true statement, so (0, 0) lies in the region 5y + 3x β₯ β25. Shade the region above y = β 3 5 x β 5. y-axis Donβt forget: You need to graph both lines on the same axes. y β x Β£ β5 0 β 0 Β£ β5 0 Β£ β5 This is a false statement, so (0, 0) doesnβt lie in the region y β x Β£ β5. Shade the region below y = x β 5. The required region is the area where the shading overlaps. β3 β2 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 β7 β8 β-axis 5 + 3 y β25 x β₯ and x Β£ β β5 y = β3 5x β 5 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 223223223223223 Guided Practice In each of Exercises 1β2, use a set of axes spanning from β5 to 7 on the x-axis and β6 to 6 on the y-axis. For each exercise, shade the region containing all solution points for both inequalities. 1. 2x + 3 |
y < 6 and y β 2x < 2 2. y β x β₯ 4 and 2x + y Β£ 5 In each of Exercises 3β6, use a set of axes spanning from β6 to 6 on the x- and y-axes. For each exercise, shade the region containing all solution points for both inequalities. 3. y < βx + 4 and y < x 5. y < x and y < βx β 2 4. y < x + 2 and y > β2x + 5 6. y Β£ 0.5x + 3 and y Β£ 2x + 2 Identifying the Inequalities Defining a Reeeeegiongiongiongiongion Identifying the Inequalities Defining a R Identifying the Inequalities Defining a R Identifying the Inequalities Defining a R Identifying the Inequalities Defining a R To identify the inequalities defining a region, you first need to establish the border line equations. Then examine a point in the region to identify the inequalities. Hereβs the method: Find the equations β use the point-slope formula to find the equations of each of the border lines. Identify the signs β choose a point in the region (but not on a line) and substitute its coordinates into each equation. Since the point does not lie on either of the lines, you will have two false statements. Replace the β=β in each equation with a β<β or β>β sign to make the statements true. If the line is solid, use β£££££β or ββ₯β₯β₯β₯β₯.β Write the inequalities which define the region. Example Example Example Example Example 33333 Find the inequalities whose simultaneous solution set is the shaded region shown on the right. Solution Solution Solution Solution Solution First line: Two points on this line are (0, 1) and (3, 2). β4 β3 β2 m1 = β 2 1 β 3 0 = 1 3 y-axis Line 2 Line 1 x-axis 1 β1 β2 β3 β4 β5 y β y1 = m(x β x1) ο¬ y β 2 = 1 3 (x β 3 224224224224224 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 Example 3 continueduedueduedued |
Example 3 contin Example 3 contin Example 3 contin Example 3 contin Second line: Two points on this line are (3, 2) and (1, β2). m1 = β β β y1 = m(x β x1) ο¬ y β 2 = 2(x β 3) ο¬ y β 2 = 2x β 6 ο¬ y = 2x β 4 So the equations of the two border lines are y = 1 3 x + 1 and y = 2x β 4. Choose a point in the shaded region, for example, (5, 4). Substitute this point into the two equations. Equation for line 1 β this is a false statement., so a > sign is needed to make it true. However, the line is solid so the sign should be β₯. So the first inequality is y β₯ 1 3 x + 1. Equation for line 2: y = 2x β 4 ο¬ 4 = 10 β 4 ο¬ 4 = 6 β this is a false statement. 4 < 6, so a < sign is needed to make it true. The line is dashed, so the < sign is correct. So the second inequality is y < 2x β 4. Therefore the inequalities defining the shaded region are y β₯β₯β₯β₯β₯ and y < 2x β 4. 1 3 x + 1 Guided Practice In Exercises 7β10, find inequalities whose simultaneous solution defines each of the shaded regions. 7. 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 β2 2 β3 + β5 β0.5 x + 1 x y+3 =9 1 2 3 4 5 6 7 8. β5 β4 x + y β β2 β3 = β2 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 225225225225225 y-axis 10. 1 2 3 4 5 6 7 x-axis β5 β4 β3 β2 9. β5 β4 β3 β2 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 y-axis 1 2 3 4 5 6 7 x-axis 6 5 4 3 2 1 0 β1 β1 β2 β3 β |
4 β5 β6 Independent Practice In Exercises 1β4, use the graph opposite to determine if the given point is included in the solution set. 1. (0, 0) 2. (2, 6) 3. (6, 2) 4. (4, β2) β6 β5 β4 β3 β2 5. Determine whether the point (β4, β3) lies in the solution region of both 3x β 4y Β£ 2 and x β 2y β₯ 1. y-axis 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β-axis 6. Determine whether the point (0, 0) lies in the solution region of both 3x β 4y Β£ 2 and x β 2y β₯ 1. 7. Determine whether the point (β3, 1) lies in the solution region of both 3x β 4y Β£ 2 and x β 2y β₯ 1. 8. On axes spanning from β3 to 9 on the x-axis and β6 to 6 on the y-axis, graph the solution set that satisfies both the inequalities y + 2x β₯ 4 and y β₯ 1.5x β 2. 9. On axes spanning from β3 to 9 on the x-axis and β8 to 4 on the y-axis, graph the solution set that satisfies both the inequalities y > x β 6 and y > βx + 2. In each of Exercises 10β11, show on axes spanning from β5 to 7 on the x-axis and β6 to 6 on the y-axis the region defined by the set of inequalities: 10. 4y β 3x < 12, 2x + y < 0, and y β₯ 1 11. 3x + 5y > 10, x β y < 2, and y Β£ 2 12. The points (2, 6.6) and (7, 8.1) lie on a line bounding a region. Find the equation of the line. 13. The region in Exercise 12 is also bounded by the lines y = 0.3x and y = |1.5x β 6|. If the points (2, 3), (5, 1.5), and (4, 4) all lie within the region, draw and shade the region on axes spanning from β2 to 12 on the x- and y-axes. ound Up ound Up RRRRRound Up |
ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up When youβre graphing a system of linear inequalities, donβt forget that you still have to pay attention to whether the lines should be solid or dashed. 226226226226226 Section 4.5 Section 4.5 Section 4.5 β Inequalities Section 4.5 Section 4.5 Chapter 4 Investigation ee Grooooowthwthwthwthwth ee Gr ee Gr TTTTTrrrrree Gr ee Gr TTTTTrrrrree Gr ee Grooooowthwthwthwthwth ee Gr ee Gr ee Gr This Chapter was all about turning raw data into graphs that show you patterns in the data. The loss of forest areas (deforestation) is a huge environmental problem. Many countries are cutting down trees at a faster rate than they can be replaced. If you look at a cross section of a tree, youβll notice a pattern of rings. Each ring is a layer of wood that took one year to grow. 26 pine trees have been felled. The number of rings on each stump and the diameter are recorded below. diameter )ega(sgnirforebmuN 32 96 51 4 64 )mc(retemaiD 6.03 1.26 5.52 2.11 4.14 91 2.9 02 35 86 23 72 23 23 7.02 0.24 1.74 1.11 4.02 0.72 8.13 )ega(sgnirforebmuN 05 73 11 84 8 34 24 36 )mc(retemaiD 3.94 0.04 7.31 3.41 8.4 9.03 4.33 5.35 61 8.4 35 46 75 85 7.33 0.53 5.54 5.25 Draw a scatter diagram of the data. Put βAge (years)β on the x-axis and βDiameter (cm)β on the y-axis. Describe the correlation between the age of the tree and the diameter of its trunk. Why do you think that the correlation is not perfect? Draw a best-fit line for the data. A best-fit line is one that passes as close to as many of the points as possible. About half the points should be on each side of the line. Would you expect this line |
to pass through the origin? Why? What is the slope of the best-fit line? What does the slope of your graph represent? Find the equation of the best-fit line. a) How old would you expect a tree to be if its diameter is 23 cm? b) What would you expect the diameter of a 6-year-old tree to be? Extension Itβs difficult to measure the diameter of a living tree. Itβs much easier to measure the circumference of the tree instead. 1) Write an equation to link a treeβs circumference with its likely age. How old is a tree likely to be if its circumference is 157 cm? 2) What is the average annual increase in a treeβs circumference? Open-ended Extension 1) Record the number of pages and the thickness of each book (excluding the cover). (For this investigation you will need a selection of books of different types.) Produce a scatter diagram of the data and draw a best-fit line. Find the slope and equation of your line. What does the slope mean? Measure the thickness of some different books. Use your equation to predict the number of pages. How accurate are your predictions? Why do you think this is? 2) Think of two sets of data that you suspect may have a linear relationship. You should choose data sets that you can easily collect. Investigate the relationship using a scatter diagram. ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up You can work out patterns in any collection of data, as long as the data is all about the same thing. estigaaaaationtiontiontiontion β Tree Growth estigestig estig pter 4 Invvvvvestig pter 4 In ChaChaChaChaChapter 4 In pter 4 In pter 4 In 227227227227227 Chapter 5 Systems of Equations Section 5.1 Systems of Equations........................... 229 Section 5.2 The Elimination Method........................ 244 Section 5.3 Applications of Systems of Equations.. 249 Investigation Breaking Even in an Egg Business....... 261 228228228228228 TTTTTopicopicopicopicopic 5.1.15.1.1 5.1.15.1.1 5.1.1 Section 5.1 phing Method phing Method he Graaaaaphing Method he Gr he Gr TTTTThe Gr |
phing Method phing Method he Gr he Graaaaaphing Method TTTTThe Gr phing Method phing Method he Gr he Gr phing Method he Gr phing Method California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa le to le to and are ae ae ae ae abbbbble to and ar and ar algebraically and ar le to le to and ar et the answwwwwererererer et the ans interprprprprpret the ans et the ans inter inter et the ans inter inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: Youβll solve systems of linear equations by graphing the lines and working out where they intersect. Key words: system of linear equations simultaneous equations In Section 4.5 you graphed two inequalities to find the region of points that satisfied both inequalities. Plotting two linear equations on a graph involves fewer steps, and it means you can show the solution to both equations graphically. tions tions Linear Equa Linear Equa Systems of Systems of tions Linear Equations Systems of Linear Equa tions Linear Equa Systems of Systems of A system of linear equations consists of two or more linear equations in the same variables. For example, 3x + 2y = 7 and x β 3y = β5 form a system of linear equations in two variables β x and y. The solution of a system of linear equations in two variables is a pair of values like x and y, or (x, y), that satisfies each of the equations in the system. For example, x = 1, y = 2 or (1, 2) is the solution of the system of equations 3x + 2y = 7 and x β 3y = β5, since it satisfies both equations. Equations in a |
system are often called simultaneous equations because any solution has to satisfy the equations simultaneously (at the same time). The equations canβt be solved independently of one another. y Graaaaaphing phing phing y Gr y Gr tions b tions b Equa Equa Solving Systems of Solving Systems of phing tions by Gr Equations b Solving Systems of Equa phing y Gr tions b Equa Solving Systems of Solving Systems of A system of two linear equations can be solved graphically, by graphing both equations in the same coordinate plane. Every point on the line of an equation is a solution of that equation. The point at which the two lines cross lies on both lines and so is the solution of both equations. Check it out: The point of intersection is the point where the lines cross. The solution of a system of linear equations in two variables is the point of intersection (x, y) of their graphs. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 229229229229229 Check it out: Use whichever method you find easiest to plot the graphs. The tables of values methodβs been used in Example 1. Example Example Example Example Example 11111 Solve this system of equations by graphing: 2x β 3y = 7 β2x + y = β1 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: 2x β 3y = 7 3y = 2x β β y 1β 3β The line goes through the points (2, β1) and (β1, β3). Line of second equation: β2x + y = β1 y = 2x β 1 x 0 1 y 1β 1 The line goes through the points (0, β1) and (1, 1). Now you can draw the graph: 6 5 4 3 2 1 β6 β5 β4 β3 β2 (β1, β3) 0 β1 β1 β2 β3 β4 β5 β6 y-axis β -axis Step 2: Read off the coordinates of the point of intersection. The point of intersection is (β1, β3). Step 3: Check whether your coordinates give true statements when they are substituted into each equation. 2x β 3y = 7 ο¬ 2(β1) β 3(β3) = 7 |
ο¬ 7 = 7 β True statement β2x + y = β1 ο¬ β2(β1) + (β3) = β1 ο¬ β1 = β1 β True statement Therefore x = β1, y = β3 is the solution of the system of equations. 230230230230230 Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 Guided Practice Solve each system of equations in Exercises 1β6 by graphing on x- and y-axes spanning from β6 to 6. 1. y + x = 2 and y = 2 3 x + 2 3. y = x β 3 and y + 2x = 3 2. y + x = 3 and 3y β x = 5 4. y β 3 2 x = 1 and y + 1 2 x = β3 5. y β x = 3 and y + x = β1 6. 2y β x = β6 and y + 1 2 x = β3 Independent Practice Solve each system of equations in Exercises 1β6 by graphing on x- and y-axes spanning from β6 to 6. 1. 2x + y = 7 and y = x + 1 3. y = β3 and x β y = 2 5. 2y + 4x = 4 and y = βx + 3 2. x + y = 0 and y = β2x 4. x β y = 4 and x + 4y = β1 6. y = βx and y = 4x Determine the solution to the systems of equations graphed in Exercises 7 and 8. 7. y = β x β6 β5 β4 β3 β1 β1 β2 β3 β4 β5 β6 y-axis 8-axis β6 β5 β4 β3 β2 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 y-axis β -axis Solve each system of equations in Exercises 9β16 by graphing on x- and y-axes spanning from β6 to 6. 9. x β y = 6 and x + y = 0 11. 4x β 3y = 0 and 4x + y = 16 10. y = 2x β 1 and x + y = 8 12. x β y = 0 and x + y = |
8 13. y = βx + 6 and x β y = β4 14. x β y = 1 and x + y = β3 15. x + y = 1 and x β 2y = 1 16. 2x + y = β8 and 3x + y = β13 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up Thereβs something very satisfying about taking two long linear equations and coming up with just a one-coordinate-pair solution. You should always substitute your solution back into the original equations, to check that youβve got the correct answer. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 231231231231231 TTTTTopicopicopicopicopic 5.1.25.1.2 5.1.25.1.2 5.1.2 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aicallyyyyy and are able to aicall aicall aicall interpret the answer graphically..... Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: Youβll learn about the substitution method and use it to solve systems of linear equations. Key words: substitution system of linear equations simultaneous equations Check it out: This is the same problem that you saw in Example 1 in the previous Topic β but this time itβs been solved with the substitution method. The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method The Substitution Method Graphing can work well if you have integer solutions, but it can be difficult to read off fractional |
solutions, and even integer solutions if the scale of your graph is small. e Graaaaaphsphsphsphsphs e Gr he Substitution Method Doesnβββββt Int Int Int Int Invvvvvolvolvolvolvolve Gr e Gr he Substitution Method Doesn TTTTThe Substitution Method Doesn he Substitution Method Doesn e Gr he Substitution Method Doesn The substitution method involves a bit more algebra than graphing, but will generally give you more accurate solutions. Substitution Method Take one of the equations and solve for one of the variables (x or y). Substitute the expression for the variable into the other equation. This gives an equation with just one variable. Solve this equation to find the value of the variable. Substitute this value into one of the earlier equations and solve it to find the value of the second variable. Example Example Example Example Example 11111 Solve this system of equations using substitution: 2x β 3y = 7 (Equation 1) β2x + y = β1 (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. In this case, itβs easiest to solve for y in Equation 2 because it has a coefficient of 1: β2x + y = β1 ο¬ y = 2x β 1 (Equation 3) Now you have y expressed in terms of x. Step 2: Substitute 2x β 1 for y in Equation 1. Then solve to find the value of x. 2x β 3y = 7 2x β 3(2x β 1) = 7 2x β 6x + 3 = 7 β4x + 3 = 7 β4x = 4 x = β1 232232232232232 Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Step 3: Substitute β1 for x into an equation to find y. Equation 3 is the best one to use here as y is already isolated β so you donβt have to do any rearranging. y = 2x β 1 ο¬ y = 2(β1) β 1 ο¬ y = β3 Therefore x = β1, y = β3 is the solution of the |
system of equations. Itβs a good idea to check that the solution is correct by substituting it into the original equations. 2x β 3y = 7 ο¬ 2(β1) β 3(β3) = 7 ο¬ β2 + 9 = 7 ο¬ 7 = 7 β True statement β2x + y = β1 ο¬ β2(β1) + (β3) = β1 ο¬ 2 β 3 = β1 ο¬ β1 = β1 β True statement The solution makes both of the original equations true statements, so it must be correct. Guided Practice Solve each system of equations in Exercises 1β6 by the substitution method. 1. y = 2x β 3 and β3y + 2x = β15 2. x + 2y = 7 and 3x β 2y = 5 4. w + z = 13 and w β 2z = 4 3. a + b = 2 and 5a β 2b = β4 6. m + 6n = 25 and n β 3m = β18 5. y + 2x = 5 and 2x β 3y = 1 Independent Practice Solve by the substitution method: 1. y = 3x and x + 21 = β2y 3. 7x β 4y = 27 and βx + 4y = 3 2. x + y = 5 and 5x + 2y = 16 4. 9y β 6x = 0 and 13x + 9 = 24y Solve each system of three equations by the substitution method: 5. 3x + y = 10, 4x β z = 7, 7y + 2z = 17 6 =, 2a + b β 2c = β11, β4a β 4b + c = β32 7 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up With the substitution method, it doesnβt matter which equation you choose to rearrange, or which variable you choose to solve for β you will get the same answer as long as you follow the steps correctly and donβt make any mistakes along the way. The important thing is to keep the algebra as simple as you possibly can. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 233233233 |
233233 TTTTTopicopicopicopicopic 5.1.35.1.3 5.1.35.1.3 5.1.3 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: Youβll show that the graphing and substitution methods both give the same solution to a system of linear equations. Key words: substitution system of linear equations simultaneous equations phing phing Comparing the Graaaaaphing Comparing the Gr Comparing the Gr phing phing Comparing the Gr Comparing the Gr Comparing the Graaaaaphing phing phing Comparing the Gr Comparing the Gr phing phing Comparing the Gr Comparing the Gr and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods and Substitution Methods Thereβs nothing new in this Topic β youβll just see that the graphing and substitution methods always give the same answer. An Example Using the Graaaaaphing Method phing Method phing Method An Example Using the Gr An Example Using the Gr phing Method phing Method An Example Using the Gr An Example Using the Gr Example Example Example Example Example 11111 Solve this system of equations by graphing: |
2y + x = 8 2y + 2x = 10 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: 2y + x = 8 ο¬ 2y = β The line goes through the points (0, 4) and (2, 3). Line of second equation: 2y + 2x = 10 ο¬ 2y = β2x + 10 ο¬ y = βx + 5 x 0 3 y 5 2 The line goes through the points (0, 5) and (3, 2). Step 2: Read off the coordinates of the point of intersection. The point of intersection is (2, 3). y = β0.6 β5 β4 β3 β2 0 β1 β1 β2 y-axis -ax s i Step 3: Check whether your coordinates give true statements when substituted into each of the equations. 2y + x = 8 ο¬ 2(3) + 2 = 8 ο¬ 8 = 8 β True statement 2y + 2x = 10 ο¬ 2(3) + 2(2) = 10 ο¬ 10 = 10 β True statement Therefore x = 2, y = 3 is the solution of the system of equations. 234234234234234 Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 The Same Example Using the Substitution Method The Same Example Using the Substitution Method The Same Example Using the Substitution Method The Same Example Using the Substitution Method The Same Example Using the Substitution Method Example Example Example Example Example 22222 Solve this system of equations using substitution: 2y + x = 8 (Equation 1) 2y + 2x = 10 (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. In this case, itβs easiest to solve for x in Equation 1 because it has a coefficient of 1: 2y + x = 8 ο¬ x = β2y + 8 (Equation 3) Now you have x expressed in terms of y. Step 2: Substitute β2y + 8 for x in Equation 2. Then solve to find the value of y. 2y + 2x = 10 2y + 2(β2y + 8) = 10 2y β 4y + 16 = 10 β2 |
y = β6 y = 3 Step 3: Substitute 3 for y in an equation to find x. Equation 3 is the best one to use here because x is already isolated β so you donβt have to do any rearranging. x = β2y + 8 x = β2(3) + 8 x = 2 So x = 2, y = 3, or (2, 3), is the solution of the system of equations. Check by substituting the solution in the original equations. 2y + x = 8 2(3) + 2 = 8 8 = 8 β True statement 2y + 2x = 10 2(3) + 2(2) = 10 10 = 10 β True statement The solution makes both of the original equations true statements, so it must be correct. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 235235235235235 Check it out: The substitution method gives the same answer as the graphing method used in Example 1. Guided Practice Solve each system of equations in Exercises 1β4 by graphing on xand y-axes spanning from β6 to 6. 1. y = 5 and x + y = 9 2. x = 4 and x + y = 6 3. 4x + 4y = 12 and 2x + y = 5 4. 5x + y = 15 and y = 1 3 x β 1 5. Solve the systems of equations in Exercises 1β4 using the substitution method. Independent Practice In Exercises 1β2, say whether it is possible to find the exact solutions of each system of equations using their graphs. 1. β6 β5 β4 β1 β1 β2 β3 β4 β5 β6 β2 1 26 β5 β4 β3 β2 0 β1 β1 β2 β3 β4 β5 β6 1 2 3 4 5 6 β2 x β 3 y = 7 Solve Exercises 3β6 by graphing. 3. y = 2 and 3x β y = 10 5. x β y = 1 and 2x + y = β1 4. x β y = 0 and x + y = 6 6. x β 2y = 4 and y = βx + 4 7. Solve the systems of equations in Exercises 3β6 using the substitution method. Solve Exercises 8β13 |
by substitution. 8. 2x β y = 8 and y = 4 10. y = 2x β 1 and x + y = 5 12. 6x + y = β 2 and 4x β 3y = 17 13. 4x β 5y = 0 and 4x β 3y = 8 9. x + y = 0 and y = β3x 11. y = 3x and 2x + 3y = 44 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up The graphing method works well with simple equations with integer solutions, but you need to draw the graphs very carefully β if your graphs arenβt accurate enough, youβll get the wrong solution. If the equations look complicated, the substitution method is likely to give more accurate results. 236236236236236 Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 TTTTTopicopicopicopicopic 5.1.45.1.4 5.1.45.1.4 5.1.4 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: Youβll learn how to spot systems of equations that |
have no solutions. Key words: inconsistent point of intersection parallel substitution system of linear equations Check it out: Equations with identical slopes, but different intercepts, result in parallel lines. So you can tell if the system is inconsistent without even drawing the graphs, by rearranging the equations into the slope-intercept form. Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems Inconsistent Systems So far, thereβs been one solution to each system of linear equations β but it doesnβt always work out like that. tions tions Linear Equa Linear Equa Systems of Systems of tions Linear Equations Systems of Linear Equa tions Linear Equa Systems of Systems of The systems so far in this Section have all had one solution. These are called independent systems. Some systems of two linear equations have no solutions β in other words, thereβs no ordered pair (x, y) that satisfies both of the equations in the system. A system of equations with no solutions is called an inconsistent system. sect sect t Inter Inconsistent System Lines Donβββββt Inter t Inter Inconsistent System Lines Don Inconsistent System Lines Don sect t Intersect sect t Inter Inconsistent System Lines Don Inconsistent System Lines Don Example Example Example Example Example 11111 Solve this system of equations by graphing Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equationβ 1β The line goes through (0, 1) and (β2, β1). Line of second equationβ 0 The line goes through (0, β3) and (3, 0). y-axis -axis 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 β2 β3 1 β6 β5 β4 y = x + Step 2: Read off the coordinates of the point of intersection. The lines are parallel, so thereβs no point of intersection. So the system of equations has no solutions β the system is inconsistent. The lines of equations in an inconsistent system are parallel. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 237237237237237 tement tement alse Sta alse Sta es a F he Substitution Method G |
ivvvvves a F es a F he Substitution Method Gi TTTTThe Substitution Method Gi he Substitution Method Gi tement alse Statement es a False Sta tement alse Sta es a F he Substitution Method Gi Hereβs the same problem you saw in Example 1, but this time using the substitution method. Example Example Example Example Example 22222 Solve this system of equations by substitutionEquation 1) (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. Equation 2 already expresses y in terms of x: y = x β 3 Step 2: Substitute x β 3 for y in Equation 1. Then solve to find x. Check it out You canβt substitute any value of x into Equation 3 and make a true statement β so there are no solutions. y β x = 1 (x β 3) β x = 1 (Equation 3) x β 3 β x = 1 β3 = 1, which is a false statement. Therefore Equation 3 doesnβt hold for any value of x. So the system of equations has no solutions β the system is inconsistent. Guided Practice 1. How many solutions does an independent system have? 2. How many solutions does an inconsistent system have? 3. Convert y = 3x β 4 and β9x + 3y = 6 to slope-intercept form. 4. Say why you can now tell that the equations from Exercise 3 must be inconsistent. 5. Use the substitution method to say whether y = 3x β 4 and β9x + 3y = 7 are inconsistent. Independent Practice In Exercises 1β4, graph each pair of equations and say whether they are independent or inconsistent. 1. y = x β 2 and y β x = β6 3. 2x β y = β1 and y = 2x + 2 2. 2y + x = 5 and y = βx + 1 4. 3x + 4y = 3 and y = β x β 4 3 4 Use the substitution method to show that each of the pairs of equations in Exercises 5β8 are inconsistent. 5. x + y = β5 and x + y = 3 7. 2x + y = β3 and 2x + y = 1 6. x + y = β6 and y = 4 β x 8. x + 2y = β2 and |
2x = 10 β 4y ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRRound Up ound Up ound Up From this Topic youβve seen that it doesnβt matter whether you use the graphing or substitution method β if the system is inconsistent, you wonβt get any solutions. 238238238238238 Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 TTTTTopicopicopicopicopic 5.1.55.1.5 5.1.55.1.5 5.1.5 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: Youβll learn how to spot systems of equations with infinitely many solutions. Key words: dependent coincide infinite substitution system of linear equations Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems Dependent Systems In Topic 5.1.4 you saw systems of equations with no solution. In this Topic youβll see systems that have infinitely many solutions. A Dependent System Has Infinitely Many Solutions A Dependent System Has Infinitely Many Solutions A Dependent System Has Infinitely |
Many Solutions A Dependent System Has Infinitely Many Solutions A Dependent System Has Infinitely Many Solutions Some systems of two linear equations have an infinite number of solutions β in other words, there are an infinite number of points (x, y) that satisfy both of the equations in the system. Every solution of one equation is also a solution of the other equation. A system of equations with an infinite number of solutions is said to be dependent. pendent System β the Lines Coincide pendent System β the Lines Coincide phing a De GrGrGrGrGraaaaaphing a De phing a De pendent System β the Lines Coincide phing a Dependent System β the Lines Coincide pendent System β the Lines Coincide phing a De Example Example Example Example Example 11111 Solve this system of equations by graphing: y + x = 4 2y + 2x = 8 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: y + x = 4 y = βx + 4 x 0 2 y 4 2 The line goes through (0, 4) and (2, 2). Line of second equation: 2y + 2x = 8 2y = β2x + 8 y = βx + 4 x 0 2 y 4 2 The line goes through (0, 4) and (2, 2). y-axis -axis 6 β5 β4 β3 β2 0 β1 β1 β2 β3 β4 β5 β6 Check it out: Every point on the line y + x = 4 is on the line 2y + 2x = 8. Step 2: Read off the coordinates of the point of intersection. The two equations represent the same line, so there are infinitely many points of intersection. Therefore the system of equations has infinitely many solutions β the system is dependent. The lines of equations in a dependent system coincide. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 239239239239239 tement tement ue Sta es a TTTTTrrrrrue Sta ue Sta es a he Substitution Method Givvvvves a es a he Substitution Method Gi TTTTThe Substitution Method Gi he Substitution Method Gi tement ue Statement tement ue Sta es a he Substitution Method Gi Hereβs the same problem you saw in Example 1, |
but this time using the substitution method. Example Example Example Example Example 22222 Solve this system of equations by substitution: y + x = 4 (Equation 1) 2y + 2x = 8 (Equation 2) Solution Solution Solution Solution Solution Step 1: Rearrange one equation so that one of the variables is expressed in terms of the other. Here, itβs easiest to solve for x or y in Equation 1. y + x = 4 y = βx + 4 (Equation 3) Step 2: Substitute βx + 4 for y in Equation 2. Then solve to find x. 2y + 2x = 8 2(βx + 4) + 2x = 8 (Equation 4) β2x + 8 + 2x = 8 8 = 8, which is a true statement. Therefore Equation 4 holds for any value of x. So the system of equations has infinitely many solutions β the system is dependent. Guided Practice 1. How many solutions does a dependent system have? Say whether the systems in Exercises 2β5 are dependent. 2. x β y = 5 and 2x = 2y + 10 3. 2x + 2y = 2 and y = βx + 1 4. 4y β 5x = 12 and y = 5 4 x + 3 5. β2x + 7y = β14 and y = 2 7 x + 1 Independent Practice By graphing, say whether the systems in Exercises 1β4 are dependent. 1. 3x β y = β1 and β3x + y = 1 2. y β 3x = β1 and β2y + 6x = 2 3. y = β2x + 6 and y = 1 2 x β 4 4. 4y + 8x = 8 and y + 2x = 2 Check it out: You can substitute any value of x into Equation 4 and make a true statement because x disappears. So there are infinitely many solutions. Use substitution to say whether the systems in Exercises 5β8 are dependent. 5. x + y = 2 and 2x + 2y = 4 7 and 2y + x = 7 6. x + y = 57 and y = 6x + 120 8. y = β 4 3 x β 4 and 3y + 4x = β12 ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up RRRRR |
ound Up ound Up ound Up It would be crazy to try to list the solutions to dependent systems, because there are infinitely many. You can actually rearrange the equations and show that theyβre really saying the same thing. 240240240240240 Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 TTTTTopicopicopicopicopic 5.1.65.1.6 5.1.65.1.6 5.1.6 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall y and are ae ae ae ae abbbbblelelelele y and ar y and ar aicall aicall aically and ar y and ar aicall et the answwwwwererererer et the ans et the ans to interprprprprpret the ans to inter to inter et the ans to inter to inter gggggrrrrraaaaaphicall phicallyyyyy..... Students are phicall phicall phicall able to solve a system of two linear inequalities in two variables and to sketch the solution sets. What it means for you: Youβll get more practice solving systems of equations and interpreting the solutions. Key words: dependent independent inconsistent system of linear equations Check it out: βInterpreting the solutionβ means stating whether the system is independent (1 solution), inconsistent (0 solutions), or dependent (infinitely many solutions). Check it out: The lines on the graph have identical slopes and different intercepts, so theyβre parallel. tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations tions Systems of Equa Equa tions Systems of Systems of ther Examples ther Examples |
β Furβ Furβ Furβ Furβ Further Examples ther Examples ther Examples β Furβ Furβ Furβ Furβ Further Examples ther Examples ther Examples ther Examples ther Examples Thereβs nothing new to learn in this Topic β just practice at using the methods youβve learned to solve systems of equations. tions tions Equa Equa t Solving Systems of t Solving Systems of actice a e Pre Practice a actice a e Pre Pr MorMorMorMorMore Pr tions Equations t Solving Systems of Equa actice at Solving Systems of tions Equa t Solving Systems of actice a Example Example Example Example Example 11111 Find and interpret the solution of this system of equations2 Solution Solution Solution Solution Solution Step 1: Graph both equations in the same coordinate plane. Line of first equation: y + x = 2 y = βx + 2 x 0 4 y 2 2β The line goes through (0, 2) and (4, β2). Line of second equation: y + x = β2 y = βx β 2 x 0 2β y 2β 0 The line goes through (0, β2) and (β2, 0). Draw the graph-axis 1 2 3 4 5 6 x-axis β x β 2 β6 β5 β4 β3 β2 6 5 4 3 2 1 0 β1 β1 β2 β3 β4 β5 β6 Step 2: Read off the coordinates of the point of intersection. The lines are parallel, so thereβs no point of intersection. The system of equations has no solutions β the system is inconsistent. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 241241241241241 Example Example Example Example Example 22222 Solve this system of equations and state whether the system is independent, inconsistent, or dependent: y β 3x = β5 (Equation 1) 3x β 4y = β7 (Equation 2) Solution Solution Solution Solution Solution Step 1: Solve for y in Equation 1. y β 3x = β5 y = 3x β 5 (Equation 3) Step 2: Substitute 3x β 5 for y in Equation 2. Then solve to find the value of x. 3x β 4y = β7 3x β 4(3x β 5) = β7 3x β 12x + 20 = β7 β9 |
x = β27 x = 3 Step 3: Substitute 3 for x in Equation 3 to find y. y = 3x β 5 y = 3(3 So x = 3, y = 4 is the solution of the system of equations. Check by substituting the solution in the original equations. y β 3x = β5 4 β 3(3) = β5 4 β 9 = β5 β5 = β 5 β True statement 3x β 4y = β7 3(3) β 4(4) = β7 9 β 16 = β7 β7 = β7 β True statement The solution is correct. The system is independent since it has one solution. 242242242242242 Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 Guided Practice Solve each system of equations in Exercises 1β6 by either substitution or graphing. If a system isnβt independent then say whether it is inconsistent or dependent. 1. 2a β 7b = β12 and βa + 4b = 7 2. 3x β y = β2 and x β y = 0 3. y + x = 6 and 3y β x = 6 4. y + 2x = β1 and 3y + 4x = 3 5. 5y β 4x = 20 and y β 4 5 x = 4 6. 5y + 3x = β5 and y β x = β1 Independent Practice Solve each system of equations in Exercises 1β20 by either substitution or graphing. If a system isnβt independent then say whether it is inconsistent or dependent. y = 3x + 6 and 1 2 2. 2x β y = β2 and β2x + y = 2 3. 2x + y = 5 and 2x + y = 1 4. x = 2y + 5 and β2x + 4y = 2 5. 2x β 4y = 4 and 2y β 2x = β5 6. 7x β y = 2 and 2y β 14x = β4 7. β3x + 9y = β39 and 2x β y = 21 8. 4x + 5y = 2 and y = β 4 5 x β 10 9. 5x β y = β13 and β3x + 8y = 67 10. β5x + 9y = β4 and β10x |
+ 18y = β1 11. 6x + 6y = β6 and 11y β x = 25 12. β4x + 7y = 15 and 8x β 14y = 37 13. β7x + 5y = 20 and 14x β 10y = β40 14. 5x + 6y = 5 and β5x + 3y = 1 15. β4x + 7y = 9 and 5x β 2y = β18 16. 5x β y = β15 and β3x + 9y = 93 17. β4x + 9y = 57 and 5x + 5y = 75 18. β2x + 5y = 33 and 10x β 25y = 69 19. β3x + 4y = 41 and β12x + 16y = β1 20. β2x β 5y = 75 and 4x + 10y = β150 ound Up ound Up RRRRRound Up ound Up ound Up RRRRRound Up ound Up ound Up ound Up ound Up That was a long Section, with lots of different methods. In Section 5.2 youβll learn one more way of solving systems of equations algebraically, and then in Section 5.3 youβll use all these methods to solve some real-life problems. Section 5.1 Section 5.1 Section 5.1 β Systems of Equations Section 5.1 Section 5.1 243243243243243 TTTTTopicopicopicopicopic 5.2.15.2.1 5.2.15.2.1 5.2.1 California Standards: Students solve ae ae ae ae a 9.0:9.0:9.0:9.0:9.0: Students solv Students solv Students solv Students solv system of tw tw tw tw two linear o linear o linear system of system of o linear system of o linear system of tions in two vo vo vo vo variaariaariaariaariabbbbbleslesleslesles tions in tw tions in tw equa equa equations in tw equa tions in tw equa algalgalgalgalgeeeeebrbrbrbrbraicall aically y y y y and are able to aicall aicall aicall interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to |
sketch the solution sets. What it means for you: Youβll learn about the elimination method and then use it to solve systems of linear equations in two variables. Key words: system of linear equations elimination method Check it out: The size (absolute value) is the important thing here. It doesnβt matter if the coefficients have opposite signs β for example, β5 and 5. Section 5.2 tions tions Equa Equa Systems of Systems of tions Equations Systems of Equa tions Equa Systems of Systems of tions tions Equa Equa Systems of Systems of Equations Systems of Equa tions Systems of Equa tions Systems of β Eliminationtiontiontiontion β Elimina β Elimina β Elimina β Elimina β Eliminationtiontiontiontion β Elimina β Elimina β Elimina β Elimina Hereβs one final method of solving systems of linear equations. tion Method tion Method Solving Using the Elimina Solving Using the Elimina tion Method Solving Using the Elimination Method tion Method Solving Using the Elimina Solving Using the Elimina The graphing method used in Section 5.1 isnβt generally very accurate when the numbers involved are large or are fractions or decimals. Itβs pretty difficult to get exact fractions or decimals from a graph. In these cases, itβs simpler to use the elimination method. The Elimination Method Multiply or divide one or both equations so that one variable has coefficients of the same size in both equations. Get rid of (eliminate) this variable by either adding the equations or subtracting one equation from the other. Example Example Example Example Example 11111 Solve this system of equations: β3y + 4x = 11 5y β 2x = 5 Solution Solution Solution Solution Solution Step 1: Make the coefficients of one variable have the same absolute value in each equation. In this case it is easier to make the x-coefficients the same size, rather than the y-coefficients. Multiplying the second equation by 2 will give you a β4x to match the 4x in the first equation. 2(5y β 2x = 5) 10y β 4x = 10 Now you have two equations whose x-coefficients have the same absolute value: β3y + 4x = 11 10y β 4x = 10 Step 2: Add the two equations OR subtract one from the other to eliminate |
a variable. In this case, the coefficients of x are opposites of each other, so adding the equations will eliminate x. β3y + 4x = 11 + 10y β 4x = 10 7y = 21 ο¬ y = 3 244244244244244 Section 5.2 Section 5.2 Section 5.2 β The Elimination Method Section 5.2 Section 5.2 Example 1 continueduedueduedued Example 1 contin Example 1 contin Example 1 contin Example 1 contin Step 3: Now choose either of the original equations and substitute in the value you have found. Substitute 3 for y, and solve for x. β3y + 4x = 11 β3(3) + 4x = 11 β9 + 4x = 11 4x = 20 x = 5 So x = 5, y = 3 is the solution to the system of equations. Itβs always a good idea to check your solution in the other equation too. 5y β 2x = 5 5(3) β 2(5) = 5 15 β 10 = 5 5 = 5 β True statement Guided Practice Solve the following systems of equations by elimination. 1. 5y β 3x = 19 4y + 3x = β1 3. 3y β 2x = 13 2y + x = 4 5. 3y + 2x = 23 4y β x = 16 7. 4y + 3x = 1 5y β 2x = β16 9. 7y β 3x = β11 2y + x = β5 2. 5y + 3x = β5 y β x = β1 4. 3y β 2x = β4 2y + 3x = β7 6. 7y β 2x = β29 2y + 5x = 14 8. 6y β 5x = 9 5y + 3x = 29 10. 8y + 5x = β14 5y β 3x = β21 tion Method tion Method actice Using the Elimina PrPrPrPrPractice Using the Elimina actice Using the Elimina tion Method actice Using the Elimination Method tion Method actice Using the Elimina In this example, youβve got to subtract the equations to eliminate a variable, rather than add them. Example Example Example Example Example 22222 Solve this system of equations: 5a + 3b = 19 3a + 2b = 12 Solution Solution Solution Solution Solution Step 1: Make the coefficients |
of one variable the same in both equations. To make the coefficients of variable b the same, multiply the first equation by 2 and the second equation by 3. 2(5a + 3b = 19) ο¬ 10a + 6b = 38 3(3a + 2b = 12) ο¬ 9a + 6b = 36 Now you have two equations that have the same coefficient of b. Section 5.2 Section 5.2 Section 5.2 β The Elimination Method Section 5.2 Section 5.2 245245245245245 Example 2 continueduedueduedued Example 2 contin Example 2 contin Example 2 contin Example 2 contin Step 2: Add or subtract the two equations to eliminate a variable. In this case, the coefficients of b are both positive, so subtracting one equation from the other will eliminate b. 10a + 6b = 38 β 9a + 6b = 36 a = 2 Step 3: Now choose either of the original equations and substitute in the value you found. Substitute 2 for a, and solve for b. 3a + 2b = 12 3(2) + 2b = 12 6 + 2b = 12 2b = 6 b = 3 Check it out: Itβs always a good idea to write both parts of the solution together clearly. Therefore a = 2, b = 3 is the solution to the system of equations. Check your solution in the other equation: 5a + 3b = 19 5(2) + 3(3) = 19 10 + 9 = 19 19 = 19 β True statement You have a few choices to make when youβre using the elimination method β such as which variable to eliminate, which equation to subtract from the other, and which of the original equations to substitute your result into. Always try to make things easy for yourself by keeping the algebra as simple as possible. Example Example Example Example Example 33333 Solve: 3y + 6x = β6 4y β 2x = 7 Solution Solution Solution Solution Solution Step 1: Make the x-coefficients the same size by multiplying the second equation by 3. 3(4y β 2x = 7) 12y β 6x = 21 Now you have two equations with x-coefficients of 6 and β6: 3y + 6x = β6 and 12y β 6x = 21. 246246246246246 Section 5.2 Section 5.2 Section 5.2 β The Elimination Method Section 5.2 |
Section 5.2 Example 3 continueduedueduedued Example 3 contin Example 3 contin Example 3 contin Example 3 contin Step 2: Add these equations together to eliminate x. 3y + 6x = β6 + 12y β 6x = 21 15y = 15 ο¬ y = 1 Step 2: Substitute 1 for y in one of the original equations and solve for x. 4y β 2x = 7 4(1) β 2x = 7 4 β 2x = 7 β2x = 3 x = β1.5 So x = β1.5, y = 1 is the solution to the system of equations. Check the solution in the other equation: 3y + 6x = β6 3(1) + 6(β1.5) = β6 3 β 9 = β6 β True Example Example Example Example Example 44444 Solve this system of equations: β a = b and 15 Solution Solution Solution Solution Solution This example has fractional coefficients. To make the equations easier to solve, first multiply each equation by the LCM of the denominators to convert the fractional coefficients to integers. Check it out: LCM of 8, 2, and 4 is 8. LCM of 3, 5, and 15 is 15. 8 β βββ β 5 8 a 1 bβ 2 = 1 4 β βββ ο¬ 5a β 4b = 2 β 15 β βββ β 2 3 a 3 bβ = 5 2 15 β βββ ο¬ 10a β 9b = 2 β Now you can solve these equations using the same method as the previous examples: Step 1: Make the a-coefficients the same by multiplying the first equation by 2. 2(5a β 4b = 2) 10a β 8b = 4 Now youβve got two equations that have the same a-coefficient: 10a β 8b = 4 and 10a β 9b = 2. Step 2: Subtract one equation from the other to eliminate a. 10a β 8b = 4 β 10a β 9b = 2 b = 2 Section 5.2 Section 5.2 Section 5.2 β The Elimination Method Section 5.2 Section 5.2 247247247247247 Check it |
out: Subtracting the second equation from the first means that there are no negative terms in the resulting equation. Check it out: You could also substitute into one of the equations in which the fractional coefficients have been converted to integers. Example 4 continueduedueduedued Example 4 contin Example 4 contin Example 4 contin Example 4 contin Step 3: Substitute 2 for b in one of the original equations and solve for a 5a = 10 ο¬ a = 2 So a = 2, b = 2 is the solution to the system of equations. Check the solution in the other equation: 2 3 a 3 bβ = 5 2 15 4 β β = 3 6 5 2 15 β 20 15 18 β = 15 2 15 β True Guided Practice Solve the following systems of equations by elimination. 11. 2a β 5b = 3 4a β 4b = β12 13. 2y = βx β 11 2y β x = β5 15. 7a + 2b = 10 3a + b = 1 12. 4y β 5x = 7 2y β 3x = β1 14. 7y β 2x = 3 6y β 5x = β 4 16. 5a β 2b = 3 3a β 5b = 17 17 15 18 Independent Practice Solve the following systems of equations by elimination. 1. 10y β 2x = 3 6y + 4x = 7 2. 3m β 4c = 1 6m β 6c = 5 3. 0.4m β 0.9c = β0.1 0.3m + 0.2c = 0.8 7. 5y β 4x = β6.1 8y + 5x = β6.625 9. 0.7y β 0.4x = β4.7 0.9y + 0.5x = β3 a a b + = 4. y β 0.4x = β1.8 y + 0.2x = β0.6 8. 0.7y β 0.3x = 2.7 0.4y + 0.3x = 0.6 10. 0.3c + 0.1d = 0.2 0.4c + 0.6d = β0.2 11. Find x and y, by first finding a and c using the following systems of equations: 0.3a β 0.7b = β0.5 and |