1. © 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 7 Algebra: Graphs, Functions, and Linear Systems

2. © 2010 Pearson Prentice Hall. All rights reserved. 2 7.3 Systems of Linear Equations in Two Variables

3. © 2010 Pearson Prentice Hall. All rights reserved. Objectives 1.Decide whether an ordered pair is a solution of a linear system. 2.Solve linear systems by graphing. 3.Solve linear systems by substitution. 4.Solve linear systems by addition. 5.Identify systems that do not have exactly one ordered-pair solution. 6.Solve problems using systems of linear equations. 3

4. © 2010 Pearson Prentice Hall. All rights reserved. Systems of Linear Equations & Their Solutions Two linear equations are called a system of linear equations or a linear system. A solution to a system of linear equations in two variables is an ordered pair that satisfies both equations in the system. 4

5. © 2010 Pearson Prentice Hall. All rights reserved. Determine whether (1,2) is a solution of the system: 2x – 3y = −4 2x + y = 4 Example 1: Determining Whether an Ordered Pair is a Solution of a System Solution: Because 1 is the x-coordinate and 2 is the y-coordinate of (1,2), we replace x with 1 and y with 2. 5

6. © 2010 Pearson Prentice Hall. All rights reserved. 2x – 3y = − 4 2x + y = 4 2(1) – 3(2) = −42(1) + 2 = 4 2 – 6 = − 4 2 + 2 = 4 − 4 = − 4, TRUE 4 = 4, TRUE The pair (1,2) satisfies both equations; it makes each equation true. Thus, the pair is a solution of the system. Example 1 continued ? ? ? ? 6

7. © 2010 Pearson Prentice Hall. All rights reserved. Graphing 2 lines 2x – 3y = -4 2x + 4 = 3y y = (2/3)x + (4/3) 2x + y = 4 y = -2x + 4 1

8. © 2010 Pearson Prentice Hall. All rights reserved. Solving Linear Systems by Graphing For a system with one solution, the coordinates of the point of intersection of the lines is the system’s solution. 8

9. © 2010 Pearson Prentice Hall. All rights reserved. Example 2: Solving Linear Systems by Graphing Solve by graphing: x + 2y = 2 x – 2y = 6. Solution: We find the solution by graphing both x + 2y = 2 and x – 2y = 6 in the same rectangular coordinate system. We will use intercepts to graph each equation. 9

10. © 2010 Pearson Prentice Hall. All rights reserved. x-intercept: Set y = 0. x + 2 · 0 = 2 x = 2 The line passes through (2,0). Example 2 Continued y-intercept: Set x = 0. 0 + 2y = 2 2y = 2 y = 1 The line passes through (0,1). We will graph x + 2y = 2 as a blue line. x + 2y = 2: 10

11. © 2010 Pearson Prentice Hall. All rights reserved. x-intercept: Set y = 0. x – 2 · 0 = 6 x = 6 The line passes through (6,0). Example 2 Continued y-intercept: Set x = 0. 0 – 2y = 6 −2y = 6 y = −3 The line passes through (0,−3). We will graph x – 2y = 6 as a red line. x – 2y = 6: 11

12. © 2010 Pearson Prentice Hall. All rights reserved. We see the two graphs intersect at (4,−1). Hence, this is the solution to the system. We can check this by substituting in (4,−1) into each equation and verifying That the solution is true for both equations. We leave this to the student. Example 2 Continued 12

13. © 2010 Pearson Prentice Hall. All rights reserved. Solving Linear Systems by the Substitution Method 13

14. © 2010 Pearson Prentice Hall. All rights reserved. Solve by the substitution method: y = −x – 1 4x – 3y = 24. Solution: Step 1 Solve either of the equations for one variable in terms of the other. This step has been done for us. The first equation, y = −x – 1, is solved for y in terms of x. Example 3: Solving a System by Substitution 14

15. © 2010 Pearson Prentice Hall. All rights reserved. Step 2 Substitute the expression from step 1 into the other equation. This gives us an equation in one variable, namely 4x – 3(−x – 1) = 24. The variable y has been eliminated. Example 3 continued 15

16. © 2010 Pearson Prentice Hall. All rights reserved. Step 3 Solve the resulting equation containing one variable.4x – 3(-x – 1) = 24 4x + 3x + 3 = 24 7x + 3 = 24 7x = 21 x = 3 Step 4 Back-substitute the obtained value into the equation from step1. Since we found x = 3 in step 3, then we back-substitute the x-value into the equation from step 1 to find the y-coordinate. Example 3 continued 16

17. © 2010 Pearson Prentice Hall. All rights reserved. Step 4 (cont.) With x = 3 and y = −4, the proposed solution is (3,−4). Step 5 Check. Use this ordered pair to verify that this solution makes each equation true. We leave this to the student. Example 3 continued 17

18. © 2010 Pearson Prentice Hall. All rights reserved. Your Turn Solve the following system of equations by the substitution method. y = 2x + 7 2x – y = -5 -4x + y = -11 2x – 3y = 3 1

19. © 2010 Pearson Prentice Hall. All rights reserved. Solving Linear Systems by the Addition Method 19

20. © 2010 Pearson Prentice Hall. All rights reserved. Solve by the addition method: 3x + 2y = 48 9x – 8y = −24. Example 5: Solving a System by the Addition Method Solution: Step 1 Rewrite both equations in the form Ax + By = C. Both equations are already in this form. Variable terms appear on the left and constants appear on the right. Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y- coefficients is 0. 20

21. © 2010 Pearson Prentice Hall. All rights reserved. Example 5 continued 3x + 2y = 48 − 9x – 6y = − 144 9x – 8y = −24 9x – 8y = − 24 Step 3 Add the equations. −14y = −168 Multiply by −3 No Change Step 4 Solve the equation in one variable. We solve −14y = −168 by dividing both sides by −14. 21

22. © 2010 Pearson Prentice Hall. All rights reserved. Step 5 Back-substitute and find the value for the other variable. 3x + 2y = 48 3x + 2(12) = 48 3x + 24 = 48 3x = 24 x = 8 Example 5 continued Step 6 Check. The solution to the system is (8,12). We can check this by verifying that the solution is true for both equations. We leave this to the student. 22

23. © 2010 Pearson Prentice Hall. All rights reserved. Linear Systems Having No Solution or Infinitely Many Solutions The number of solutions to a system of two linear equations in two variables is given by one of the following: Number of SolutionsWhat This Means Graphically Exactly one ordered-pair solutionThe two lines intersect at one point. No SolutionThe two lines are parallel. Infinitely many solutionsThe two lines are identical. 23

24. © 2010 Pearson Prentice Hall. All rights reserved. Solve the system: 4x + 6y = 12 6x + 9y = 12. Solution: Because no variable is isolated, we will use the addition method. 4x + 6y = 12 6x + 9y = 12 The false statement 0 = 12 indicates that the system has no solution. The solution is the empty set, Ø. Example 7: A System with no Solution Multiply by 3 Multiply by -2 Add: 24

25. © 2010 Pearson Prentice Hall. All rights reserved. Solve the system: y = 3x – 2 15x – 5y = 10. Solution: Because the variable y is isolated in y = 3x – 2, the first equation, we will use the substitution method. Example 8: A System with Infinitely Many Solutions 25

26. © 2010 Pearson Prentice Hall. All rights reserved. The statement 10 = 10 is true. Hence, this indicates that the system has infinitely many solutions. Example 8 continued 26

27. © 2010 Pearson Prentice Hall. All rights reserved. Modeling with Systems of Equations: Making Money (and Losing It) Revenue and Cost Functions A company produces and sells x units of a product. Revenue Function R(x) = (price per unit sold)x Cost Function C(x) = fixed cost + (cost per unit produced)x The point of intersection of the graphs of the revenue and cost functions is called the break-even point. 27

28. © 2010 Pearson Prentice Hall. All rights reserved. A company is planning to manufacture radically different wheelchairs. Fixed cost will be $500,000 and it will cost $400 to produce each wheelchair. Each wheelchair will be sold for $600. a.Write the cost function, C, of producing x wheelchairs. b.Write the revenue function, R, from the sale of x wheelchairs. c.Determine the break-even point. Describe what this means. Example 9: Finding a Break-Even Point 28

29. © 2010 Pearson Prentice Hall. All rights reserved. Solution: a. The cost function is the sum of the fixed cost and the variable cost. b.The revenue function is the money generated from the sale of x wheelchairs. Example 9 Continued 29

30. © 2010 Pearson Prentice Hall. All rights reserved. c.The break even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system C(x) = 500,000 + 400x R(x) = 600x, Or y = 500,000 + 400x y = 600x. Example 9 continued 30

31. © 2010 Pearson Prentice Hall. All rights reserved. Using substitution, we substitute 600x in for y in the first equation: 600x = 500,000 + 400x 200x = 500,000 x = 2500 Back-substituting 2500 for x in either of the system’s equations (or functions), we obtain Example 9 continued 31

32. © 2010 Pearson Prentice Hall. All rights reserved. The break-even point is (2500, 1,500,000). This means that the company will break even if it produces and sells 2500 wheelchairs for $1,500,000. Example 9 continued 32

33. © 2010 Pearson Prentice Hall. All rights reserved. The Profit Function The profit, P(x), generated after producing and selling x units of a product is given by the profit function P(x) = R(x) – C(x), where R and C are the revenue and cost, respectively. The profit function, P(x), for the previous example is P(x) = R(x) – C(x) = 600x – (500,000 + 400x) = 200x – 500,000. 33