1. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 2 Systems of Linear Equations and Inequalities Chapter 4

3. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 3 4.2 Solving Systems of Linear Equations by Substitution

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 4 4.2 Solving Systems of Linear Equations by Substitution Objectives 1.Solve linear systems by substitution. 2.Solve special systems by substitution. 3.Solve linear systems with fractions and decimals by substitution.

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 5 Example 1 Solve the system by the substitution method. Solving Linear Systems by Substitution 5x + 2y = 2 y = – 3x 5x + 2y = 2 5x + 2(– 3x) = 2 5x + – 6x = 2 – 1x = 2 x = – 2 Let y = – 3x. Multiply. Combine like terms. Multiply by – 1. The second equation is already solved for y. This equation says that y = – 3x. Substituting – 3x for y in the first equation gives 4.2 Solving Systems of Linear Equations by Substitution

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 6 Solving Linear Systems by Substitution y = – 3x = – 3(– 2) = 6 Check that the solution of the given system is (– 2, 6) by substituting – 2 for x and 6 for y in both equations. 5x + 2y = 2 Let x = – 2. y = – 3x Example 1 (cont.) Because x = – 2, we find y from the equation y = – 3x by substituting – 2 for x. 5(– 2) + 2(6) = 2 2 = 2 6 = – 3(– 2) 6 = 6 The solution set of the system is {(– 2, 6)}. ?? True 4.2 Solving Systems of Linear Equations by Substitution

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 7 Example 2 Solve the system by the substitution method. Solving Linear Systems by Substitution 3x + 4y = 4 x = 2y + 18 3x + 4y = 4 3(2y + 18) + 4y = 4 6y + 54 + 4y = 4 10y + 54 = 4 Let x = 2y + 18. Distributive property Combine like terms. The second equation gives x in terms of y. Substitute 2y + 18 for x in the first equation. 10y = – 50Subtract 54. y = – 5Divide by 10. 4.2 Solving Systems of Linear Equations by Substitution

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 8 Solving Linear Systems by Substitution x = 2y + 18 = 2(– 5) + 18 = 8 Check that the solution of the given system is (8, – 5) by substituting 8 for x and – 5 for y in both equations.. 3x + 4y = 4 Let y = – 5. x = 2y + 18 Example 2 (continued) Because y = – 5, we find x from the equation x = 2y + 18 by substituting – 5 for y. 3(8) + 4(– 5) = 4 4 = 4 8 = 2(– 5) + 18 8 = 8 The solution set of the system is {(8, – 5)}. ?? True 4.2 Solving Systems of Linear Equations by Substitution

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 9 Solving a Linear System by Substitution Step 1 Solve one equation for either variable. If one of the variables has a coefficient of 1 or – 1, choose it, since it usually makes the substitution easier. Step 2 Substitute for that variable in the other equation. The result should be an equation with just one variable. Step 3 Solve the equation from Step 2. Step 4 Substitute the result from Step 3 into the equation from Step 1 to find the value of the other variable. Step 5 Check the solution in both of the original equations. Then write the solution set. Solving Linear Systems by Substitution 4.2 Solving Systems of Linear Equations by Substitution

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 10 Example 3 Use substitution to solve the system. Solving Linear Systems by Substitution 3x + 2y = 26 5x – y = 13 (1) (2) Step 1 For the substitution method, we must solve one of the equations for either x or y. Because the coefficient of y in equation (2) is – 1, we choose equation (2) and solve for y. 5x – y – 5x = 13 – 5xSubtract 5x. – y = 13 – 5xCombine like terms. y = – 13 + 5xMultiply by – 1. 4.2 Solving Systems of Linear Equations by Substitution

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 11 Example 3 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution 3x + 2y = 26 5x – y = 13 3x + 2y = 26 (1) (2) (1) Step 2 Now substitute – 13 + 5x for y in equation (1). 3x + 2(– 13 + 5x) = 26Let y = – 13 + 5x. 4.2 Solving Systems of Linear Equations by Substitution

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 12 Example 3 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution 3x + 2y = 26 5x – y = 13 (1) (2) Step 3 Now solve the equation from Step 2. 3x + 2(– 13 + 5x) = 26From Step 2. 3x – 26 + 10x = 26Distributive property 13x – 26 = 26Combine like terms. 13x – 26 + 26 = 26 + 26Add 26. 13x = 52Combine like terms. x = 4Divide by 13. 4.2 Solving Systems of Linear Equations by Substitution

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 13 Example 3 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution 3x + 2y = 26 5x – y = 13 (1) (2) Step 4 Since y = – 13 + 5x and x = 4, y = – 13 + 5(4) = 7. Step 5 Check that (4, 7) is the solution. 3x + 2y = 265x – y = 13 3(4) + 2(7) = 26 12 + 14 = 26 26 = 26 5(4) – 7 = 13 20 – 7 = 13 13 = 13 Since both results are true, the solution set of the system is {(4, 7)}. ?? True ?? 4.2 Solving Systems of Linear Equations by Substitution

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 14 Example 4 Use substitution to solve the system. Solving Linear Systems by Substitution 3x – 2y = 0 5x – 4y = 6 (1) (2) Step 1 To use the substitution method, we must solve one of the equations for one of the variables. We choose equation (2) and solve for x. 5x – 4y + 4y = 6 + 4yAdd 4y. 5x = 6 + 4yCombine like terms. Divide by 5. x = + y 6 5 4 5 4.2 Solving Systems of Linear Equations by Substitution

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 15 Example 4 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution 3x – 2y = 0 (1) (2) (1) 3x – 2y = 0 5x – 4y = 6 Step 2 Now substitute for x in equation (1). + y 6 5 4 5 3 – 2y = 0 + y 6 5 4 5 Let x =. + y 6 5 4 5 4.2 Solving Systems of Linear Equations by Substitution

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 16 Example 4 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution (1) (2) 3x – 2y = 0 5x – 4y = 6 Step 3 Now solve the equation from Step 2. 3 – 2y = 0 + y 6 5 4 5 – 2y = 0 + y 18 5 12 5 Distributive property Combine like terms. = 0 + y 18 5 2 5 From Step 2. 4.2 Solving Systems of Linear Equations by Substitution

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 17 Example 4 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution (1) (2) 3x – 2y = 0 5x – 4y = 6 Subtract. 18 5 = 0 – + y – 18 5 2 5 5 5 Combine like terms. y 2 5 = – 18 5 From previous slide. = 0 + y 18 5 2 5 Step 3 (continued) 4.2 Solving Systems of Linear Equations by Substitution

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 18 Example 4 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution (1) (2) 3x – 2y = 0 5x – 4y = 6 From previous slide. y 2 5 = – 18 5 y = – 9 Multiply by. 5 2 Step 3 (continued) 4.2 Solving Systems of Linear Equations by Substitution

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 19 Example 4 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution (1) (2) 3x – 2y = 0 5x – 4y = 6 Step 4 Find x by substituting – 9 for y in + y. 6 5 4 5 x = + (– 9) = – 6 6 5 4 5 4.2 Solving Systems of Linear Equations by Substitution

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 20 Example 4 (continued) Use substitution to solve the system. Solving Linear Systems by Substitution (1) (2) 3x – 2y = 0 5x – 4y = 6 Step 5 Check that (– 6, – 9) is the solution. 3x – 2y = 05x – 4y = 6 3(– 6) – 2(– 9) = 0 (– 18) – (– 18) = 0 – 18 + 18 = 0 0 = 0 5(– 6) – 4(– 9) = 6 (– 30) – (– 36) = 6 (– 30) + 36 = 6 6 = 6 Both results are true, so the solution set of the system is {(– 6, – 9)}. 4.2 Solving Systems of Linear Equations by Substitution

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 21 Example 5 Use substitution to solve the system. Solving Special Systems by Substitution y = 2x – 7 3y – 6x = – 6 3y – 6x = 6 (1) (2) Substitute 2x – 7 for y in equation (2). 3(2x – 7) – 6x = – 6Let y = 2x – 7. 6x – 21 – 6x = – 6Distributive property – 21 = – 6False. This false result means that the equations in the system have graphs that are parallel lines. The system is inconsistent and the solution set is 4.2 Solving Systems of Linear Equations by Substitution

22. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 22 x y Solving Special Systems by Substitution y = 2x – 7 3y – 6x = – 6. (1) (2) (1) 4.2 Solving Systems of Linear Equations by Substitution Example 5 (continued)

23. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 23 Example 6 Solve the system by the substitution method. Solving Special Systems by Substitution 3 = 5x – y – 4y – 12 = – 20x (1) (2) Begin by solving equation (1) for y to get y = 5x – 3. Substitute 5x – 3 for y in equation (2) and solve the resulting equation. – 4(5x – 3) – 12 = – 20xLet y = 5x – 3. – 20x + 12 – 12 = – 20xDistributive property 0 = 0Add 20x; combine terms. This true result means that every solution of one equation is also a solution of the other, so the system has an infinite number of solutions. 4.2 Solving Systems of Linear Equations by Substitution

24. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 24 x y Example 6 (continued) (1) (2) 3 = 5x – y – 4y – 12 = – 20x The system has an infinite number of solutions – all the ordered pairs corresponding to points that lie on the common graph. 4.2 Solving Systems of Linear Equations by Substitution Solving Special Systems by Substitution

25. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 25 Example 7 Solve the system by the substitution method. Solving Linear Systems with Fractions and Decimals (1) (2) Clear equation (1) of fractions by multiplying each side by 3. x + 5y = 1 2 3 x – y = – 1 4 1 8 13 8 Multiply by 4. x + 5y = 1 2 3 33 2 3 333 Distributive property 2x + 15y = 3(3) 4.2 Solving Systems of Linear Equations by Substitution

26. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 26 Example 7 (cont.) Solve the system by the substitution method. Solving Linear Systems with Fractions and Decimals (1) (2) Now, clear equation (2) of fractions by multiplying each side by 8. x + 5y = 1 2 3 x – y = – 1 4 1 8 13 8 Multiply by 8. Distributive property 2x – y = – 13(4) x – y = – 1 4 1 8 13 8 88 x – y = – 1 4 1 8 13 8 888 4.2 Solving Systems of Linear Equations by Substitution

27. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 27 Example 7 (cont.) The given system of equations has been simplified to the equivalent system. Solving Linear Systems with Fractions and Decimals To solve this system by substitution, equation (4) can be solved for y. (1) (2) x + 5y = 1 2 3 x – y = – 1 4 1 8 13 8 Subtract 2x. 2x – y = – 13 2x + 15y = 3(3) (4) 2x – y = – 13(4) 2x – y – 2x = – 13 – 2x Combine like terms.– y = – 13 – 2x Multiply by – 1.y = 13 + 2x 4.2 Solving Systems of Linear Equations by Substitution

28. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 28 Example 7 (cont.) The given system of equations has been simplified to the equivalent system. Solving Linear Systems with Fractions and Decimals Now substitute 13 + 2x for y in equation (3). (1) (2) x + 5y = 1 2 3 x – y = – 1 4 1 8 13 8 Let y = 13 + 2x. 2x – y = – 13 2x + 15y = 3(3) (4) 2x + 15y = 3(3) 2x + 15(13 + 2x) = 3 Distributive property2x + 195 + 30x = 3 Combine like terms.32x + 195 = 3 4.2 Solving Systems of Linear Equations by Substitution

29. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 29 Example 7 (cont.) The given system of equations has been simplified to the equivalent system. Solving Linear Systems with Fractions and Decimals Now substitute 13 + 2x for y in equation (3). (1) (2) x + 5y = 1 2 3 x – y = – 1 4 1 8 13 8 2x – y = – 13 2x + 15y = 3(3) (4) Combine like terms.32x + 195 = 3 Subtract 195.32x + 195 – 195 = 3 – 195 Combine like terms.32x = – 192 Divide by 32.x = – 6 4.2 Solving Systems of Linear Equations by Substitution

30. Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.2 – Slide 30 Example 7 (cont.) The given system of equations has been simplified to the equivalent system. Solving Linear Systems with Fractions and Decimals Substitute – 6 for x in y = 13 + 2x to get (1) (2) x + 5y = 1 2 3 x – y = – 1 4 1 8 13 8 2x – y = – 13 2x + 15y = 3(3) (4) Check by substituting – 6 for x and 1 for y in both of the original equations. The solution set is {(– 6, 1)}. y = 13 + 2(– 6) = 1. 4.2 Solving Systems of Linear Equations by Substitution