1. Slide 7 - 1 Copyright © 2009 Pearson Education, Inc. 7.2 Solving Systems of Equations by the Substitution and Addition Methods

2. Slide 7 - 2 Copyright © 2009 Pearson Education, Inc. Procedure for Solving a System of Equations Using the Substitution Method Solve one of the equations for one of the variables. If possible, solve for a variable with a coefficient of 1. Substitute the expression found in step 1 into the other equation. Solve the equation found in step 2 for the variable. Substitute the value found in step 3 into the equation, rewritten in step 1, and solve for the remaining variable.

3. Slide 7 - 3 Copyright © 2009 Pearson Education, Inc. Example: Substitution Method Solve the following system of equations by substitution. 5x  y = 5 4x  y = 3 Solve the first equation for y. 5x  y = 5 y = 5x  5 Then we substitute 5x  5 for y in the other equation to give an equation in one variable. 4x  y = 3 4x  (5x  5) = 3

4. Slide 7 - 4 Copyright © 2009 Pearson Education, Inc. Example: Substitution Method continued Now solve the equation for x. 4x – 5x + 5 = 3 –x + 5 = 3 –x = –2 x = 2 Substitute x = 2 in the equation solved for y and determine the y value. y = 5x  5 y = 5(2) – 5 y = 10 – 5 y = 5 Thus, the solution is the ordered pair (2, 5).

5. Slide 7 - 5 Copyright © 2009 Pearson Education, Inc. Practice problem: Solve by substitution y = 2x – 1 3x – y = -1

6. Slide 7 - 6 Copyright © 2009 Pearson Education, Inc. Addition Method If neither of the equations in a system of linear equations has a variable with the coefficient of 1, it is generally easier to solve the system by using the addition (or elimination) method. To use this method, it is necessary to obtain two equations whose sum will be a single equation containing only one variable.

7. Slide 7 - 7 Copyright © 2009 Pearson Education, Inc. Procedure for Solving a System of Equations by the Addition Method 1.If necessary, rewrite the equations so that the variables appear on one side of the equal sign and the constant appears on the other side of the equal sign. 2.If necessary, multiply one or both equations by a constant(s) so that when you add the equations, the result will be an equation containing only one variable. 3.Add the equations to obtain a single equation in one variable.

8. Slide 7 - 8 Copyright © 2009 Pearson Education, Inc. Procedure for Solving a System of Equations by the Addition Method continued 4.Solve the equation in step 3 for the variable. 5.Substitute the value found in step 4 into either of the original equations and solve for the other variable.

9. Slide 7 - 9 Copyright © 2009 Pearson Education, Inc. Example: Multiplying Both Equations Solve the system using the elimination method. 6x + 2y = 4 10x + 7y =  8 Solution: In this system, we cannot eliminate a variable by multiplying only one equation and then adding. To eliminate the variable x, we can multiply the first equation by 5 and the second equation by  3. Then we will be able to eliminate the x variable.

10. Slide 7 - 10 Copyright © 2009 Pearson Education, Inc. Continued, 10x + 7y =  8 6x + 2y = 4 30x + 10y = 20  30x  21y = 24  11y = 44 y =  4 We can now find x by substituting  4 for y in either of the original equations. Substituting: 6x + 2y = 4 6x + 2(  4) = 4 6x  8 = 4 6x = 12 x = 2 The solution is (2,  4).

11. Slide 7 - 11 Copyright © 2009 Pearson Education, Inc. Practice problem: Solve by Addition Method x+ y = 12 x – y = 2

12. Slide 7 - 12 Copyright © 2009 Pearson Education, Inc. Practice problem: Solve by Addition Method x + 3y = 7 2x – y = 7

13. Slide 7 - 13 Copyright © 2009 Pearson Education, Inc. Homework: p. 401 # 3 – 36 (x3) Show all work