## Algebra 1

Published by Prentice Hall

# Chapter 6 - Systems of Equations and Inequalities - 6-3 Solving Systems Using Elimination - Practice and Problem-Solving Exercises - Page 379: 42

(5,0)

#### Work Step by Step

Our two equations are: 1. 2x=5(2-y) 2. y=3(-x+5) First lets simplify by distributing the 5 in the first equation and the 3 in the second equation. Our two equations look this: 1. 2x=10-5y 2. y=-3x+15 For the first equation lets bring all of the x and y terms on one side of the equation. 2x=10-5y (add 5y on both sides) 2x+5y=10 Now do the same for the second equation. y=-3x+15 (add 3x to both sides) 3x+y=15 Now our two equations look like this: 1. 2x+5y=10 2. 3x+y=15 Now lets multiply the bottom (aka. the second) equation by 5. 1. 2x+5y=10 5(3x+y=15) 2. 15x+5y=75 Now subtract the second equation from the first equation in order to eliminate the y terms -13x=-65 Solve for x by dividing both sides by -13. You get: x=5 Now that we know the value of x, substitute into our simplified first equation to find the value of y. 2(5)+5y=10 10+5y=10 (subtract 10 from both sides) 5y=0 (divide both sides by 5) y=0 So our ordered pair is (5, 0)

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