## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by Substitution - Exercise Set: 45

x = 1 y = -3

#### Work Step by Step

simplify to first equation-5y + 6y = 3x + 2 ( x - 5 ) -3x + 5 y = 3x + 2x - 10 - 3x + 5 y = 2x -5 simplify to second equation 4 ( x + y ) -x + y = -12 4x + 4y -x + y = -12 3x + 5y = -12 here substitude y = 2x - 5 3x + 5 ( 2x - 5 ) = -12 3x + 10x -25 = -12 use the distributive property 13x -25 = -12 simplify 13x = -12 + 25 13x = 13 divide both sides by 13 $\frac{13x}{13}$ = $\frac{13}{13}$ x = 1 substitude x =1 to y = 2x - 5 y = $2\times1$ - 5 y = 2 - 5 y = -3

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