## College Algebra (6th Edition)

Published by Pearson

# Chapter 1 - Equations and Inequalities - Exercise Set 1.2 - Page 119: 87

#### Answer

$-7$

#### Work Step by Step

Solving for $x$ $\frac{3(x+3)}{5}= 2x+6$ Multiplying both sides by $5$ $5(\frac{3(x+3)}{5})=5( 2x+6)$ $3x+9= 10x+30$ $3x-10x=30-9$ $-7x= 21$ $x=-3$ Solving for $y$ $-2y-10= 5y+18$ $-2y- 5y=10+18$ $-7y=28$ $y=-4$ Substituting $x$ and $y$ values in $x^{2}-(xy-y) = (-3)^{2}-((-3 \times -4)-(-4))$ $= 9-(12+4)$ $= 9-16$ $=-7$

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