## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 818: 27

#### Answer

The required value is $x=1,y=-2$

#### Work Step by Step

We have subtract three times the second equation from two times the first equation to obtain, Substitute the value of $y=-2$ in the first equation and solve for x as: \begin{align} & 3x-4\left( -2 \right)=11 \\ & 3x+8=11 \\ & 3x=3 \\ & x=1 \end{align}

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