Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 818: 26


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

Work Step by Step

We have to multiply the first equation with 2 and then subtract this from the second equation: $\begin{align} & 6x+5y-2\left( 3x-7y \right)=7-2\left( 13 \right) \\ & 6x+5y-6x+14y=7-26 \\ & 19y=-19 \\ & y=-1 \end{align}$ Substitute the value of $y=-1$ in the first equation to obtain, $\begin{align} & 3x-7\left( -1 \right)=13 \\ & 3x+7=13 \\ & 3x=6 \\ & x=2 \end{align}$ Thus, the value of $x=2$ and $y=-1$.
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