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: 23

Answer

The required value is $x=-4,y=3$

Work Step by Step

We have to multiply the first equation with four and then add this to the second equation. $\begin{align} & 4\left( x+2y \right)+\left( -4x+3y \right)=4\left( 2 \right)+25 \\ & 4x+8y-4x+3y=8+25 \\ & 11y=33 \\ & y=3 \end{align}$ Substitute the value of $y=3$ in the first equation and solve for x as follows: $\begin{align} & x+2\left( 3 \right)=2 \\ & x+6=2 \\ & x=2-6 \\ & x=-4 \end{align}$ Thus, the value of $x=-4$ and $y=3$.
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