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: 48


The solution of the system of equations is $\left( x,y \right)=\left( 8,-1 \right)$.

Work Step by Step

We have to simplify the system of equations: $\begin{align} & \frac{x-y}{3}=\frac{x+y}{2}-\frac{1}{2} \\ & 2x-2y=3x+3y-3 \end{align}$ That is, $x+5y=3$ (I) And, $\begin{align} & \frac{x+2}{2}-4=\frac{y+4}{3} \\ & 3x+6-24=2y+8 \end{align}$ And, $3x-2y=26$ (II) And multiply equation (I) by −3 to obtain: $-3x-15y=-9$ (III) And add equations (II) and (III): $\begin{align} & 3x-2y-3x-15y=26-9 \\ & -17y=17 \\ & y=-1 \end{align}$ Put the value $y=-1$ in equation (I): $\begin{align} & x-5=3 \\ & x=8 \end{align}$
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