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: 40


The solution set is $\left\{ \left( -1,-1 \right) \right\}$.

Work Step by Step

By rewriting the first equation, we get: $\begin{align} & \frac{x}{6}-\frac{3y}{6}=\frac{2}{6} \\ & x-3y=2 \end{align}$ Therefore, the system can be rewritten as: $\begin{align} & x-3y=2 \\ & x+2y=-3 \end{align}$ And subtract the second equation from the first equation: $\begin{align} & x-3y-x-2y=2-\left( -3 \right) \\ & -5y=2+3 \\ & -5y=5 \\ & y=-1 \end{align}$ Substitute the value in the first equation and solve for x: $\begin{align} & x-3\left( -1 \right)=2 \\ & x+3=2 \\ & x=-1 \end{align}$. Thus, The solution set is $\left\{ \left( -1,-1 \right) \right\}$.
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