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: 29

Answer

The required value is $x=\frac{7}{25},y=-\frac{1}{25}$

Work Step by Step

We know that the system can be rewritten as: $\begin{align} & 3x-4y=1 \\ & 4x+3y=1 \end{align}$ And multiply the first equation by 3 and the second equation by $4$ and add them: $\begin{align} & 3\left( 3x-4y \right)+4\left( 4x+3y \right)=3\left( 1 \right)+4\left( 1 \right) \\ & 9x-12y+16x-12y=3+4 \\ & 25x=7 \\ & x=\frac{7}{25} \end{align}$ Substitute the value of $x=\frac{7}{25}$ in the first equation: $\begin{align} & 3\left( \frac{7}{25} \right)-4y=1 \\ & \frac{21}{25}-4y=1 \\ & 4y=-\frac{4}{25} \\ & y=-\frac{1}{25} \end{align}$
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