Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 11 - Systems of Equations and Inequalities - 11.1 Systems of Linear Equations: Substitution and Elimination - 11.1 Assess Your Understanding - Page 715: 37



Work Step by Step

We are given the system of equations: $\begin{cases} \dfrac{1}{2}x+\dfrac{1}{3}y=3\\ \dfrac{1}{4}x-\dfrac{2}{3}y=-1 \end{cases}$ Use the elimination method. Multiply the first equation by $2$ and add it to the second equation to eliminate $y$ and determine $x$: $\begin{cases} 2\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)=2(3)\\ \dfrac{1}{4}x-\dfrac{2}{3}y=-1 \end{cases}$ $\begin{cases} x+\dfrac{2}{3}y=6\\ \dfrac{1}{4}x-\dfrac{2}{3}y=-1 \end{cases}$ $x+\dfrac{2}{3}y+\dfrac{1}{4}x-\dfrac{2}{3}y=6-1$ $\dfrac{5}{4}x=5$ $x=4$ Determine $y$ using the first equation: $\dfrac{1}{2}x+\dfrac{1}{3}y=3$ $\dfrac{1}{2}(4)+\dfrac{1}{3}y=3$ $2+\dfrac{1}{3}y=3$ $\dfrac{1}{3}y=3-2$ $\dfrac{1}{3}y=1$ $y=3$ The solution set of the system is: $\left\{\left(4,3\right)\right\}$
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