Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Review - Page 251: 47

Answer

$\left( \dfrac{7}{3},-\dfrac{8}{3} \right)$

Work Step by Step

Substituting $y=x-5$ in the second equation results to \begin{array}{l}\require{cancel} y=-2x+2 \\\\ x-5=-2x+2 \\\\ x+2x=2+5 \\\\ 3x=7 \\\\ x=\dfrac{7}{3} .\end{array} Substituting $x=\dfrac{7}{3}$ into the first equation results to \begin{array}{l}\require{cancel} y=x-5 \\\\ y=\dfrac{7}{3}-5 \\\\ y=\dfrac{7}{3}-\dfrac{15}{3} \\\\ y=-\dfrac{8}{3} .\end{array} Hence, the solution set is $ \left( \dfrac{7}{3},-\dfrac{8}{3} \right) .$
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