Intermediate Algebra (6th Edition)

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

Chapter 4 - Section 4.1 - Solving Systems of Linear Equations in Two Variables - Exercise Set: 98

Answer

$\left(\dfrac{1}{4}, -\dfrac{1}{3} \right)$

Work Step by Step

In the given system, \begin{cases} \dfrac{2}{x}+\dfrac{3}{y}=-1 \\\\ \dfrac{3}{x}-\dfrac{2}{y}=18 ,\end{cases} multiplying the first equation by $ 2 $ and the second equation by $ 3 $ results to \begin{cases} \dfrac{4}{x}+\dfrac{6}{y}=-2 \\\\ \dfrac{9}{x}-\dfrac{6}{y}=54 .\end{cases} Adding the two equations results to $ \dfrac{13}{x}=52 .$ By cross-multiplication and by the properties of equality, then \begin{array}{l}\require{cancel} 13=52x \\\\ \dfrac{13}{52}=x \\\\ x=\dfrac{1}{4} .\end{array} Substituting $x=\dfrac{1}{4}$ into the first equation, $ \dfrac{2}{x}+\dfrac{3}{y}=-1 ,$ results to \begin{array}{l}\require{cancel} \dfrac{2}{1/4}+\dfrac{3}{y}=-1 \\\\ 8+\dfrac{3}{y}=-1 \\\\ \dfrac{3}{y}=-1-8 \\\\ \dfrac{3}{y}=-9 .\end{array} By cross-multiplication and by the properties of equality, then \begin{array}{l}\require{cancel} -9y=3 \\\\ y=\dfrac{3}{-9} \\\\ y=-\dfrac{1}{3} .\end{array} Hence, the solution to the given system is the point $ \left(\dfrac{1}{4}, -\dfrac{1}{3} \right) .$
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