## Intermediate Algebra (6th Edition)

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