#### 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)
.$