Answer
$\left\{\left(4,3\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
\dfrac{4}{x}-\dfrac{3}{y}=0\\
\dfrac{6}{x}+\dfrac{3}{2y}=2
\end{cases}$
Note:
$u=\dfrac{1}{x}$
$v=\dfrac{1}{y}$
Rewrite the system in terms of $u$ and $v$:
$\begin{cases}
4u-3v=0\\
6u+1.5v=2
\end{cases}$
Use the elimination method. Multiply the second equation by 2 and add it to the first equation to eliminate $v$ and determine $u$:
$\begin{cases}
4u-3v=0\\
2(6u+1.5v)=2(2)
\end{cases}$
$\begin{cases}
4u-3v=0\\
12u+3v=4
\end{cases}$
$4u-3v+12u+3v=0+4$
$16u=4$
$u=\dfrac{1}{4}$
Determine $v$ using the first equation:
$4u-3v=0$
$4\left(\dfrac{1}{4}\right)=3v$
$1=3v$
$v=\dfrac{1}{3}$
Determine $u$ and $v$:
$\dfrac{1}{u}=\dfrac{1}{4}\Rightarrow u=4$
$\dfrac{1}{v}=\dfrac{1}{3}\Rightarrow v=3$
The solution set of the system is:
$\left\{\left(4,3\right)\right\}$
