## Precalculus (10th Edition)

$\left\{\left(4,3\right)\right\}$
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: $\left\{\left(4,3\right)\right\}$