Answer
$x=\dfrac{3}{2}$
$y=3$
Work Step by Step
$\begin{cases}\dfrac{x}{3}+\dfrac{y}{6}=1\\\dfrac{x}{2}-\dfrac{y}{4}=0\end{cases}$
Multiply the first equation by $6$ and the second equation by $4$ to remove the fractions:
$\begin{cases}2x+y=6\\2x-y=0\end{cases}$
Add the two equations to remove the $y$:
$4x=6$
Solve for $x$:
$x=\dfrac{6}{4}=\dfrac{3}{2}$
Substitute $x=\dfrac{3}{2}$ in the first equation:
$2\Big(\dfrac{3}{2}\Big)+y=6$
Solve for $y$:
$3+y=6$
$y=6-3=3$
Our two solutions are:
$x=\dfrac{3}{2}$
$y=3$
