Answer
$\{(4,3)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
\dfrac{1}{2}x+\dfrac{1}{3}y=3\\
\dfrac{1}{4}x-\dfrac{2}{3}y=-1
\end{cases}$
Use the elimination method. Multiply the first equation by $2$ and add it to the second equation to eliminate $y$ and determine $x$:
$\begin{cases}
2\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)=2(3)\\
\dfrac{1}{4}x-\dfrac{2}{3}y=-1
\end{cases}$
$\begin{cases}
x+\dfrac{2}{3}y=6\\
\dfrac{1}{4}x-\dfrac{2}{3}y=-1
\end{cases}$
$x+\dfrac{2}{3}y+\dfrac{1}{4}x-\dfrac{2}{3}y=6-1$
$\dfrac{5}{4}x=5$
$x=4$
Determine $y$ using the first equation:
$\dfrac{1}{2}x+\dfrac{1}{3}y=3$
$\dfrac{1}{2}(4)+\dfrac{1}{3}y=3$
$2+\dfrac{1}{3}y=3$
$\dfrac{1}{3}y=3-2$
$\dfrac{1}{3}y=1$
$y=3$
The solution set of the system is:
$\left\{\left(4,3\right)\right\}$
