Answer
$(2, \frac{1}{2})$
Work Step by Step
Step 1. Based on the given system, we have
$D=\begin{vmatrix} 1 & 2 \\ 3 & -4 \end{vmatrix}=1(-4)-2(3)=-10$
Step 2. We have
$D_x=\begin{vmatrix} 3 & 2 \\ 4 & -4 \end{vmatrix}=3(-4)-2(4)=-20$
Step 3. We have
$D_y=\begin{vmatrix} 1 & 3 \\ 3 & 4 \end{vmatrix}=1(4)-3(3)=-5$
Step 4. Using Cramer's Rule, we have $x=\frac{D_x}{D}=\frac{-20}{-10}=2$ and $y=\frac{D_y}{D}=\frac{-5}{-10}=\frac{1}{2}$. Thus the solution set is $(2, \frac{1}{2})$
