Answer
$(-7,-28)$
Work Step by Step
Solving $AX=B,\ \qquad X=A^{-1}B$.
This problem is connected to exercise $34$, as the coefficient matrix is
$A=\left[\begin{array}{rr}{-4}&{1}\\{6}&{-2}\end{array}\right]$, and $A^{-1}=\displaystyle \left[\begin{array}{ll}{-1}&{-\displaystyle \frac{1}{2}}\\{-3}&{-2}\end{array}\right]$
Writing the system $\left\{\begin{aligned}-4x+y&=0\\6x-2y&=14\end{aligned}\right.$
in matrix form, $AX=B, $ the solution is
$X=A^{-1}B=\left[\begin{array}{rr}
{1}&{-1/2}\\
{-3}&{-2}\end{array}\right]\left[\begin{array}{l}
0\\
14
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
0-7\\
0-28
\end{array}\right]=\left[\begin{array}{l}
-7\\
-28
\end{array}\right]$
Solution: $(x,y)=(-7,-28)$
