Given a system of equation:
[tex]\begin{bmatrix}8x+7y=39 \\ 4x-14y=-68\end{bmatrix}[/tex]Step 1: Isolate the equation 1 for x
[tex]\begin{gathered} 8x+7y=39 \\ 8x=39-7y \\ x=\frac{39-7y}{8} \end{gathered}[/tex]Step 2: Substitute the value of x in equation 2
[tex]\begin{gathered} 4x-14y=-68 \\ 4(\frac{39-7y}{8})-14y=-68 \\ \frac{39-7y}{2}-14y=-68 \\ \text{ multiply through by 2} \\ 39-7y-28y=-126 \\ 39-35y=-126 \\ -35y=-136-39 \\ -35y=-175 \\ y=-\frac{175}{-35} \\ y=5 \end{gathered}[/tex]Step 3: Substitute the value of y
[tex]\begin{gathered} \mathrm{Substitute\: }y=5 \\ x=\frac{39-7y}{8} \\ x=\frac{39-7(5)}{8} \\ x=\frac{39-35}{8} \\ x=\frac{4}{8} \\ x=\frac{1}{2} \end{gathered}[/tex]Hence the correct answer for the system of equations are
[tex]x=\frac{1}{2},y=5[/tex]