Let:
[tex]\begin{gathered} 5x+4y=-12_{\text{ }}(1) \\ x=-8_{\text{ }}(2) \end{gathered}[/tex]Replace (2) into (1):
[tex]\begin{gathered} 5(-8)+4y=-12 \\ -40+4y=-12 \\ solve_{\text{ }}for_{\text{ }}y\colon \\ 4y=40-12 \\ 4y=28 \\ y=\frac{28}{4} \\ y=7 \end{gathered}[/tex]So:
x = -8
y = 7