We have the linear system of equations:
[tex]\begin{gathered} 3x-8y=-50 \\ 6y=x \end{gathered}[/tex]and we have to solve it by substitution.
We can replace the value of x, that is 6y, in the first equation and solve for y:
[tex]\begin{gathered} 3x-8y=-50 \\ 3(6y)-8y=-50 \\ 18y-8y=-50 \\ 10y=-50 \\ y=\frac{-50}{10} \\ y=-5 \end{gathered}[/tex]Now we can calculate x as:
[tex]x=6y=6(-5)=-30[/tex]Answer: x=-30 and y=-5