The second equation already has the value of x isolated, so in order to solve the system by substitution, we can apply this value of x from the second equation in the first one:
[tex]\begin{gathered} x=-5y \\ \\ -3x-6y=45 \\ -3\cdot(-5y)-6y=45 \\ 15y-6y=45 \\ 9y=45 \\ y=\frac{45}{9}=5 \end{gathered}[/tex]Now that we have the value of y, we can find the value of x:
[tex]x=-5y=-5(5)=-25[/tex]So we have that x = -25 and y = 5.
