We are given the following system of equations:
[tex]\begin{gathered} 6x-4y=-8,(1) \\ y=-6x+2,(2) \end{gathered}[/tex]To solve this system by substitution we will replace the value of "y" from equation (2) in equation (1)
[tex]6x-4(-6x+2)=-8[/tex]Now we use the distributive property:
[tex]6x+24x-8=-8[/tex]Now we add like terms:
[tex]30x-8=-8[/tex]Now we add 8 to both sides:
[tex]30x-8+8=-8+8[/tex]Solving the operations:
[tex]30x=0[/tex]Dividing by 30:
[tex]x=\frac{0}{30}=0[/tex]Therefore x = 0. Now we replace the value of "x" in equation (2):
[tex]\begin{gathered} y=-6x+2 \\ y=-6(0)+2 \\ y=2 \end{gathered}[/tex]Therefore, the solution of the system is:
[tex](x,y)=(0,2)[/tex]