The system of equations is:
[tex]\begin{gathered} x=-7y+32 \\ 2x-2y=16 \end{gathered}[/tex]To solve by substitution, first we need to isolate one of the variables. However, this is already done in the first equation.
Next, we want to substitute x into the other equation, so let's do that:
[tex]2(-7y+32)-2y=16[/tex]Notice that we divide both sides by two to simplify our work:
[tex]\begin{gathered} \frac{2(-7y+32)-2y}{2}=\frac{16}{2} \\ \frac{2(-7y+32)}{2}-\frac{2y}{2}=\frac{16}{2} \\ -7y+32-y=8 \end{gathered}[/tex]Now we isolate y:
[tex]\begin{gathered} -7y+32-y=8 \\ -8y+32=8 \\ -8y=8-32 \\ -8y=-24 \\ y=\frac{-24}{-8}=3 \end{gathered}[/tex]Now that we have y, we substitute in either equations to find x. Let's do it in the first one:
[tex]\begin{gathered} x=-7y+32 \\ x=-7\cdot3+32=-21+32=11 \end{gathered}[/tex]So, y = 3 and x = 11.
