To solve a system of equations using the substitution method
We use one equation to find x in terms of y, then
Substitute x in the second equation by the result of the first step
Since the given system is
[tex]\begin{gathered} x-4y=20\rightarrow(1) \\ -2x+2y=-16\rightarrow(2) \end{gathered}[/tex]
We will add equation (1) by 4y (each side) to find x in terms of y
[tex]\begin{gathered} x-4y+4y=20+4y \\ x=20+4y\rightarrow(3) \end{gathered}[/tex]
Now, substitute x in equation (2) by equation (3)
[tex]-2(20+4y)+2y=-16[/tex]
Simplify the left side
[tex]\begin{gathered} -2(20)-2(4y)+2y=-16 \\ -40-8y+2y=-16 \end{gathered}[/tex]
Add the like terms on the left side
[tex]\begin{gathered} -40+(-8y+2y)=-16 \\ -40+(-6y)=-16 \\ -40-6y=-16 \end{gathered}[/tex]
Add 40 to each side
[tex]\begin{gathered} -40+40-6y=-16+40 \\ -6y=24 \end{gathered}[/tex]
Divide both sides by -6 to find y
[tex]\begin{gathered} \frac{-6y}{-6}=\frac{24}{-6} \\ y=-4 \end{gathered}[/tex]
Substitute y in equation (3) by -4
[tex]\begin{gathered} x=20+4(-4) \\ x=20+(-16) \\ x=20-16 \\ x=4 \end{gathered}[/tex]
The answer is (x, y) = (4, -4)
x = 4
y = -4
