[tex]\fbox{Solution by using Matrix}[/tex]
[tex]\text{Linear Equation} = -8x-8y=0 , -8x+2y=-20[/tex]
[tex]\text{Rewrite the linear equations above as a matrix}[/tex]
[tex] \left[\begin{array}{ccc}8&-8&0\\-8&2&-20\\\end{array}\right] [/tex]
[tex]\text{Apply to Row2 : Row2 + Row1}[/tex]
[tex] \left[\begin{array}{ccc}8&-8&0\\-8&-6&-20\\\end{array}\right] [/tex]
[tex]\text{ Simplify rows}[/tex]
[tex] \left[\begin{array}{ccc}8&-8&0\\0&1&10/3\\\end{array}\right] [/tex]
[tex]\text{Note: The matrix is now in echelon form.}[/tex][tex]\text{The steps below are for back substitution.}[/tex]
[tex]\text{Apply to Row1 : Row1 + 8 Row2}[/tex]
[tex] \left[\begin{array}{ccc}8&0&80/30\\0&1&10/3\\\end{array}\right] [/tex]
[tex]\text{ Simplify rows}[/tex]
[tex] \left[\begin{array}{ccc}1&0&10/3\\0&1&10/3\\\end{array}\right] [/tex]
[tex]\text{Therefore, the solution is}[/tex]
[tex]x= \dfrac{10}{3} \ \text{and} \ \ y=\dfrac{10}{3}[/tex]