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