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Solve the following system by any method -8x-8y=0 -8x+2y=-20

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\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= (10)/(3) \ \text{and} \ \ y=(10)/(3)
User Aryak Sengupta
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