Solve Using a Matrix by Elimination x-5y=-5 -4x-2y=20

$x - 5 y = - 5$ $- 4 x - 2 y = 20$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & - 5 & - 5 \\ - 4 & - 2 & 20 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{2} = 4 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = 4 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - 5 & - 5 \\ 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = 4 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = 4 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - 5 & - 5 \\ (4) \cdot (1) - 4 & (4) \cdot (- 5) - 2 & (4) \cdot (- 5) + 20 \end{array}]$

$R_{2} = 4 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - 5 & - 5 \\ 0 & - 22 & 0 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{22} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - \frac{1}{22} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - 5 & - 5 \\ - \frac{1}{22} R_{2} & - \frac{1}{22} R_{2} & - \frac{1}{22} R_{2} \end{array}]$

$R_{2} = - \frac{1}{22} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{22} R_{2}$ .

$[\begin{array}{c} 1 & - 5 & - 5 \\ (- \frac{1}{22}) \cdot (0) & (- \frac{1}{22}) \cdot (- 22) & (- \frac{1}{22}) \cdot (0) \end{array}]$

$R_{2} = - \frac{1}{22} R_{2}$

Simplify $R_{2}$ (row $2$ ).

$[\begin{array}{c} 1 & - 5 & - 5 \\ 0 & 1 & 0 \end{array}]$

Perform the row operation $R_{1} = 5 \cdot R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = 5 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 5 \cdot R_{2} + R_{1} & 5 \cdot R_{2} + R_{1} & 5 \cdot R_{2} + R_{1} \\ 0 & 1 & 0 \end{array}]$

$R_{1} = 5 \cdot R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = 5 \cdot R_{2} + R_{1}$ .

$[\begin{array}{c} (5) \cdot (0) + 1 & (5) \cdot (1) - 5 & (5) \cdot (0) - 5 \\ 0 & 1 & 0 \end{array}]$

$R_{1} = 5 \cdot R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

$[\begin{array}{c} 1 & 0 & - 5 \\ 0 & 1 & 0 \end{array}]$

Step 3

Use the result matrix to declare the final solutions to the system of equations.

$x = - 5$

$y = 0$

Step 4

The solution is the set of ordered pairs that makes the system true.

$(- 5, 0)$

Solve Using a Matrix by Elimination x-5y=-5 -4x-2y=20

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