Solve Using a Matrix by Elimination 4x+8y=20 -4x+2y=-30

$4 x + 8 y = 20$ $- 4 x + 2 y = - 30$

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & 8 & 20 \\ - 4 & 2 & - 30 \end{array}]$

Find the reduced row echelon form of the matrix.

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

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

$[\begin{array}{c} \frac{1}{4} R_{1} & \frac{1}{4} R_{1} & \frac{1}{4} R_{1} \\ - 4 & 2 & - 30 \end{array}]$

$R_{1} = \frac{1}{4} R_{1}$

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

$[\begin{array}{c} (\frac{1}{4}) \cdot (4) & (\frac{1}{4}) \cdot (8) & (\frac{1}{4}) \cdot (20) \\ - 4 & 2 & - 30 \end{array}]$

$R_{1} = \frac{1}{4} R_{1}$

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

$[\begin{array}{c} 1 & 2 & 5 \\ - 4 & 2 & - 30 \end{array}]$

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 $0$ .

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

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

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

$[\begin{array}{c} 1 & 2 & 5 \\ (4) \cdot (1) - 4 & (4) \cdot (2) + 2 & (4) \cdot (5) - 30 \end{array}]$

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

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

$[\begin{array}{c} 1 & 2 & 5 \\ 0 & 10 & - 10 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{10} R_{2}$ 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 $R_{2} = \frac{1}{10} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 2 & 5 \\ \frac{1}{10} R_{2} & \frac{1}{10} R_{2} & \frac{1}{10} R_{2} \end{array}]$

$R_{2} = \frac{1}{10} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{10} R_{2}$ .

$[\begin{array}{c} 1 & 2 & 5 \\ (\frac{1}{10}) \cdot (0) & (\frac{1}{10}) \cdot (10) & (\frac{1}{10}) \cdot (- 10) \end{array}]$

$R_{2} = \frac{1}{10} R_{2}$

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

$[\begin{array}{c} 1 & 2 & 5 \\ 0 & 1 & - 1 \end{array}]$

Perform the row operation $R_{1} = - 2 \cdot R_{2} + R_{1}$ 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 $R_{1} = - 2 \cdot R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

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

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

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

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

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

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

$[\begin{array}{c} 1 & 0 & 7 \\ 0 & 1 & - 1 \end{array}]$

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

$x = 7$

$y = - 1$

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

$(7, - 1)$

Solve Using a Matrix by Elimination 4x+8y=20 -4x+2y=-30

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