Solve Using Matrices by Elimination x plus y equals nine , three x minus y equals seven

$x + y = 9$ , $3 x - y = 7$

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & 9 \\ 3 & - 1 & 7 \end{array}]$

Row reduce.

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Perform the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 1 & 9 \\ - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} & - 3 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & 1 & 9 \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (1) - 1 & (- 3) \cdot (9) + 7 \end{array}]$

$R_{2} = - 3 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & 1 & 9 \\ 0 & - 4 & - 20 \end{array}]$

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

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

$[\begin{array}{c} 1 & 1 & 9 \\ - \frac{1}{4} R_{2} & - \frac{1}{4} R_{2} & - \frac{1}{4} R_{2} \end{array}]$

$R_{2} = - \frac{1}{4} R_{2}$

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

$[\begin{array}{c} 1 & 1 & 9 \\ (- \frac{1}{4}) \cdot (0) & (- \frac{1}{4}) \cdot (- 4) & (- \frac{1}{4}) \cdot (- 20) \end{array}]$

$R_{2} = - \frac{1}{4} R_{2}$

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

$[\begin{array}{c} 1 & 1 & 9 \\ 0 & 1 & 5 \end{array}]$

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

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

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

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

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

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (5) + 9 \\ 0 & 1 & 5 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 4 \\ 0 & 1 & 5 \end{array}]$

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

$x = 4$

$y = 5$

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

$(4, 5)$

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