Solve Using a Matrix by Elimination 2x-7y=2 , 14x+7y=22

$2 x - 7 y = 2$ , $14 x + 7 y = 22$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & - 7 & 2 \\ 14 & 7 & 22 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ 14 & 7 & 22 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (- 7) & (\frac{1}{2}) \cdot (2) \\ 14 & 7 & 22 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{7}{2} & 1 \\ 14 & 7 & 22 \end{array}]$

Perform the row operation $R_{2} = - 14 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 14 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & - \frac{7}{2} & 1 \\ - 14 \cdot R_{1} + R_{2} & - 14 \cdot R_{1} + R_{2} & - 14 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 14 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 14 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{2} & 1 \\ (- 14) \cdot (1) + 14 & (- 14) \cdot (- \frac{7}{2}) + 7 & (- 14) \cdot (1) + 22 \end{array}]$

$R_{2} = - 14 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & - \frac{7}{2} & 1 \\ 0 & 56 & 8 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{56} R_{2}$ 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 $R_{2} = \frac{1}{56} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{7}{2} & 1 \\ \frac{1}{56} R_{2} & \frac{1}{56} R_{2} & \frac{1}{56} R_{2} \end{array}]$

$R_{2} = \frac{1}{56} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{56} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{2} & 1 \\ (\frac{1}{56}) \cdot (0) & (\frac{1}{56}) \cdot (56) & (\frac{1}{56}) \cdot (8) \end{array}]$

$R_{2} = \frac{1}{56} R_{2}$

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

$[\begin{array}{c} 1 & - \frac{7}{2} & 1 \\ 0 & 1 & \frac{1}{7} \end{array}]$

Perform the row operation $R_{1} = \frac{7}{2} R_{2} + R_{1}$ 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 $R_{1} = \frac{7}{2} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{7}{2} R_{2} + R_{1} & \frac{7}{2} R_{2} + R_{1} & \frac{7}{2} R_{2} + R_{1} \\ 0 & 1 & \frac{1}{7} \end{array}]$

$R_{1} = \frac{7}{2} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{7}{2} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{7}{2}) \cdot (0) + 1 & (\frac{7}{2}) \cdot (1) - \frac{7}{2} & (\frac{7}{2}) \cdot (\frac{1}{7}) + 1 \\ 0 & 1 & \frac{1}{7} \end{array}]$

$R_{1} = \frac{7}{2} R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & \frac{3}{2} \\ 0 & 1 & \frac{1}{7} \end{array}]$

Step 3

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

$x = \frac{3}{2}$

$y = \frac{1}{7}$

Step 4

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

$(\frac{3}{2}, \frac{1}{7})$

Solve Using a Matrix by Elimination 2x-7y=2 , 14x+7y=22

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