Solve Using a Matrix by Elimination 2x+3y=-6 , 4x-2y=3

$2 x + 3 y = - 6$ , $4 x - 2 y = 3$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & 3 & - 6 \\ 4 & - 2 & 3 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{1}{2} R_{1} & \frac{1}{2} R_{1} & \frac{1}{2} R_{1} \\ 4 & - 2 & 3 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (3) & (\frac{1}{2}) \cdot (- 6) \\ 4 & - 2 & 3 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{3}{2} & - 3 \\ 4 & - 2 & 3 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 3 \\ - 4 \cdot R_{1} + R_{2} & - 4 \cdot R_{1} + R_{2} & - 4 \cdot R_{1} + R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{3}{2} & - 3 \\ (- 4) \cdot (1) + 4 & (- 4) \cdot (\frac{3}{2}) - 2 & (- 4) \cdot (- 3) + 3 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{3}{2} & - 3 \\ 0 & - 8 & 15 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{8} R_{2}$ 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 $R_{2} = - \frac{1}{8} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 3 \\ - \frac{1}{8} R_{2} & - \frac{1}{8} R_{2} & - \frac{1}{8} R_{2} \end{array}]$

$R_{2} = - \frac{1}{8} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{8} R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & - 3 \\ (- \frac{1}{8}) \cdot (0) & (- \frac{1}{8}) \cdot (- 8) & (- \frac{1}{8}) \cdot (15) \end{array}]$

$R_{2} = - \frac{1}{8} R_{2}$

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

$[\begin{array}{c} 1 & \frac{3}{2} & - 3 \\ 0 & 1 & - \frac{15}{8} \end{array}]$

Perform the row operation $R_{1} = - \frac{3}{2} R_{2} + R_{1}$ 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 $R_{1} = - \frac{3}{2} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} & - \frac{3}{2} R_{2} + R_{1} \\ 0 & 1 & - \frac{15}{8} \end{array}]$

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

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

$[\begin{array}{c} (- \frac{3}{2}) \cdot (0) + 1 & (- \frac{3}{2}) \cdot (1) + \frac{3}{2} & (- \frac{3}{2}) \cdot (- \frac{15}{8}) - 3 \\ 0 & 1 & - \frac{15}{8} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & - \frac{3}{16} \\ 0 & 1 & - \frac{15}{8} \end{array}]$

Step 3

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

$x = - \frac{3}{16}$

$y = - \frac{15}{8}$

Step 4

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

$(- \frac{3}{16}, - \frac{15}{8})$

Solve Using a Matrix by Elimination 2x+3y=-6 , 4x-2y=3

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