Solve Using a Matrix by Elimination 4x-2y=-6 3x-6y=-18

$4 x - 2 y = - 6$ $3 x - 6 y = - 18$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 4 & - 2 & - 6 \\ 3 & - 6 & - 18 \end{array}]$

Step 2

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 $1$ .

$[\begin{array}{c} \frac{1}{4} R_{1} & \frac{1}{4} R_{1} & \frac{1}{4} R_{1} \\ 3 & - 6 & - 18 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{4}) \cdot (4) & (\frac{1}{4}) \cdot (- 2) & (\frac{1}{4}) \cdot (- 6) \\ 3 & - 6 & - 18 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 3 & - 6 & - 18 \end{array}]$

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 $0$ .

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 $0$ .

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

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

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ (- 3) \cdot (1) + 3 & (- 3) \cdot (- \frac{1}{2}) - 6 & (- 3) \cdot (- \frac{3}{2}) - 18 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ 0 & - \frac{9}{2} & - \frac{27}{2} \end{array}]$

Perform the row operation $R_{2} = - \frac{2}{9} R_{2}$ 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 $R_{2} = - \frac{2}{9} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

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

$R_{2} = - \frac{2}{9} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{2}{9} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - \frac{3}{2} \\ (- \frac{2}{9}) \cdot (0) & (- \frac{2}{9}) \cdot (- \frac{9}{2}) & (- \frac{2}{9}) \cdot (- \frac{27}{2}) \end{array}]$

$R_{2} = - \frac{2}{9} R_{2}$

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

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

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

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

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

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

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

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

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

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

Step 3

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

$x = 0$

$y = 3$

Step 4

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

$(0, 3)$

Solve Using a Matrix by Elimination 4x-2y=-6 3x-6y=-18

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