Solve Using a Matrix by Elimination 2x-3y=8 1x+2y=-3

$2 x - 3 y = 8$ $1 x + 2 y = - 3$

Simplify the left side.

Multiply $x$ by $1$ .

$x + 2 y = - 3$

$2 x - 3 y = 8$

$x + 2 y = - 3$

$2 x - 3 y = 8$

Write the system of equations in matrix form.

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

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 $1$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$[\begin{array}{c} 1 & - \frac{3}{2} & 4 \\ 0 & \frac{7}{2} & - 7 \end{array}]$

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

$[\begin{array}{c} 1 & - \frac{3}{2} & 4 \\ \frac{2}{7} R_{2} & \frac{2}{7} R_{2} & \frac{2}{7} R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{3}{2} & 4 \\ (\frac{2}{7}) \cdot (0) & (\frac{2}{7}) \cdot (\frac{7}{2}) & (\frac{2}{7}) \cdot (- 7) \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{3}{2} & 4 \\ 0 & 1 & - 2 \end{array}]$

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 $0$ .

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

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

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

$[\begin{array}{c} (\frac{3}{2}) \cdot (0) + 1 & (\frac{3}{2}) \cdot (1) - \frac{3}{2} & (\frac{3}{2}) \cdot (- 2) + 4 \\ 0 & 1 & - 2 \end{array}]$

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

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

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

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

$x = 1$

$y = - 2$

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

$(1, - 2)$

Solve Using a Matrix by Elimination 2x-3y=8 1x+2y=-3

Related Questions

Support LCMGCF.com by disabling your adblocker.