Solve Using a Matrix by Elimination 3x+5y=6 , -4x+2y=5

$3 x + 5 y = 6$ , $- 4 x + 2 y = 5$

Write the system of equations in matrix form.

$[\begin{array}{c} 3 & 5 & 6 \\ - 4 & 2 & 5 \end{array}]$

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} \frac{1}{3} R_{1} & \frac{1}{3} R_{1} & \frac{1}{3} R_{1} \\ - 4 & 2 & 5 \end{array}]$

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

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

$[\begin{array}{c} (\frac{1}{3}) \cdot (3) & (\frac{1}{3}) \cdot (5) & (\frac{1}{3}) \cdot (6) \\ - 4 & 2 & 5 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 2 \\ - 4 & 2 & 5 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 2 \\ 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} & 4 \cdot R_{1} + R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 2 \\ (4) \cdot (1) - 4 & (4) \cdot (\frac{5}{3}) + 2 & (4) \cdot (2) + 5 \end{array}]$

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

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

$[\begin{array}{c} 1 & \frac{5}{3} & 2 \\ 0 & \frac{26}{3} & 13 \end{array}]$

Perform the row operation $R_{2} = \frac{3}{26} R_{2}$ 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 $R_{2} = \frac{3}{26} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 2 \\ \frac{3}{26} R_{2} & \frac{3}{26} R_{2} & \frac{3}{26} R_{2} \end{array}]$

$R_{2} = \frac{3}{26} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{3}{26} R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 2 \\ (\frac{3}{26}) \cdot (0) & (\frac{3}{26}) \cdot (\frac{26}{3}) & (\frac{3}{26}) \cdot (13) \end{array}]$

$R_{2} = \frac{3}{26} R_{2}$

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

$[\begin{array}{c} 1 & \frac{5}{3} & 2 \\ 0 & 1 & \frac{3}{2} \end{array}]$

Perform the row operation $R_{1} = - \frac{5}{3} R_{2} + R_{1}$ 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 $R_{1} = - \frac{5}{3} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} - \frac{5}{3} R_{2} + R_{1} & - \frac{5}{3} R_{2} + R_{1} & - \frac{5}{3} R_{2} + R_{1} \\ 0 & 1 & \frac{3}{2} \end{array}]$

$R_{1} = - \frac{5}{3} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = - \frac{5}{3} R_{2} + R_{1}$ .

$[\begin{array}{c} (- \frac{5}{3}) \cdot (0) + 1 & (- \frac{5}{3}) \cdot (1) + \frac{5}{3} & (- \frac{5}{3}) \cdot (\frac{3}{2}) + 2 \\ 0 & 1 & \frac{3}{2} \end{array}]$

$R_{1} = - \frac{5}{3} R_{2} + R_{1}$

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

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

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

$x = - \frac{1}{2}$

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

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

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

Solve Using a Matrix by Elimination 3x+5y=6 , -4x+2y=5

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