Solve Using a Matrix by Elimination 5x-3y=18 , 2x+7y=-1

$5 x - 3 y = 18$ , $2 x + 7 y = - 1$

Write the system of equations in matrix form.

$[\begin{array}{c} 5 & - 3 & 18 \\ 2 & 7 & - 1 \end{array}]$

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} \frac{1}{5} R_{1} & \frac{1}{5} R_{1} & \frac{1}{5} R_{1} \\ 2 & 7 & - 1 \end{array}]$

$R_{1} = \frac{1}{5} R_{1}$

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

$[\begin{array}{c} (\frac{1}{5}) \cdot (5) & (\frac{1}{5}) \cdot (- 3) & (\frac{1}{5}) \cdot (18) \\ 2 & 7 & - 1 \end{array}]$

$R_{1} = \frac{1}{5} R_{1}$

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

$[\begin{array}{c} 1 & - \frac{3}{5} & \frac{18}{5} \\ 2 & 7 & - 1 \end{array}]$

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

$[\begin{array}{c} 1 & - \frac{3}{5} & \frac{18}{5} \\ - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} & - 2 \cdot R_{1} + R_{2} \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{3}{5} & \frac{18}{5} \\ (- 2) \cdot (1) + 2 & (- 2) \cdot (- \frac{3}{5}) + 7 & (- 2) \cdot (\frac{18}{5}) - 1 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{3}{5} & \frac{18}{5} \\ 0 & \frac{41}{5} & - \frac{41}{5} \end{array}]$

Perform the row operation $R_{2} = \frac{5}{41} R_{2}$ 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 $R_{2} = \frac{5}{41} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{3}{5} & \frac{18}{5} \\ \frac{5}{41} R_{2} & \frac{5}{41} R_{2} & \frac{5}{41} R_{2} \end{array}]$

$R_{2} = \frac{5}{41} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{5}{41} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{3}{5} & \frac{18}{5} \\ (\frac{5}{41}) \cdot (0) & (\frac{5}{41}) \cdot (\frac{41}{5}) & (\frac{5}{41}) \cdot (- \frac{41}{5}) \end{array}]$

$R_{2} = \frac{5}{41} R_{2}$

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

$[\begin{array}{c} 1 & - \frac{3}{5} & \frac{18}{5} \\ 0 & 1 & - 1 \end{array}]$

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

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

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

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

$[\begin{array}{c} (\frac{3}{5}) \cdot (0) + 1 & (\frac{3}{5}) \cdot (1) - \frac{3}{5} & (\frac{3}{5}) \cdot (- 1) + \frac{18}{5} \\ 0 & 1 & - 1 \end{array}]$

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

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

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

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

$x = 3$

$y = - 1$

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

$(3, - 1)$

Solve Using a Matrix by Elimination 5x-3y=18 , 2x+7y=-1

Related Questions

Support LCMGCF.com by disabling your adblocker.