Solve Using a Matrix by Elimination 2x-5y=-2 , -1/4x+2y=7

$2 x - 5 y = - 2$ , $- \frac{1}{4} x + 2 y = 7$

Step 1

Simplify the left side.

Combine $x$ and $\frac{1}{4}$ .

$- \frac{x}{4} + 2 y = 7$

$2 x - 5 y = - 2$

$- \frac{x}{4} + 2 y = 7$

$2 x - 5 y = - 2$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & - 5 & - 2 \\ - \frac{1}{4} & 2 & 7 \end{array}]$

Step 3

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 $1$ .

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

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

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

$[\begin{array}{c} (\frac{1}{2}) \cdot (2) & (\frac{1}{2}) \cdot (- 5) & (\frac{1}{2}) \cdot (- 2) \\ - \frac{1}{4} & 2 & 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{5}{2} & - 1 \\ - \frac{1}{4} & 2 & 7 \end{array}]$

Perform the row operation $R_{2} = \frac{1}{4} R_{1} + R_{2}$ 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 $R_{2} = \frac{1}{4} R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

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

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

$[\begin{array}{c} 1 & - \frac{5}{2} & - 1 \\ (\frac{1}{4}) \cdot (1) - \frac{1}{4} & (\frac{1}{4}) \cdot (- \frac{5}{2}) + 2 & (\frac{1}{4}) \cdot (- 1) + 7 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{5}{2} & - 1 \\ 0 & \frac{11}{8} & \frac{27}{4} \end{array}]$

Perform the row operation $R_{2} = \frac{8}{11} R_{2}$ 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 $R_{2} = \frac{8}{11} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & - 1 \\ \frac{8}{11} R_{2} & \frac{8}{11} R_{2} & \frac{8}{11} R_{2} \end{array}]$

$R_{2} = \frac{8}{11} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{8}{11} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{2} & - 1 \\ (\frac{8}{11}) \cdot (0) & (\frac{8}{11}) \cdot (\frac{11}{8}) & (\frac{8}{11}) \cdot (\frac{27}{4}) \end{array}]$

$R_{2} = \frac{8}{11} R_{2}$

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

$[\begin{array}{c} 1 & - \frac{5}{2} & - 1 \\ 0 & 1 & \frac{54}{11} \end{array}]$

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

$[\begin{array}{c} \frac{5}{2} R_{2} + R_{1} & \frac{5}{2} R_{2} + R_{1} & \frac{5}{2} R_{2} + R_{1} \\ 0 & 1 & \frac{54}{11} \end{array}]$

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

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

$[\begin{array}{c} (\frac{5}{2}) \cdot (0) + 1 & (\frac{5}{2}) \cdot (1) - \frac{5}{2} & (\frac{5}{2}) \cdot (\frac{54}{11}) - 1 \\ 0 & 1 & \frac{54}{11} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & \frac{124}{11} \\ 0 & 1 & \frac{54}{11} \end{array}]$

Step 4

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

$x = \frac{124}{11}$

$y = \frac{54}{11}$

Step 5

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

$(\frac{124}{11}, \frac{54}{11})$

Solve Using a Matrix by Elimination 2x-5y=-2 , -1/4x+2y=7

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