Solve Using a Matrix by Elimination 7x-2y=3 , -1/2x+3y=8

$7 x - 2 y = 3$ , $- \frac{1}{2} x + 3 y = 8$

Step 1

Simplify the left side.

Combine $x$ and $\frac{1}{2}$ .

$- \frac{x}{2} + 3 y = 8$

$7 x - 2 y = 3$

$- \frac{x}{2} + 3 y = 8$

$7 x - 2 y = 3$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} 7 & - 2 & 3 \\ - \frac{1}{2} & 3 & 8 \end{array}]$

Step 3

Find the reduced row echelon form of the matrix.

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

$[\begin{array}{c} \frac{1}{7} R_{1} & \frac{1}{7} R_{1} & \frac{1}{7} R_{1} \\ - \frac{1}{2} & 3 & 8 \end{array}]$

$R_{1} = \frac{1}{7} R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{1}{7} R_{1}$ .

$[\begin{array}{c} (\frac{1}{7}) \cdot (7) & (\frac{1}{7}) \cdot (- 2) & (\frac{1}{7}) \cdot (3) \\ - \frac{1}{2} & 3 & 8 \end{array}]$

$R_{1} = \frac{1}{7} R_{1}$

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

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{3}{7} \\ - \frac{1}{2} & 3 & 8 \end{array}]$

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

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

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

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

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{3}{7} \\ (\frac{1}{2}) \cdot (1) - \frac{1}{2} & (\frac{1}{2}) \cdot (- \frac{2}{7}) + 3 & (\frac{1}{2}) \cdot (\frac{3}{7}) + 8 \end{array}]$

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

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

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{3}{7} \\ 0 & \frac{20}{7} & \frac{115}{14} \end{array}]$

Perform the row operation $R_{2} = \frac{7}{20} R_{2}$ 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 $R_{2} = \frac{7}{20} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{3}{7} \\ \frac{7}{20} R_{2} & \frac{7}{20} R_{2} & \frac{7}{20} R_{2} \end{array}]$

$R_{2} = \frac{7}{20} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{7}{20} R_{2}$ .

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{3}{7} \\ (\frac{7}{20}) \cdot (0) & (\frac{7}{20}) \cdot (\frac{20}{7}) & (\frac{7}{20}) \cdot (\frac{115}{14}) \end{array}]$

$R_{2} = \frac{7}{20} R_{2}$

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

$[\begin{array}{c} 1 & - \frac{2}{7} & \frac{3}{7} \\ 0 & 1 & \frac{23}{8} \end{array}]$

Perform the row operation $R_{1} = \frac{2}{7} R_{2} + R_{1}$ 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 $R_{1} = \frac{2}{7} R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} \frac{2}{7} R_{2} + R_{1} & \frac{2}{7} R_{2} + R_{1} & \frac{2}{7} R_{2} + R_{1} \\ 0 & 1 & \frac{23}{8} \end{array}]$

$R_{1} = \frac{2}{7} R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = \frac{2}{7} R_{2} + R_{1}$ .

$[\begin{array}{c} (\frac{2}{7}) \cdot (0) + 1 & (\frac{2}{7}) \cdot (1) - \frac{2}{7} & (\frac{2}{7}) \cdot (\frac{23}{8}) + \frac{3}{7} \\ 0 & 1 & \frac{23}{8} \end{array}]$

$R_{1} = \frac{2}{7} R_{2} + R_{1}$

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

$[\begin{array}{c} 1 & 0 & \frac{5}{4} \\ 0 & 1 & \frac{23}{8} \end{array}]$

Step 4

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

$x = \frac{5}{4}$

$y = \frac{23}{8}$

Step 5

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

$(\frac{5}{4}, \frac{23}{8})$

Solve Using a Matrix by Elimination 7x-2y=3 , -1/2x+3y=8

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