Solve Using a Matrix by Elimination x+3y=-8 21x+6y=3

$x + 3 y = - 8$ $21 x + 6 y = 3$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 3 & - 8 \\ 21 & 6 & 3 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

Perform the row operation $R_{2} = - 21 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 21 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 3 & - 8 \\ - 21 \cdot R_{1} + R_{2} & - 21 \cdot R_{1} + R_{2} & - 21 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 21 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 21 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & 3 & - 8 \\ (- 21) \cdot (1) + 21 & (- 21) \cdot (3) + 6 & (- 21) \cdot (- 8) + 3 \end{array}]$

$R_{2} = - 21 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & 3 & - 8 \\ 0 & - 57 & 171 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{57} R_{2}$ 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 $R_{2} = - \frac{1}{57} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 3 & - 8 \\ - \frac{1}{57} R_{2} & - \frac{1}{57} R_{2} & - \frac{1}{57} R_{2} \end{array}]$

$R_{2} = - \frac{1}{57} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{57} R_{2}$ .

$[\begin{array}{c} 1 & 3 & - 8 \\ (- \frac{1}{57}) \cdot (0) & (- \frac{1}{57}) \cdot (- 57) & (- \frac{1}{57}) \cdot (171) \end{array}]$

$R_{2} = - \frac{1}{57} R_{2}$

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

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

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

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

$R_{1} = - 3 \cdot R_{2} + R_{1}$

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

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

$R_{1} = - 3 \cdot R_{2} + R_{1}$

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

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

Step 3

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

$x = 1$

$y = - 3$

Step 4

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

$(1, - 3)$

Solve Using a Matrix by Elimination x+3y=-8 21x+6y=3

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