Solve Using a Matrix by Elimination 2x-3y=4 -4x+6y=-8

$2 x - 3 y = 4$ $- 4 x + 6 y = - 8$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 2 & - 3 & 4 \\ - 4 & 6 & - 8 \end{array}]$

Step 2

Row reduce to eliminate one of the variables.

Perform the row operation $R_{2} = \frac{1}{6} R_{2}$ 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 $R_{2} = \frac{1}{6} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 2 & - 3 & 4 \\ \frac{1}{6} R_{2} & \frac{1}{6} R_{2} & \frac{1}{6} R_{2} \end{array}]$

$R_{2} = \frac{1}{6} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{6} R_{2}$ .

$[\begin{array}{c} 2 & - 3 & 4 \\ (\frac{1}{6}) \cdot (- 4) & (\frac{1}{6}) \cdot (6) & (\frac{1}{6}) \cdot (- 8) \end{array}]$

$R_{2} = \frac{1}{6} R_{2}$

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

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

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 $0$ .

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

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

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

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

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

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

$[\begin{array}{c} 0 & 0 & 0 \\ - \frac{2}{3} & 1 & - \frac{4}{3} \end{array}]$

Step 3

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

$- \frac{2 x}{3} + y = - \frac{4}{3}$

Step 4

Add $\frac{2 x}{3}$ to both sides of the equation.

$y = - \frac{4}{3} + \frac{2 x}{3}$

Step 5

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

$(x, - \frac{4}{3} + \frac{2 x}{3})$

Solve Using a Matrix by Elimination 2x-3y=4 -4x+6y=-8

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