Solve Using an Augmented Matrix -3x+7y=2 , -2x+9y=4

$- 3 x + 7 y = 2$ , $- 2 x + 9 y = 4$

Step 1

Write the system as a matrix.

$[\begin{array}{c} - 3 & 7 & 2 \\ - 2 & 9 & 4 \end{array}]$

Step 2

Find the reduced row echelon form.

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 7 & - \frac{1}{3} \cdot 2 \\ - 2 & 9 & 4 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & - \frac{2}{3} \\ - 2 & 9 & 4 \end{array}]$

Perform the row operation $R_{2} = R_{2} + 2 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & - \frac{2}{3} \\ - 2 + 2 \cdot 1 & 9 + 2 (- \frac{7}{3}) & 4 + 2 (- \frac{2}{3}) \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & - \frac{2}{3} \\ 0 & \frac{13}{3} & \frac{8}{3} \end{array}]$

Multiply each element of $R_{2}$ by $\frac{3}{13}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & - \frac{2}{3} \\ \frac{3}{13} \cdot 0 & \frac{3}{13} \cdot \frac{13}{3} & \frac{3}{13} \cdot \frac{8}{3} \end{array}]$

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{3} & - \frac{2}{3} \\ 0 & 1 & \frac{8}{13} \end{array}]$

Perform the row operation $R_{1} = R_{1} + \frac{7}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{7}{3} \cdot 0 & - \frac{7}{3} + \frac{7}{3} \cdot 1 & - \frac{2}{3} + \frac{7}{3} \cdot \frac{8}{13} \\ 0 & 1 & \frac{8}{13} \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{10}{13} \\ 0 & 1 & \frac{8}{13} \end{array}]$

Step 3

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

$x = \frac{10}{13} y = \frac{8}{13}$

Step 4

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

$(\frac{10}{13}, \frac{8}{13})$

Solve Using an Augmented Matrix -3x+7y=2 , -2x+9y=4

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