Solve Using a Matrix by Elimination 3p+9q=6 , 5p-5q=30

$3 p + 9 q = 6$ , $5 p - 5 q = 30$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 3 & 9 & 6 \\ 5 & - 5 & 30 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{9}{3} & \frac{6}{3} \\ 5 & - 5 & 30 \end{array}]$

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 3 & 2 \\ 5 & - 5 & 30 \end{array}]$

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 3 & 2 \\ 5 - 5 \cdot 1 & - 5 - 5 \cdot 3 & 30 - 5 \cdot 2 \end{array}]$

Simplify $R_{2}$ .

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

Multiply each element of $R_{2}$ by $- \frac{1}{20}$ to make the entry at $2, 2$ a $1$ .

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

Simplify $R_{2}$ .

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

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

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

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 5 \\ 0 & 1 & - 1 \end{array}]$

Step 3

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

$p = 5 q = - 1$

Step 4

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

$(5, - 1)$