Precalculus Examples

$- x - 4 y = - 3$ , $3 x - 8 y = - 9$

Step 1

Write the system as a matrix.

$[\begin{array}{c} - 1 & - 4 & - 3 \\ 3 & - 8 & - 9 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} - - 1 & - - 4 & - - 3 \\ 3 & - 8 & - 9 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 4 & 3 \\ 3 & - 8 & - 9 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 4 & 3 \\ 3 - 3 \cdot 1 & - 8 - 3 \cdot 4 & - 9 - 3 \cdot 3 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 4 & 3 \\ 0 & - 20 & - 18 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 & 4 & 3 \\ - \frac{1}{20} \cdot 0 & - \frac{1}{20} \cdot - 20 & - \frac{1}{20} \cdot - 18 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 4 & 3 \\ 0 & 1 & \frac{9}{10} \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} - 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} - 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 4 \cdot 0 & 4 - 4 \cdot 1 & 3 - 4 (\frac{9}{10}) \\ 0 & 1 & \frac{9}{10} \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{3}{5} \\ 0 & 1 & \frac{9}{10} \end{array}]$

Step 3

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

$x = - \frac{3}{5}$

$y = \frac{9}{10}$

Step 4

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

$(- \frac{3}{5}, \frac{9}{10})$