Precalculus Examples

$7 x - 7 y = - 7$ , $- 3 x + 3 y = 3$

Step 1

Write the system as a matrix.

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

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{7}{7} & \frac{- 7}{7} & \frac{- 7}{7} \\ - 3 & 3 & 3 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

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

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 $0$ .

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

Step 2.2.2

Simplify $R_{2}$ .

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

Step 3

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

$x - y = - 1$

$0 = 0$

Step 4

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

$(- 1 + y, y)$