Precalculus Examples

$2 x + 2 y = 13$ , $12 x + 12 y = 78$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 2 & 2 & 13 \\ 12 & 12 & 78 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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

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Step 2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{2}{2} & \frac{13}{2} \\ 12 & 12 & 78 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & \frac{13}{2} \\ 12 & 12 & 78 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 12 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 12 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & \frac{13}{2} \\ 12 - 12 \cdot 1 & 12 - 12 \cdot 1 & 78 - 12 (\frac{13}{2}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

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

Step 3

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

$x + y = \frac{13}{2}$

$0 = 0$

Step 4

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

$(\frac{13}{2} - y, y)$