Finite Math Examples

$4 x + 7 y = - 12$ , $11 x + 6 y = 20$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 4 & 7 & - 12 \\ 11 & 6 & 20 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{7}{4} & \frac{- 12}{4} \\ 11 & 6 & 20 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{7}{4} & - 3 \\ 11 & 6 & 20 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 11 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 11 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{4} & - 3 \\ 11 - 11 \cdot 1 & 6 - 11 (\frac{7}{4}) & 20 - 11 \cdot - 3 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{4} & - 3 \\ 0 & - \frac{53}{4} & 53 \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $- \frac{4}{53}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $- \frac{4}{53}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{7}{4} & - 3 \\ - \frac{4}{53} \cdot 0 & - \frac{4}{53} (- \frac{53}{4}) & - \frac{4}{53} \cdot 53 \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{4} & - 3 \\ 0 & 1 & - 4 \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} - \frac{7}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} - \frac{7}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{7}{4} \cdot 0 & \frac{7}{4} - \frac{7}{4} \cdot 1 & - 3 - \frac{7}{4} \cdot - 4 \\ 0 & 1 & - 4 \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

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

Step 3

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

$x = 4$

$y = - 4$

Step 4

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

$(4, - 4)$