Finite Math Examples

$7 y + 6 z = - 2$ , $7 x - 4 y - 5 z = 1$ , $10 x - 5 z = - 4$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 0 & 7 & 6 & - 2 \\ 7 & - 4 & - 5 & 1 \\ 10 & 0 & - 5 & - 4 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 7 & - 4 & - 5 & 1 \\ 0 & 7 & 6 & - 2 \\ 10 & 0 & - 5 & - 4 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} \frac{7}{7} & \frac{- 4}{7} & \frac{- 5}{7} & \frac{1}{7} \\ 0 & 7 & 6 & - 2 \\ 10 & 0 & - 5 & - 4 \end{array}]$

Step 2.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 7 & 6 & - 2 \\ 10 & 0 & - 5 & - 4 \end{array}]$

Step 2.3

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 2.3.1

Perform the row operation $R_{3} = R_{3} - 10 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 7 & 6 & - 2 \\ 10 - 10 \cdot 1 & 0 - 10 (- \frac{4}{7}) & - 5 - 10 (- \frac{5}{7}) & - 4 - 10 (\frac{1}{7}) \end{array}]$

Step 2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 7 & 6 & - 2 \\ 0 & \frac{40}{7} & \frac{15}{7} & - \frac{38}{7} \end{array}]$

Step 2.4

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

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Step 2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{7}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ \frac{0}{7} & \frac{7}{7} & \frac{6}{7} & \frac{- 2}{7} \\ 0 & \frac{40}{7} & \frac{15}{7} & - \frac{38}{7} \end{array}]$

Step 2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 1 & \frac{6}{7} & - \frac{2}{7} \\ 0 & \frac{40}{7} & \frac{15}{7} & - \frac{38}{7} \end{array}]$

Step 2.5

Perform the row operation $R_{3} = R_{3} - \frac{40}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{40}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 1 & \frac{6}{7} & - \frac{2}{7} \\ 0 - \frac{40}{7} \cdot 0 & \frac{40}{7} - \frac{40}{7} \cdot 1 & \frac{15}{7} - \frac{40}{7} \cdot \frac{6}{7} & - \frac{38}{7} - \frac{40}{7} (- \frac{2}{7}) \end{array}]$

Step 2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 1 & \frac{6}{7} & - \frac{2}{7} \\ 0 & 0 & - \frac{135}{49} & - \frac{186}{49} \end{array}]$

Step 2.6

Multiply each element of $R_{3}$ by $- \frac{49}{135}$ to make the entry at $3, 3$ a $1$ .

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Step 2.6.1

Multiply each element of $R_{3}$ by $- \frac{49}{135}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 1 & \frac{6}{7} & - \frac{2}{7} \\ - \frac{49}{135} \cdot 0 & - \frac{49}{135} \cdot 0 & - \frac{49}{135} (- \frac{135}{49}) & - \frac{49}{135} (- \frac{186}{49}) \end{array}]$

Step 2.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 1 & \frac{6}{7} & - \frac{2}{7} \\ 0 & 0 & 1 & \frac{62}{45} \end{array}]$

Step 2.7

Perform the row operation $R_{2} = R_{2} - \frac{6}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 2.7.1

Perform the row operation $R_{2} = R_{2} - \frac{6}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 - \frac{6}{7} \cdot 0 & 1 - \frac{6}{7} \cdot 0 & \frac{6}{7} - \frac{6}{7} \cdot 1 & - \frac{2}{7} - \frac{6}{7} \cdot \frac{62}{45} \\ 0 & 0 & 1 & \frac{62}{45} \end{array}]$

Step 2.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & - \frac{5}{7} & \frac{1}{7} \\ 0 & 1 & 0 & - \frac{22}{15} \\ 0 & 0 & 1 & \frac{62}{45} \end{array}]$

Step 2.8

Perform the row operation $R_{1} = R_{1} + \frac{5}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 2.8.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{7} \cdot 0 & - \frac{4}{7} + \frac{5}{7} \cdot 0 & - \frac{5}{7} + \frac{5}{7} \cdot 1 & \frac{1}{7} + \frac{5}{7} \cdot \frac{62}{45} \\ 0 & 1 & 0 & - \frac{22}{15} \\ 0 & 0 & 1 & \frac{62}{45} \end{array}]$

Step 2.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{7} & 0 & \frac{71}{63} \\ 0 & 1 & 0 & - \frac{22}{15} \\ 0 & 0 & 1 & \frac{62}{45} \end{array}]$

Step 2.9

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

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Step 2.9.1

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

$[\begin{array}{c} 1 + \frac{4}{7} \cdot 0 & - \frac{4}{7} + \frac{4}{7} \cdot 1 & 0 + \frac{4}{7} \cdot 0 & \frac{71}{63} + \frac{4}{7} (- \frac{22}{15}) \\ 0 & 1 & 0 & - \frac{22}{15} \\ 0 & 0 & 1 & \frac{62}{45} \end{array}]$

Step 2.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{13}{45} \\ 0 & 1 & 0 & - \frac{22}{15} \\ 0 & 0 & 1 & \frac{62}{45} \end{array}]$

Step 3

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

$x = \frac{13}{45}$

$y = - \frac{22}{15}$

$z = \frac{62}{45}$

Step 4

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

$(\frac{13}{45}, - \frac{22}{15}, \frac{62}{45})$