Finite Math Examples

$- 25 - 9 y - z = - 16$ , $- 15 - 15 y - 2 z = 26$ , $20 x + 9 y + z = 6$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Move all terms not containing a variable to the right side of the equation.

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Step 1.1.1

Add $25$ to both sides of the equation.

$- 9 y - z = - 16 + 25$

$- 15 - 15 y - 2 z = 26$

$20 x + 9 y + z = 6$

Step 1.1.2

Add $- 16$ and $25$ .

$- 9 y - z = 9$

$- 15 - 15 y - 2 z = 26$

$20 x + 9 y + z = 6$

$- 9 y - z = 9$

$- 15 - 15 y - 2 z = 26$

$20 x + 9 y + z = 6$

Step 1.2

Move all terms not containing a variable to the right side of the equation.

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Step 1.2.1

Add $15$ to both sides of the equation.

$- 9 y - z = 9$

$- 15 y - 2 z = 26 + 15$

$20 x + 9 y + z = 6$

Step 1.2.2

Add $26$ and $15$ .

$- 9 y - z = 9$

$- 15 y - 2 z = 41$

$20 x + 9 y + z = 6$

$- 9 y - z = 9$

$- 15 y - 2 z = 41$

$20 x + 9 y + z = 6$

$- 9 y - z = 9$

$- 15 y - 2 z = 41$

$20 x + 9 y + z = 6$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 0 & - 9 & - 1 & 9 \\ 0 & - 15 & - 2 & 41 \\ 20 & 9 & 1 & 6 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Swap $R_{3}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 20 & 9 & 1 & 6 \\ 0 & - 15 & - 2 & 41 \\ 0 & - 9 & - 1 & 9 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} \frac{20}{20} & \frac{9}{20} & \frac{1}{20} & \frac{6}{20} \\ 0 & - 15 & - 2 & 41 \\ 0 & - 9 & - 1 & 9 \end{array}]$

Step 3.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 & - 15 & - 2 & 41 \\ 0 & - 9 & - 1 & 9 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{1}{15}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ - \frac{1}{15} \cdot 0 & - \frac{1}{15} \cdot - 15 & - \frac{1}{15} \cdot - 2 & - \frac{1}{15} \cdot 41 \\ 0 & - 9 & - 1 & 9 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 & 1 & \frac{2}{15} & - \frac{41}{15} \\ 0 & - 9 & - 1 & 9 \end{array}]$

Step 3.4

Perform the row operation $R_{3} = R_{3} + 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{3} = R_{3} + 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 & 1 & \frac{2}{15} & - \frac{41}{15} \\ 0 + 9 \cdot 0 & - 9 + 9 \cdot 1 & - 1 + 9 (\frac{2}{15}) & 9 + 9 (- \frac{41}{15}) \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 & 1 & \frac{2}{15} & - \frac{41}{15} \\ 0 & 0 & \frac{1}{5} & - \frac{78}{5} \end{array}]$

Step 3.5

Multiply each element of $R_{3}$ by $5$ to make the entry at $3, 3$ a $1$ .

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Step 3.5.1

Multiply each element of $R_{3}$ by $5$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 & 1 & \frac{2}{15} & - \frac{41}{15} \\ 5 \cdot 0 & 5 \cdot 0 & 5 (\frac{1}{5}) & 5 (- \frac{78}{5}) \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 & 1 & \frac{2}{15} & - \frac{41}{15} \\ 0 & 0 & 1 & - 78 \end{array}]$

Step 3.6

Perform the row operation $R_{2} = R_{2} - \frac{2}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{2} = R_{2} - \frac{2}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 - \frac{2}{15} \cdot 0 & 1 - \frac{2}{15} \cdot 0 & \frac{2}{15} - \frac{2}{15} \cdot 1 & - \frac{41}{15} - \frac{2}{15} \cdot - 78 \\ 0 & 0 & 1 & - 78 \end{array}]$

Step 3.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{9}{20} & \frac{1}{20} & \frac{3}{10} \\ 0 & 1 & 0 & \frac{23}{3} \\ 0 & 0 & 1 & - 78 \end{array}]$

Step 3.7

Perform the row operation $R_{1} = R_{1} - \frac{1}{20} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{20} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{20} \cdot 0 & \frac{9}{20} - \frac{1}{20} \cdot 0 & \frac{1}{20} - \frac{1}{20} \cdot 1 & \frac{3}{10} - \frac{1}{20} \cdot - 78 \\ 0 & 1 & 0 & \frac{23}{3} \\ 0 & 0 & 1 & - 78 \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{9}{20} & 0 & \frac{21}{5} \\ 0 & 1 & 0 & \frac{23}{3} \\ 0 & 0 & 1 & - 78 \end{array}]$

Step 3.8

Perform the row operation $R_{1} = R_{1} - \frac{9}{20} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.8.1

Perform the row operation $R_{1} = R_{1} - \frac{9}{20} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{9}{20} \cdot 0 & \frac{9}{20} - \frac{9}{20} \cdot 1 & 0 - \frac{9}{20} \cdot 0 & \frac{21}{5} - \frac{9}{20} \cdot \frac{23}{3} \\ 0 & 1 & 0 & \frac{23}{3} \\ 0 & 0 & 1 & - 78 \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{3}{4} \\ 0 & 1 & 0 & \frac{23}{3} \\ 0 & 0 & 1 & - 78 \end{array}]$

Step 4

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

$x = \frac{3}{4}$

$y = \frac{23}{3}$

$z = - 78$

Step 5

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

$(\frac{3}{4}, \frac{23}{3}, - 78)$