Finite Math Examples

$3 x + 2 y - 1 z = 2$ , $2 x - 3 y + 1 z = - 2$ , $1 x - 1 y - 1 z = 4$

Step 1

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

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Step 1.1

Rewrite $- 1 z$ as $- z$ .

$3 x + 2 y - z = 2$

$2 x - 3 y + 1 z = - 2$

$1 x - 1 y - 1 z = 4$

Step 1.2

Multiply $z$ by $1$ .

$3 x + 2 y - z = 2$

$2 x - 3 y + z = - 2$

$1 x - 1 y - 1 z = 4$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $x$ by $1$ .

$3 x + 2 y - z = 2$

$2 x - 3 y + z = - 2$

$x - 1 y - 1 z = 4$

Step 1.3.2

Rewrite $- 1 y$ as $- y$ .

$3 x + 2 y - z = 2$

$2 x - 3 y + z = - 2$

$x - y - 1 z = 4$

Step 1.3.3

Rewrite $- 1 z$ as $- z$ .

$3 x + 2 y - z = 2$

$2 x - 3 y + z = - 2$

$x - y - z = 4$

$3 x + 2 y - z = 2$

$2 x - 3 y + z = - 2$

$x - y - z = 4$

$3 x + 2 y - z = 2$

$2 x - 3 y + z = - 2$

$x - y - z = 4$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 3 & 2 & - 1 & 2 \\ 2 & - 3 & 1 & - 2 \\ 1 & - 1 & - 1 & 4 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} \frac{3}{3} & \frac{2}{3} & \frac{- 1}{3} & \frac{2}{3} \\ 2 & - 3 & 1 & - 2 \\ 1 & - 1 & - 1 & 4 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 2 & - 3 & 1 & - 2 \\ 1 & - 1 & - 1 & 4 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 2 - 2 \cdot 1 & - 3 - 2 (\frac{2}{3}) & 1 - 2 (- \frac{1}{3}) & - 2 - 2 (\frac{2}{3}) \\ 1 & - 1 & - 1 & 4 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & - \frac{13}{3} & \frac{5}{3} & - \frac{10}{3} \\ 1 & - 1 & - 1 & 4 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & - \frac{13}{3} & \frac{5}{3} & - \frac{10}{3} \\ 1 - 1 & - 1 - \frac{2}{3} & - 1 + \frac{1}{3} & 4 - \frac{2}{3} \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & - \frac{13}{3} & \frac{5}{3} & - \frac{10}{3} \\ 0 & - \frac{5}{3} & - \frac{2}{3} & \frac{10}{3} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $- \frac{3}{13}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $- \frac{3}{13}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ - \frac{3}{13} \cdot 0 & - \frac{3}{13} (- \frac{13}{3}) & - \frac{3}{13} \cdot \frac{5}{3} & - \frac{3}{13} (- \frac{10}{3}) \\ 0 & - \frac{5}{3} & - \frac{2}{3} & \frac{10}{3} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - \frac{5}{13} & \frac{10}{13} \\ 0 & - \frac{5}{3} & - \frac{2}{3} & \frac{10}{3} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} + \frac{5}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} + \frac{5}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - \frac{5}{13} & \frac{10}{13} \\ 0 + \frac{5}{3} \cdot 0 & - \frac{5}{3} + \frac{5}{3} \cdot 1 & - \frac{2}{3} + \frac{5}{3} (- \frac{5}{13}) & \frac{10}{3} + \frac{5}{3} \cdot \frac{10}{13} \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - \frac{5}{13} & \frac{10}{13} \\ 0 & 0 & - \frac{17}{13} & \frac{60}{13} \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $- \frac{13}{17}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $- \frac{13}{17}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - \frac{5}{13} & \frac{10}{13} \\ - \frac{13}{17} \cdot 0 & - \frac{13}{17} \cdot 0 & - \frac{13}{17} (- \frac{17}{13}) & - \frac{13}{17} \cdot \frac{60}{13} \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & - \frac{5}{13} & \frac{10}{13} \\ 0 & 0 & 1 & - \frac{60}{17} \end{array}]$

Step 3.7

Perform the row operation $R_{2} = R_{2} + \frac{5}{13} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{2} = R_{2} + \frac{5}{13} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 + \frac{5}{13} \cdot 0 & 1 + \frac{5}{13} \cdot 0 & - \frac{5}{13} + \frac{5}{13} \cdot 1 & \frac{10}{13} + \frac{5}{13} (- \frac{60}{17}) \\ 0 & 0 & 1 & - \frac{60}{17} \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & - \frac{60}{17} \end{array}]$

Step 3.8

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

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Step 3.8.1

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

$[\begin{array}{c} 1 + \frac{1}{3} \cdot 0 & \frac{2}{3} + \frac{1}{3} \cdot 0 & - \frac{1}{3} + \frac{1}{3} \cdot 1 & \frac{2}{3} + \frac{1}{3} (- \frac{60}{17}) \\ 0 & 1 & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & - \frac{60}{17} \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & 0 & - \frac{26}{51} \\ 0 & 1 & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & - \frac{60}{17} \end{array}]$

Step 3.9

Perform the row operation $R_{1} = R_{1} - \frac{2}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.9.1

Perform the row operation $R_{1} = R_{1} - \frac{2}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{2}{3} \cdot 0 & \frac{2}{3} - \frac{2}{3} \cdot 1 & 0 - \frac{2}{3} \cdot 0 & - \frac{26}{51} - \frac{2}{3} (- \frac{10}{17}) \\ 0 & 1 & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & - \frac{60}{17} \end{array}]$

Step 3.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{2}{17} \\ 0 & 1 & 0 & - \frac{10}{17} \\ 0 & 0 & 1 & - \frac{60}{17} \end{array}]$

Step 4

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

$x = - \frac{2}{17}$

$y = - \frac{10}{17}$

$z = - \frac{60}{17}$

Step 5

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

$(- \frac{2}{17}, - \frac{10}{17}, - \frac{60}{17})$