Finite Math Examples

$- 10 x - 15 y - Z = - 8$ , $- 6 x - 25 y - 2 z = 39$ , $8 x + 15 y + z = - 2$

Step 1

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

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Step 1.1

Move $- 15 y$ .

$- 10 x - Z - 15 y = - 8$

$- 6 x - 25 y - 2 z = 39$

$8 x + 15 y + z = - 2$

Step 1.2

Reorder $- 10 x$ and $- Z$ .

$- Z - 10 x - 15 y = - 8$

$- 6 x - 25 y - 2 z = 39$

$8 x + 15 y + z = - 2$

$- Z - 10 x - 15 y = - 8$

$- 6 x - 25 y - 2 z = 39$

$8 x + 15 y + z = - 2$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 1 & - 10 & - 15 & 0 & - 8 \\ 0 & - 6 & - 25 & - 2 & 39 \\ 0 & 8 & 15 & 1 & - 2 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} - - 1 & - - 10 & - - 15 & - 0 & - - 8 \\ 0 & - 6 & - 25 & - 2 & 39 \\ 0 & 8 & 15 & 1 & - 2 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 & - 6 & - 25 & - 2 & 39 \\ 0 & 8 & 15 & 1 & - 2 \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot - 25 & - \frac{1}{6} \cdot - 2 & - \frac{1}{6} \cdot 39 \\ 0 & 8 & 15 & 1 & - 2 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 & 1 & \frac{25}{6} & \frac{1}{3} & - \frac{13}{2} \\ 0 & 8 & 15 & 1 & - 2 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 & 1 & \frac{25}{6} & \frac{1}{3} & - \frac{13}{2} \\ 0 - 8 \cdot 0 & 8 - 8 \cdot 1 & 15 - 8 (\frac{25}{6}) & 1 - 8 (\frac{1}{3}) & - 2 - 8 (- \frac{13}{2}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 & 1 & \frac{25}{6} & \frac{1}{3} & - \frac{13}{2} \\ 0 & 0 & - \frac{55}{3} & - \frac{5}{3} & 50 \end{array}]$

Step 3.4

Multiply each element of $R_{3}$ by $- \frac{3}{55}$ to make the entry at $3, 3$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{3}$ by $- \frac{3}{55}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 & 1 & \frac{25}{6} & \frac{1}{3} & - \frac{13}{2} \\ - \frac{3}{55} \cdot 0 & - \frac{3}{55} \cdot 0 & - \frac{3}{55} (- \frac{55}{3}) & - \frac{3}{55} (- \frac{5}{3}) & - \frac{3}{55} \cdot 50 \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 & 1 & \frac{25}{6} & \frac{1}{3} & - \frac{13}{2} \\ 0 & 0 & 1 & \frac{1}{11} & - \frac{30}{11} \end{array}]$

Step 3.5

Perform the row operation $R_{2} = R_{2} - \frac{25}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{2} = R_{2} - \frac{25}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 - \frac{25}{6} \cdot 0 & 1 - \frac{25}{6} \cdot 0 & \frac{25}{6} - \frac{25}{6} \cdot 1 & \frac{1}{3} - \frac{25}{6} \cdot \frac{1}{11} & - \frac{13}{2} - \frac{25}{6} (- \frac{30}{11}) \\ 0 & 0 & 1 & \frac{1}{11} & - \frac{30}{11} \end{array}]$

Step 3.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 10 & 15 & 0 & 8 \\ 0 & 1 & 0 & - \frac{1}{22} & \frac{107}{22} \\ 0 & 0 & 1 & \frac{1}{11} & - \frac{30}{11} \end{array}]$

Step 3.6

Perform the row operation $R_{1} = R_{1} - 15 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} - 15 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 15 \cdot 0 & 10 - 15 \cdot 0 & 15 - 15 \cdot 1 & 0 - 15 (\frac{1}{11}) & 8 - 15 (- \frac{30}{11}) \\ 0 & 1 & 0 & - \frac{1}{22} & \frac{107}{22} \\ 0 & 0 & 1 & \frac{1}{11} & - \frac{30}{11} \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 10 & 0 & - \frac{15}{11} & \frac{538}{11} \\ 0 & 1 & 0 & - \frac{1}{22} & \frac{107}{22} \\ 0 & 0 & 1 & \frac{1}{11} & - \frac{30}{11} \end{array}]$

Step 3.7

Perform the row operation $R_{1} = R_{1} - 10 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{1} = R_{1} - 10 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 10 \cdot 0 & 10 - 10 \cdot 1 & 0 - 10 \cdot 0 & - \frac{15}{11} - 10 (- \frac{1}{22}) & \frac{538}{11} - 10 (\frac{107}{22}) \\ 0 & 1 & 0 & - \frac{1}{22} & \frac{107}{22} \\ 0 & 0 & 1 & \frac{1}{11} & - \frac{30}{11} \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{10}{11} & \frac{3}{11} \\ 0 & 1 & 0 & - \frac{1}{22} & \frac{107}{22} \\ 0 & 0 & 1 & \frac{1}{11} & - \frac{30}{11} \end{array}]$

Step 4

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

$Z - \frac{10}{11} z = \frac{3}{11}$

$x - \frac{1}{22} z = \frac{107}{22}$

$y + \frac{1}{11} z = - \frac{30}{11}$

Step 5

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

$(\frac{3}{11} + \frac{10 z}{11}, \frac{107}{22} + \frac{z}{22}, - \frac{30}{11} - \frac{z}{11}, z)$