Finite Math Examples

$x + 8 + 8 z = 8$ , $7 x + 7 y + z = 1$ , $8 x + 15 y + 9 z = - 9$

Step 1

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

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Step 1.1

Subtract $8$ from both sides of the equation.

$x + 8 z = 8 - 8$

$7 x + 7 y + z = 1$

$8 x + 15 y + 9 z = - 9$

Step 1.2

Subtract $8$ from $8$ .

$x + 8 z = 0$

$7 x + 7 y + z = 1$

$8 x + 15 y + 9 z = - 9$

$x + 8 z = 0$

$7 x + 7 y + z = 1$

$8 x + 15 y + 9 z = - 9$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 7 & 7 & 1 & 1 \\ 8 & 15 & 9 & - 9 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 7 - 7 \cdot 1 & 7 - 7 \cdot 0 & 1 - 7 \cdot 8 & 1 - 7 \cdot 0 \\ 8 & 15 & 9 & - 9 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 7 & - 55 & 1 \\ 8 & 15 & 9 & - 9 \end{array}]$

Step 3.2

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 7 & - 55 & 1 \\ 8 - 8 \cdot 1 & 15 - 8 \cdot 0 & 9 - 8 \cdot 8 & - 9 - 8 \cdot 0 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 7 & - 55 & 1 \\ 0 & 15 & - 55 & - 9 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ \frac{0}{7} & \frac{7}{7} & \frac{- 55}{7} & \frac{1}{7} \\ 0 & 15 & - 55 & - 9 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 1 & - \frac{55}{7} & \frac{1}{7} \\ 0 & 15 & - 55 & - 9 \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 1 & - \frac{55}{7} & \frac{1}{7} \\ 0 - 15 \cdot 0 & 15 - 15 \cdot 1 & - 55 - 15 (- \frac{55}{7}) & - 9 - 15 (\frac{1}{7}) \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 1 & - \frac{55}{7} & \frac{1}{7} \\ 0 & 0 & \frac{440}{7} & - \frac{78}{7} \end{array}]$

Step 3.5

Multiply each element of $R_{3}$ by $\frac{7}{440}$ to make the entry at $3, 3$ a $1$ .

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Step 3.5.1

Multiply each element of $R_{3}$ by $\frac{7}{440}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 1 & - \frac{55}{7} & \frac{1}{7} \\ \frac{7}{440} \cdot 0 & \frac{7}{440} \cdot 0 & \frac{7}{440} \cdot \frac{440}{7} & \frac{7}{440} (- \frac{78}{7}) \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 1 & - \frac{55}{7} & \frac{1}{7} \\ 0 & 0 & 1 & - \frac{39}{220} \end{array}]$

Step 3.6

Perform the row operation $R_{2} = R_{2} + \frac{55}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{2} = R_{2} + \frac{55}{7} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 + \frac{55}{7} \cdot 0 & 1 + \frac{55}{7} \cdot 0 & - \frac{55}{7} + \frac{55}{7} \cdot 1 & \frac{1}{7} + \frac{55}{7} (- \frac{39}{220}) \\ 0 & 0 & 1 & - \frac{39}{220} \end{array}]$

Step 3.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 8 & 0 \\ 0 & 1 & 0 & - \frac{5}{4} \\ 0 & 0 & 1 & - \frac{39}{220} \end{array}]$

Step 3.7

Perform the row operation $R_{1} = R_{1} - 8 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{1} = R_{1} - 8 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 8 \cdot 0 & 0 - 8 \cdot 0 & 8 - 8 \cdot 1 & 0 - 8 (- \frac{39}{220}) \\ 0 & 1 & 0 & - \frac{5}{4} \\ 0 & 0 & 1 & - \frac{39}{220} \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{78}{55} \\ 0 & 1 & 0 & - \frac{5}{4} \\ 0 & 0 & 1 & - \frac{39}{220} \end{array}]$

Step 4

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

$x = \frac{78}{55}$

$y = - \frac{5}{4}$

$z = - \frac{39}{220}$

Step 5

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

$(\frac{78}{55}, - \frac{5}{4}, - \frac{39}{220})$