Finite Math Examples

$x + y + z = 22$ , $x + \frac{y}{10} + \frac{z}{6} = 5$ , $\frac{x}{2} + \frac{y}{15} + \frac{z}{8} = \frac{77}{24}$

Step 1

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

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Step 1.1

Reorder terms.

$x + y + z = 22$

$x + \frac{1}{10} y + \frac{1}{6} z = 5$

$\frac{x}{2} + \frac{y}{15} + \frac{z}{8} = \frac{77}{24}$

Step 1.2

Reorder terms.

$x + y + z = 22$

$x + \frac{1}{10} y + \frac{1}{6} z = 5$

$\frac{1}{2} x + \frac{1}{15} y + \frac{1}{8} z = \frac{77}{24}$

$x + y + z = 22$

$x + \frac{1}{10} y + \frac{1}{6} z = 5$

$\frac{1}{2} x + \frac{1}{15} y + \frac{1}{8} z = \frac{77}{24}$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 1 & \frac{1}{10} & \frac{1}{6} & 5 \\ \frac{1}{2} & \frac{1}{15} & \frac{1}{8} & \frac{77}{24} \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 1 - 1 & \frac{1}{10} - 1 & \frac{1}{6} - 1 & 5 - 22 \\ \frac{1}{2} & \frac{1}{15} & \frac{1}{8} & \frac{77}{24} \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & - \frac{9}{10} & - \frac{5}{6} & - 17 \\ \frac{1}{2} & \frac{1}{15} & \frac{1}{8} & \frac{77}{24} \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & - \frac{9}{10} & - \frac{5}{6} & - 17 \\ \frac{1}{2} - \frac{1}{2} \cdot 1 & \frac{1}{15} - \frac{1}{2} \cdot 1 & \frac{1}{8} - \frac{1}{2} \cdot 1 & \frac{77}{24} - \frac{1}{2} \cdot 22 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & - \frac{9}{10} & - \frac{5}{6} & - 17 \\ 0 & - \frac{13}{30} & - \frac{3}{8} & - \frac{187}{24} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{10}{9}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{10}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ - \frac{10}{9} \cdot 0 & - \frac{10}{9} (- \frac{9}{10}) & - \frac{10}{9} (- \frac{5}{6}) & - \frac{10}{9} \cdot - 17 \\ 0 & - \frac{13}{30} & - \frac{3}{8} & - \frac{187}{24} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & 1 & \frac{25}{27} & \frac{170}{9} \\ 0 & - \frac{13}{30} & - \frac{3}{8} & - \frac{187}{24} \end{array}]$

Step 3.4

Perform the row operation $R_{3} = R_{3} + \frac{13}{30} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{3} = R_{3} + \frac{13}{30} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & 1 & \frac{25}{27} & \frac{170}{9} \\ 0 + \frac{13}{30} \cdot 0 & - \frac{13}{30} + \frac{13}{30} \cdot 1 & - \frac{3}{8} + \frac{13}{30} \cdot \frac{25}{27} & - \frac{187}{24} + \frac{13}{30} \cdot \frac{170}{9} \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & 1 & \frac{25}{27} & \frac{170}{9} \\ 0 & 0 & \frac{17}{648} & \frac{85}{216} \end{array}]$

Step 3.5

Multiply each element of $R_{3}$ by $\frac{648}{17}$ to make the entry at $3, 3$ a $1$ .

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Step 3.5.1

Multiply each element of $R_{3}$ by $\frac{648}{17}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & 1 & \frac{25}{27} & \frac{170}{9} \\ \frac{648}{17} \cdot 0 & \frac{648}{17} \cdot 0 & \frac{648}{17} \cdot \frac{17}{648} & \frac{648}{17} \cdot \frac{85}{216} \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & 1 & \frac{25}{27} & \frac{170}{9} \\ 0 & 0 & 1 & 15 \end{array}]$

Step 3.6

Perform the row operation $R_{2} = R_{2} - \frac{25}{27} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{2} = R_{2} - \frac{25}{27} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 - \frac{25}{27} \cdot 0 & 1 - \frac{25}{27} \cdot 0 & \frac{25}{27} - \frac{25}{27} \cdot 1 & \frac{170}{9} - \frac{25}{27} \cdot 15 \\ 0 & 0 & 1 & 15 \end{array}]$

Step 3.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 22 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 15 \end{array}]$

Step 3.7

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

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Step 3.7.1

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

$[\begin{array}{c} 1 - 0 & 1 - 0 & 1 - 1 & 22 - 15 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 15 \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 0 & 7 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 15 \end{array}]$

Step 3.8

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

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Step 3.8.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 0 - 0 & 7 - 5 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 15 \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 15 \end{array}]$

Step 4

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

$x = 2$

$y = 5$

$z = 15$

Step 5

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

$(2, 5, 15)$