Finite Math Examples

$2 x + 7 y - 32 = 0$ , $5 x - y + 5 z = 1$ , $7 x + 6 y + 2 z = 1$

Step 1

Add $32$ to both sides of the equation.

$2 x + 7 y = 32$

$5 x - y + 5 z = 1$

$7 x + 6 y + 2 z = 1$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 2 & 7 & 0 & 32 \\ 5 & - 1 & 5 & 1 \\ 7 & 6 & 2 & 1 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{7}{2} & \frac{0}{2} & \frac{32}{2} \\ 5 & - 1 & 5 & 1 \\ 7 & 6 & 2 & 1 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 5 & - 1 & 5 & 1 \\ 7 & 6 & 2 & 1 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 5 - 5 \cdot 1 & - 1 - 5 (\frac{7}{2}) & 5 - 5 \cdot 0 & 1 - 5 \cdot 16 \\ 7 & 6 & 2 & 1 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & - \frac{37}{2} & 5 & - 79 \\ 7 & 6 & 2 & 1 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & - \frac{37}{2} & 5 & - 79 \\ 7 - 7 \cdot 1 & 6 - 7 (\frac{7}{2}) & 2 - 7 \cdot 0 & 1 - 7 \cdot 16 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & - \frac{37}{2} & 5 & - 79 \\ 0 & - \frac{37}{2} & 2 & - 111 \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $- \frac{2}{37}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $- \frac{2}{37}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ - \frac{2}{37} \cdot 0 & - \frac{2}{37} (- \frac{37}{2}) & - \frac{2}{37} \cdot 5 & - \frac{2}{37} \cdot - 79 \\ 0 & - \frac{37}{2} & 2 & - 111 \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & 1 & - \frac{10}{37} & \frac{158}{37} \\ 0 & - \frac{37}{2} & 2 & - 111 \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} + \frac{37}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} + \frac{37}{2} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & 1 & - \frac{10}{37} & \frac{158}{37} \\ 0 + \frac{37}{2} \cdot 0 & - \frac{37}{2} + \frac{37}{2} \cdot 1 & 2 + \frac{37}{2} (- \frac{10}{37}) & - 111 + \frac{37}{2} \cdot \frac{158}{37} \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & 1 & - \frac{10}{37} & \frac{158}{37} \\ 0 & 0 & - 3 & - 32 \end{array}]$

Step 3.6

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

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Step 3.6.1

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

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & 1 & - \frac{10}{37} & \frac{158}{37} \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot - 32 \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & 1 & - \frac{10}{37} & \frac{158}{37} \\ 0 & 0 & 1 & \frac{32}{3} \end{array}]$

Step 3.7

Perform the row operation $R_{2} = R_{2} + \frac{10}{37} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{2} = R_{2} + \frac{10}{37} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 + \frac{10}{37} \cdot 0 & 1 + \frac{10}{37} \cdot 0 & - \frac{10}{37} + \frac{10}{37} \cdot 1 & \frac{158}{37} + \frac{10}{37} \cdot \frac{32}{3} \\ 0 & 0 & 1 & \frac{32}{3} \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{7}{2} & 0 & 16 \\ 0 & 1 & 0 & \frac{794}{111} \\ 0 & 0 & 1 & \frac{32}{3} \end{array}]$

Step 3.8

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

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Step 3.8.1

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

$[\begin{array}{c} 1 - \frac{7}{2} \cdot 0 & \frac{7}{2} - \frac{7}{2} \cdot 1 & 0 - \frac{7}{2} \cdot 0 & 16 - \frac{7}{2} \cdot \frac{794}{111} \\ 0 & 1 & 0 & \frac{794}{111} \\ 0 & 0 & 1 & \frac{32}{3} \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1003}{111} \\ 0 & 1 & 0 & \frac{794}{111} \\ 0 & 0 & 1 & \frac{32}{3} \end{array}]$

Step 4

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

$x = - \frac{1003}{111}$

$y = \frac{794}{111}$

$z = \frac{32}{3}$

Step 5

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

$(- \frac{1003}{111}, \frac{794}{111}, \frac{32}{3})$