Finite Math Examples

$7 x - 3 x + 3 z = 5$ , $7 x + 2 y - z = 12$ , $35 x - 10 y + 11 z = 32$

Step 1

Subtract $3 x$ from $7 x$ .

$4 x + 3 z = 5$

$7 x + 2 y - z = 12$

$35 x - 10 y + 11 z = 32$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 4 & 0 & 3 & 5 \\ 7 & 2 & - 1 & 12 \\ 35 & - 10 & 11 & 32 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{0}{4} & \frac{3}{4} & \frac{5}{4} \\ 7 & 2 & - 1 & 12 \\ 35 & - 10 & 11 & 32 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 7 & 2 & - 1 & 12 \\ 35 & - 10 & 11 & 32 \end{array}]$

Step 3.2

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 3.2.1

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

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 7 - 7 \cdot 1 & 2 - 7 \cdot 0 & - 1 - 7 (\frac{3}{4}) & 12 - 7 (\frac{5}{4}) \\ 35 & - 10 & 11 & 32 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 2 & - \frac{25}{4} & \frac{13}{4} \\ 35 & - 10 & 11 & 32 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 35 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 35 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 2 & - \frac{25}{4} & \frac{13}{4} \\ 35 - 35 \cdot 1 & - 10 - 35 \cdot 0 & 11 - 35 (\frac{3}{4}) & 32 - 35 (\frac{5}{4}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 2 & - \frac{25}{4} & \frac{13}{4} \\ 0 & - 10 & - \frac{61}{4} & - \frac{47}{4} \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ \frac{0}{2} & \frac{2}{2} & \frac{- \frac{25}{4}}{2} & \frac{\frac{13}{4}}{2} \\ 0 & - 10 & - \frac{61}{4} & - \frac{47}{4} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 1 & - \frac{25}{8} & \frac{13}{8} \\ 0 & - 10 & - \frac{61}{4} & - \frac{47}{4} \end{array}]$

Step 3.5

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

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Step 3.5.1

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

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 1 & - \frac{25}{8} & \frac{13}{8} \\ 0 + 10 \cdot 0 & - 10 + 10 \cdot 1 & - \frac{61}{4} + 10 (- \frac{25}{8}) & - \frac{47}{4} + 10 (\frac{13}{8}) \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 1 & - \frac{25}{8} & \frac{13}{8} \\ 0 & 0 & - \frac{93}{2} & \frac{9}{2} \end{array}]$

Step 3.6

Multiply each element of $R_{3}$ by $- \frac{2}{93}$ to make the entry at $3, 3$ a $1$ .

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Step 3.6.1

Multiply each element of $R_{3}$ by $- \frac{2}{93}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 1 & - \frac{25}{8} & \frac{13}{8} \\ - \frac{2}{93} \cdot 0 & - \frac{2}{93} \cdot 0 & - \frac{2}{93} (- \frac{93}{2}) & - \frac{2}{93} \cdot \frac{9}{2} \end{array}]$

Step 3.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 1 & - \frac{25}{8} & \frac{13}{8} \\ 0 & 0 & 1 & - \frac{3}{31} \end{array}]$

Step 3.7

Perform the row operation $R_{2} = R_{2} + \frac{25}{8} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{2} = R_{2} + \frac{25}{8} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 + \frac{25}{8} \cdot 0 & 1 + \frac{25}{8} \cdot 0 & - \frac{25}{8} + \frac{25}{8} \cdot 1 & \frac{13}{8} + \frac{25}{8} (- \frac{3}{31}) \\ 0 & 0 & 1 & - \frac{3}{31} \end{array}]$

Step 3.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{3}{4} & \frac{5}{4} \\ 0 & 1 & 0 & \frac{41}{31} \\ 0 & 0 & 1 & - \frac{3}{31} \end{array}]$

Step 3.8

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.8.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & 0 - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & \frac{5}{4} - \frac{3}{4} (- \frac{3}{31}) \\ 0 & 1 & 0 & \frac{41}{31} \\ 0 & 0 & 1 & - \frac{3}{31} \end{array}]$

Step 3.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{41}{31} \\ 0 & 1 & 0 & \frac{41}{31} \\ 0 & 0 & 1 & - \frac{3}{31} \end{array}]$

Step 4

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

$x = \frac{41}{31}$

$y = \frac{41}{31}$

$z = - \frac{3}{31}$

Step 5

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

$(\frac{41}{31}, \frac{41}{31}, - \frac{3}{31})$