Finite Math Examples

$4 x - y + 3 z = 4$ , $- 7 y + 3 z = - 16$ , $- 6 + 7 y + 6 z = 58$

Step 1

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

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Step 1.1

Add $6$ to both sides of the equation.

$4 x - y + 3 z = 4$

$- 7 y + 3 z = - 16$

$7 y + 6 z = 58 + 6$

Step 1.2

Add $58$ and $6$ .

$4 x - y + 3 z = 4$

$- 7 y + 3 z = - 16$

$7 y + 6 z = 64$

$4 x - y + 3 z = 4$

$- 7 y + 3 z = - 16$

$7 y + 6 z = 64$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 4 & - 1 & 3 & 4 \\ 0 & - 7 & 3 & - 16 \\ 0 & 7 & 6 & 64 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{4}{4} & \frac{- 1}{4} & \frac{3}{4} & \frac{4}{4} \\ 0 & - 7 & 3 & - 16 \\ 0 & 7 & 6 & 64 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 & - 7 & 3 & - 16 \\ 0 & 7 & 6 & 64 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot 3 & - \frac{1}{7} \cdot - 16 \\ 0 & 7 & 6 & 64 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 & 1 & - \frac{3}{7} & \frac{16}{7} \\ 0 & 7 & 6 & 64 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 & 1 & - \frac{3}{7} & \frac{16}{7} \\ 0 - 7 \cdot 0 & 7 - 7 \cdot 1 & 6 - 7 (- \frac{3}{7}) & 64 - 7 (\frac{16}{7}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 & 1 & - \frac{3}{7} & \frac{16}{7} \\ 0 & 0 & 9 & 48 \end{array}]$

Step 3.4

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{3}$ by $\frac{1}{9}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 & 1 & - \frac{3}{7} & \frac{16}{7} \\ \frac{0}{9} & \frac{0}{9} & \frac{9}{9} & \frac{48}{9} \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 & 1 & - \frac{3}{7} & \frac{16}{7} \\ 0 & 0 & 1 & \frac{16}{3} \end{array}]$

Step 3.5

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

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Step 3.5.1

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

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 + \frac{3}{7} \cdot 0 & 1 + \frac{3}{7} \cdot 0 & - \frac{3}{7} + \frac{3}{7} \cdot 1 & \frac{16}{7} + \frac{3}{7} \cdot \frac{16}{3} \\ 0 & 0 & 1 & \frac{16}{3} \end{array}]$

Step 3.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & \frac{3}{4} & 1 \\ 0 & 1 & 0 & \frac{32}{7} \\ 0 & 0 & 1 & \frac{16}{3} \end{array}]$

Step 3.6

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 3.6.1

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

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & - \frac{1}{4} - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & 1 - \frac{3}{4} \cdot \frac{16}{3} \\ 0 & 1 & 0 & \frac{32}{7} \\ 0 & 0 & 1 & \frac{16}{3} \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{4} & 0 & - 3 \\ 0 & 1 & 0 & \frac{32}{7} \\ 0 & 0 & 1 & \frac{16}{3} \end{array}]$

Step 3.7

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.7.1

Perform the row operation $R_{1} = R_{1} + \frac{1}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{1}{4} \cdot 0 & - \frac{1}{4} + \frac{1}{4} \cdot 1 & 0 + \frac{1}{4} \cdot 0 & - 3 + \frac{1}{4} \cdot \frac{32}{7} \\ 0 & 1 & 0 & \frac{32}{7} \\ 0 & 0 & 1 & \frac{16}{3} \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{13}{7} \\ 0 & 1 & 0 & \frac{32}{7} \\ 0 & 0 & 1 & \frac{16}{3} \end{array}]$

Step 4

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

$x = - \frac{13}{7}$

$y = \frac{32}{7}$

$z = \frac{16}{3}$

Step 5

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

$(- \frac{13}{7}, \frac{32}{7}, \frac{16}{3})$