Finite Math Examples

$3 x \cdot 1 + 8 x \cdot 2 - 18 x \cdot 3 = 32$ , $4 x \cdot 1 + 34 x \cdot 2 - 74 x \cdot 3 = 156$ , $x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1

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

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Step 1.1

Simplify each term.

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Step 1.1.1

Multiply $3$ by $1$ .

$3 x + 8 x \cdot 2 - 18 x \cdot 3 = 32$

$4 x \cdot 1 + 34 x \cdot 2 - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.1.2

Multiply $2$ by $8$ .

$3 x + 16 x - 18 x \cdot 3 = 32$

$4 x \cdot 1 + 34 x \cdot 2 - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.1.3

Multiply $3$ by $- 18$ .

$3 x + 16 x - 54 x = 32$

$4 x \cdot 1 + 34 x \cdot 2 - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

$3 x + 16 x - 54 x = 32$

$4 x \cdot 1 + 34 x \cdot 2 - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.2

Add $3 x$ and $16 x$ .

$19 x - 54 x = 32$

$4 x \cdot 1 + 34 x \cdot 2 - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.3

Subtract $54 x$ from $19 x$ .

$- 35 x = 32$

$4 x \cdot 1 + 34 x \cdot 2 - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.4

Simplify each term.

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Step 1.4.1

Multiply $4$ by $1$ .

$- 35 x = 32$

$4 x + 34 x \cdot 2 - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.4.2

Multiply $2$ by $34$ .

$- 35 x = 32$

$4 x + 68 x - 74 x \cdot 3 = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.4.3

Multiply $3$ by $- 74$ .

$- 35 x = 32$

$4 x + 68 x - 222 x = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

$- 35 x = 32$

$4 x + 68 x - 222 x = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.5

Add $4 x$ and $68 x$ .

$- 35 x = 32$

$72 x - 222 x = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.6

Subtract $222 x$ from $72 x$ .

$- 35 x = 32$

$- 150 x = 156$

$x \cdot 1 + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.7

Simplify each term.

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Step 1.7.1

Multiply $x$ by $1$ .

$- 35 x = 32$

$- 150 x = 156$

$x + 6 x \cdot 2 - 13 x \cdot 3 = 27$

Step 1.7.2

Multiply $2$ by $6$ .

$- 35 x = 32$

$- 150 x = 156$

$x + 12 x - 13 x \cdot 3 = 27$

Step 1.7.3

Multiply $3$ by $- 13$ .

$- 35 x = 32$

$- 150 x = 156$

$x + 12 x - 39 x = 27$

$- 35 x = 32$

$- 150 x = 156$

$x + 12 x - 39 x = 27$

Step 1.8

Add $x$ and $12 x$ .

$- 35 x = 32$

$- 150 x = 156$

$13 x - 39 x = 27$

Step 1.9

Subtract $39 x$ from $13 x$ .

$- 35 x = 32$

$- 150 x = 156$

$- 26 x = 27$

$- 35 x = 32$

$- 150 x = 156$

$- 26 x = 27$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 35 & 32 \\ - 150 & 156 \\ - 26 & 27 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{35}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{35}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{35} \cdot - 35 & - \frac{1}{35} \cdot 32 \\ - 150 & 156 \\ - 26 & 27 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ - 150 & 156 \\ - 26 & 27 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + 150 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} + 150 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ - 150 + 150 \cdot 1 & 156 + 150 (- \frac{32}{35}) \\ - 26 & 27 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ 0 & \frac{132}{7} \\ - 26 & 27 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} + 26 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} + 26 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ 0 & \frac{132}{7} \\ - 26 + 26 \cdot 1 & 27 + 26 (- \frac{32}{35}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ 0 & \frac{132}{7} \\ 0 & \frac{113}{35} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{7}{132}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{7}{132}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ \frac{7}{132} \cdot 0 & \frac{7}{132} \cdot \frac{132}{7} \\ 0 & \frac{113}{35} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ 0 & 1 \\ 0 & \frac{113}{35} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} - \frac{113}{35} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{113}{35} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ 0 & 1 \\ 0 - \frac{113}{35} \cdot 0 & \frac{113}{35} - \frac{113}{35} \cdot 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{32}{35} \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 3.6

Perform the row operation $R_{1} = R_{1} + \frac{32}{35} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} + \frac{32}{35} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{32}{35} \cdot 0 & - \frac{32}{35} + \frac{32}{35} \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 0$

$0 = 1$

$0 = 0$

Step 5

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

$(error = error)$