Finite Math Examples

$3 x \cdot 1 + 9 x \cdot 2 - 21 x \cdot 3 = 15$ , $2 x \cdot 1 + x \cdot 2 - 4 x \cdot 3 = 0$ , $x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

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 $1$ .

$3 x + 9 x \cdot 2 - 21 x \cdot 3 = 15$

$2 x \cdot 1 + x \cdot 2 - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.1.2

Multiply $2$ by $9$ .

$3 x + 18 x - 21 x \cdot 3 = 15$

$2 x \cdot 1 + x \cdot 2 - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.1.3

Multiply $3$ by $- 21$ .

$3 x + 18 x - 63 x = 15$

$2 x \cdot 1 + x \cdot 2 - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

$3 x + 18 x - 63 x = 15$

$2 x \cdot 1 + x \cdot 2 - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.2

Add $3 x$ and $18 x$ .

$21 x - 63 x = 15$

$2 x \cdot 1 + x \cdot 2 - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.3

Subtract $63 x$ from $21 x$ .

$- 42 x = 15$

$2 x \cdot 1 + x \cdot 2 - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.4

Simplify each term.

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Step 1.4.1

Multiply $2$ by $1$ .

$- 42 x = 15$

$2 x + x \cdot 2 - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.4.2

Move $2$ to the left of $x$ .

$- 42 x = 15$

$2 x + 2 \cdot x - 4 x \cdot 3 = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.4.3

Multiply $3$ by $- 4$ .

$- 42 x = 15$

$2 x + 2 x - 12 x = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

$- 42 x = 15$

$2 x + 2 x - 12 x = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.5

Add $2 x$ and $2 x$ .

$- 42 x = 15$

$4 x - 12 x = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.6

Subtract $12 x$ from $4 x$ .

$- 42 x = 15$

$- 8 x = 0$

$x \cdot 1 + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.7

Simplify each term.

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Step 1.7.1

Multiply $x$ by $1$ .

$- 42 x = 15$

$- 8 x = 0$

$x + x \cdot 2 - 3 x \cdot 3 = 1$

Step 1.7.2

Move $2$ to the left of $x$ .

$- 42 x = 15$

$- 8 x = 0$

$x + 2 \cdot x - 3 x \cdot 3 = 1$

Step 1.7.3

Multiply $3$ by $- 3$ .

$- 42 x = 15$

$- 8 x = 0$

$x + 2 x - 9 x = 1$

$- 42 x = 15$

$- 8 x = 0$

$x + 2 x - 9 x = 1$

Step 1.8

Add $x$ and $2 x$ .

$- 42 x = 15$

$- 8 x = 0$

$3 x - 9 x = 1$

Step 1.9

Subtract $9 x$ from $3 x$ .

$- 42 x = 15$

$- 8 x = 0$

$- 6 x = 1$

$- 42 x = 15$

$- 8 x = 0$

$- 6 x = 1$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 42 & 15 \\ - 8 & 0 \\ - 6 & 1 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{42}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{42}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{42} \cdot - 42 & - \frac{1}{42} \cdot 15 \\ - 8 & 0 \\ - 6 & 1 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ - 8 & 0 \\ - 6 & 1 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ - 8 + 8 \cdot 1 & 0 + 8 (- \frac{5}{14}) \\ - 6 & 1 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ 0 & - \frac{20}{7} \\ - 6 & 1 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} + 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} + 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ 0 & - \frac{20}{7} \\ - 6 + 6 \cdot 1 & 1 + 6 (- \frac{5}{14}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ 0 & - \frac{20}{7} \\ 0 & - \frac{8}{7} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $- \frac{7}{20}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $- \frac{7}{20}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ - \frac{7}{20} \cdot 0 & - \frac{7}{20} (- \frac{20}{7}) \\ 0 & - \frac{8}{7} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ 0 & 1 \\ 0 & - \frac{8}{7} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} + \frac{8}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} + \frac{8}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ 0 & 1 \\ 0 + \frac{8}{7} \cdot 0 & - \frac{8}{7} + \frac{8}{7} \cdot 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{14} \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 3.6

Perform the row operation $R_{1} = R_{1} + \frac{5}{14} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{14} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{14} \cdot 0 & - \frac{5}{14} + \frac{5}{14} \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

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.

$(error = error)$