Finite Math Examples

$3 x \cdot 1 - 9 x \cdot 2 + 12 x \cdot 3 = 6$ , $- 2 x \cdot 1 + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

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 $1$ .

$3 x - 9 x \cdot 2 + 12 x \cdot 3 = 6$

$- 2 x \cdot 1 + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

Step 1.1.2

Multiply $2$ by $- 9$ .

$3 x - 18 x + 12 x \cdot 3 = 6$

$- 2 x \cdot 1 + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

Step 1.1.3

Multiply $3$ by $12$ .

$3 x - 18 x + 36 x = 6$

$- 2 x \cdot 1 + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

$3 x - 18 x + 36 x = 6$

$- 2 x \cdot 1 + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

Step 1.2

Subtract $18 x$ from $3 x$ .

$- 15 x + 36 x = 6$

$- 2 x \cdot 1 + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

Step 1.3

Add $- 15 x$ and $36 x$ .

$21 x = 6$

$- 2 x \cdot 1 + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

Step 1.4

Simplify each term.

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Step 1.4.1

Multiply $- 2$ by $1$ .

$21 x = 6$

$- 2 x + 6 x \cdot 2 - 8 x \cdot 3 = - 4$

Step 1.4.2

Multiply $2$ by $6$ .

$21 x = 6$

$- 2 x + 12 x - 8 x \cdot 3 = - 4$

Step 1.4.3

Multiply $3$ by $- 8$ .

$21 x = 6$

$- 2 x + 12 x - 24 x = - 4$

$21 x = 6$

$- 2 x + 12 x - 24 x = - 4$

Step 1.5

Add $- 2 x$ and $12 x$ .

$21 x = 6$

$10 x - 24 x = - 4$

Step 1.6

Subtract $24 x$ from $10 x$ .

$21 x = 6$

$- 14 x = - 4$

$21 x = 6$

$- 14 x = - 4$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 21 & 6 \\ - 14 & - 4 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{21}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{21}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{21}{21} & \frac{6}{21} \\ - 14 & - 4 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{7} \\ - 14 & - 4 \end{array}]$

$[\begin{array}{c} 1 & \frac{2}{7} \\ - 14 & - 4 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & \frac{2}{7} \\ - 14 + 14 \cdot 1 & - 4 + 14 (\frac{2}{7}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

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

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

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

Step 4

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

$x = \frac{2}{7}$

$0 = 0$

Step 5

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

$(error = error)$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$