Finite Math Examples

$3 x \cdot 1 - 5 x \cdot 2 = - 2$ , $- 6 x \cdot 1 + 10 x \cdot 2 = 4$

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 $1$ .

$3 x - 5 x \cdot 2 = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

Step 1.1.2

Multiply $2$ by $- 5$ .

$3 x - 10 x = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

$3 x - 10 x = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

Step 1.2

Subtract $10 x$ from $3 x$ .

$- 7 x = - 2$

$- 6 x \cdot 1 + 10 x \cdot 2 = 4$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $- 6$ by $1$ .

$- 7 x = - 2$

$- 6 x + 10 x \cdot 2 = 4$

Step 1.3.2

Multiply $2$ by $10$ .

$- 7 x = - 2$

$- 6 x + 20 x = 4$

$- 7 x = - 2$

$- 6 x + 20 x = 4$

Step 1.4

Add $- 6 x$ and $20 x$ .

$- 7 x = - 2$

$14 x = 4$

$- 7 x = - 2$

$14 x = 4$

Step 2

Write the system as a matrix.

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

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

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

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 $R_{2}$ .

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

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.

$(error = error)$