Pre-Algebra Examples

$\frac{1}{5} \cdot (- 12 + 9 x) = \frac{3}{10} \cdot (6 x - 8)$

Step 1

Simplify $\frac{1}{5} \cdot (- 12 + 9 x)$ .

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Step 1.1

Rewrite.

$0 + 0 + \frac{1}{5} \cdot (- 12 + 9 x) = \frac{3}{10} \cdot (6 x - 8)$

Step 1.2

Simplify by adding zeros.

$\frac{1}{5} \cdot (- 12 + 9 x) = \frac{3}{10} \cdot (6 x - 8)$

Step 1.3

Apply the distributive property.

$\frac{1}{5} \cdot - 12 + \frac{1}{5} (9 x) = \frac{3}{10} \cdot (6 x - 8)$

Step 1.4

Combine $\frac{1}{5}$ and $- 12$ .

$\frac{- 12}{5} + \frac{1}{5} (9 x) = \frac{3}{10} \cdot (6 x - 8)$

Step 1.5

Multiply $\frac{1}{5} (9 x)$ .

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Step 1.5.1

Combine $9$ and $\frac{1}{5}$ .

$\frac{- 12}{5} + \frac{9}{5} x = \frac{3}{10} \cdot (6 x - 8)$

Step 1.5.2

Combine $\frac{9}{5}$ and $x$ .

$\frac{- 12}{5} + \frac{9 x}{5} = \frac{3}{10} \cdot (6 x - 8)$

Step 1.6

Move the negative in front of the fraction.

$- \frac{12}{5} + \frac{9 x}{5} = \frac{3}{10} \cdot (6 x - 8)$

Step 2

Simplify $\frac{3}{10} \cdot (6 x - 8)$ .

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Step 2.1

Apply the distributive property.

$- \frac{12}{5} + \frac{9 x}{5} = \frac{3}{10} (6 x) + \frac{3}{10} \cdot - 8$

Step 2.2

Cancel the common factor of $2$ .

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Step 2.2.1

Factor $2$ out of $10$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{3}{2 (5)} (6 x) + \frac{3}{10} \cdot - 8$

Step 2.2.2

Factor $2$ out of $6 x$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{3}{2 (5)} (2 (3 x)) + \frac{3}{10} \cdot - 8$

Step 2.2.3

Cancel the common factor.

$- \frac{12}{5} + \frac{9 x}{5} = \frac{3}{2 \cdot 5} (2 (3 x)) + \frac{3}{10} \cdot - 8$

Step 2.2.4

Rewrite the expression.

$- \frac{12}{5} + \frac{9 x}{5} = \frac{3}{5} (3 x) + \frac{3}{10} \cdot - 8$

Step 2.3

Combine $3$ and $\frac{3}{5}$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{3 \cdot 3}{5} x + \frac{3}{10} \cdot - 8$

Step 2.4

Multiply $3$ by $3$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9}{5} x + \frac{3}{10} \cdot - 8$

Step 2.5

Combine $\frac{9}{5}$ and $x$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} + \frac{3}{10} \cdot - 8$

Step 2.6

Cancel the common factor of $2$ .

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Step 2.6.1

Factor $2$ out of $10$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} + \frac{3}{2 (5)} \cdot - 8$

Step 2.6.2

Factor $2$ out of $- 8$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} + \frac{3}{2 \cdot 5} \cdot (2 \cdot - 4)$

Step 2.6.3

Cancel the common factor.

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} + \frac{3}{2 \cdot 5} \cdot (2 \cdot - 4)$

Step 2.6.4

Rewrite the expression.

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} + \frac{3}{5} \cdot - 4$

Step 2.7

Combine $\frac{3}{5}$ and $- 4$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} + \frac{3 \cdot - 4}{5}$

Step 2.8

Simplify the expression.

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Step 2.8.1

Multiply $3$ by $- 4$ .

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} + \frac{- 12}{5}$

Step 2.8.2

Move the negative in front of the fraction.

$- \frac{12}{5} + \frac{9 x}{5} = \frac{9 x}{5} - \frac{12}{5}$

Step 3

Move all terms containing $x$ to the left side of the equation.

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Step 3.1

Subtract $\frac{9 x}{5}$ from both sides of the equation.

$- \frac{12}{5} + \frac{9 x}{5} - \frac{9 x}{5} = - \frac{12}{5}$

Step 3.2

Combine the opposite terms in $- \frac{12}{5} + \frac{9 x}{5} - \frac{9 x}{5}$ .

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Step 3.2.1

Subtract $\frac{9 x}{5}$ from $\frac{9 x}{5}$ .

$- \frac{12}{5} + 0 = - \frac{12}{5}$

Step 3.2.2

Add $- \frac{12}{5}$ and $0$ .

$- \frac{12}{5} = - \frac{12}{5}$

Step 4

Since $- \frac{12}{5} = - \frac{12}{5}$ , the equation will always be true.

Always true