Basic Math Examples

$\frac{5}{9} \cdot (8 + c) = - 20$

Step 1

Multiply both sides of the equation by $\frac{9}{5}$ .

$\frac{9}{5} (\frac{5}{9} \cdot (8 + c)) = \frac{9}{5} \cdot - 20$

Step 2

Simplify both sides of the equation.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $\frac{9}{5} (\frac{5}{9} \cdot (8 + c))$ .

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Step 2.1.1.1

Apply the distributive property.

$\frac{9}{5} (\frac{5}{9} \cdot 8 + \frac{5}{9} c) = \frac{9}{5} \cdot - 20$

Step 2.1.1.2

Multiply $\frac{5}{9} \cdot 8$ .

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Step 2.1.1.2.1

Combine $\frac{5}{9}$ and $8$ .

$\frac{9}{5} (\frac{5 \cdot 8}{9} + \frac{5}{9} c) = \frac{9}{5} \cdot - 20$

Step 2.1.1.2.2

Multiply $5$ by $8$ .

$\frac{9}{5} (\frac{40}{9} + \frac{5}{9} c) = \frac{9}{5} \cdot - 20$

Step 2.1.1.3

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

$\frac{9}{5} (\frac{40}{9} + \frac{5 c}{9}) = \frac{9}{5} \cdot - 20$

Step 2.1.1.4

Apply the distributive property.

$\frac{9}{5} \cdot \frac{40}{9} + \frac{9}{5} \cdot \frac{5 c}{9} = \frac{9}{5} \cdot - 20$

Step 2.1.1.5

Cancel the common factor of $9$ .

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Step 2.1.1.5.1

Cancel the common factor.

$\frac{9}{5} \cdot \frac{40}{9} + \frac{9}{5} \cdot \frac{5 c}{9} = \frac{9}{5} \cdot - 20$

Step 2.1.1.5.2

Rewrite the expression.

$\frac{1}{5} \cdot 40 + \frac{9}{5} \cdot \frac{5 c}{9} = \frac{9}{5} \cdot - 20$

Step 2.1.1.6

Cancel the common factor of $5$ .

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Step 2.1.1.6.1

Factor $5$ out of $40$ .

$\frac{1}{5} \cdot (5 (8)) + \frac{9}{5} \cdot \frac{5 c}{9} = \frac{9}{5} \cdot - 20$

Step 2.1.1.6.2

Cancel the common factor.

$\frac{1}{5} \cdot (5 \cdot 8) + \frac{9}{5} \cdot \frac{5 c}{9} = \frac{9}{5} \cdot - 20$

Step 2.1.1.6.3

Rewrite the expression.

$8 + \frac{9}{5} \cdot \frac{5 c}{9} = \frac{9}{5} \cdot - 20$

Step 2.1.1.7

Cancel the common factor of $9$ .

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Step 2.1.1.7.1

Cancel the common factor.

$8 + \frac{9}{5} \cdot \frac{5 c}{9} = \frac{9}{5} \cdot - 20$

Step 2.1.1.7.2

Rewrite the expression.

$8 + \frac{1}{5} (5 c) = \frac{9}{5} \cdot - 20$

Step 2.1.1.8

Cancel the common factor of $5$ .

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Step 2.1.1.8.1

Factor $5$ out of $5 c$ .

$8 + \frac{1}{5} (5 (c)) = \frac{9}{5} \cdot - 20$

Step 2.1.1.8.2

Cancel the common factor.

$8 + \frac{1}{5} (5 c) = \frac{9}{5} \cdot - 20$

Step 2.1.1.8.3

Rewrite the expression.

$8 + c = \frac{9}{5} \cdot - 20$

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $\frac{9}{5} \cdot - 20$ .

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Step 2.2.1.1

Cancel the common factor of $5$ .

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Step 2.2.1.1.1

Factor $5$ out of $- 20$ .

$8 + c = \frac{9}{5} \cdot (5 (- 4))$

Step 2.2.1.1.2

Cancel the common factor.

$8 + c = \frac{9}{5} \cdot (5 \cdot - 4)$

Step 2.2.1.1.3

Rewrite the expression.

$8 + c = 9 \cdot - 4$

Step 2.2.1.2

Multiply $9$ by $- 4$ .

$8 + c = - 36$

Step 3

Move all terms not containing $c$ to the right side of the equation.

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Step 3.1

Subtract $8$ from both sides of the equation.

$c = - 36 - 8$

Step 3.2

Subtract $8$ from $- 36$ .

$c = - 44$