Basic Math Examples

$\frac{1}{3} \cdot (1 - \frac{1}{3}) \div \frac{5}{3} \cdot (1 + \frac{1}{3})$

Step 1

Simplify the numerator.

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Step 1.1

Write $1$ as a fraction with a common denominator.

$\frac{1}{3} (\frac{3}{3} - \frac{1}{3}) \div \frac{5}{3} \cdot (1 + \frac{1}{3})$

Step 1.2

Combine the numerators over the common denominator.

$\frac{1}{3} \cdot \frac{3 - 1}{3} \div \frac{5}{3} \cdot (1 + \frac{1}{3})$

Step 1.3

Subtract $1$ from $3$ .

$\frac{1}{3} \cdot \frac{2}{3} \div \frac{5}{3} \cdot (1 + \frac{1}{3})$

Step 2

Combine fractions.

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Step 2.1

Multiply $\frac{1}{3}$ by $\frac{2}{3}$ .

$\frac{2}{3 \cdot 3} \div \frac{5}{3} \cdot (1 + \frac{1}{3})$

Step 2.2

Multiply $3$ by $3$ .

$\frac{2}{9} \div \frac{5}{3} \cdot (1 + \frac{1}{3})$

Step 3

To divide by a fraction, multiply by its reciprocal.

$\frac{2}{9} \cdot \frac{3}{5} \cdot (1 + \frac{1}{3})$

Step 4

Cancel the common factor of $3$ .

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Step 4.1

Factor $3$ out of $9$ .

$\frac{2}{3 (3)} \cdot \frac{3}{5} \cdot (1 + \frac{1}{3})$

Step 4.2

Cancel the common factor.

$\frac{2}{3 \cdot 3} \cdot \frac{3}{5} \cdot (1 + \frac{1}{3})$

Step 4.3

Rewrite the expression.

$\frac{2}{3} \cdot \frac{1}{5} \cdot (1 + \frac{1}{3})$

Step 5

Multiply $\frac{2}{3}$ by $\frac{1}{5}$ .

$\frac{2}{3 \cdot 5} \cdot (1 + \frac{1}{3})$

Step 6

Simplify the expression.

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Step 6.1

Multiply $3$ by $5$ .

$\frac{2}{15} \cdot (1 + \frac{1}{3})$

Step 6.2

Write $1$ as a fraction with a common denominator.

$\frac{2}{15} \cdot (\frac{3}{3} + \frac{1}{3})$

Step 6.3

Combine the numerators over the common denominator.

$\frac{2}{15} \cdot \frac{3 + 1}{3}$

Step 6.4

Add $3$ and $1$ .

$\frac{2}{15} \cdot \frac{4}{3}$

Step 7

Multiply $\frac{2}{15} \cdot \frac{4}{3}$ .

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Step 7.1

Multiply $\frac{2}{15}$ by $\frac{4}{3}$ .

$\frac{2 \cdot 4}{15 \cdot 3}$

Step 7.2

Multiply $2$ by $4$ .

$\frac{8}{15 \cdot 3}$

Step 7.3

Multiply $15$ by $3$ .

$\frac{8}{45}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{8}{45}$

Decimal Form:

$0.1\overline{7}$