Solve (frac{2}3^2+8/9):(2-1/3)=

$$( \frac { 2 } { 3 } ^ { 2 } + \frac { 8 } { 9 } ) : ( 2 - \frac { 1 } { 3 } ) =$$

Evaluate

$$\frac{4}{5}=0.8$$

Solution Steps

Calculate \frac{2}{3} to the power of 2 and get \frac{4}{9}.

$$\frac{\frac{4}{9}+\frac{8}{9}}{2-\frac{1}{3}}$$

Since \frac{4}{9} and \frac{8}{9} have the same denominator, add them by adding their numerators.

$$\frac{\frac{4+8}{9}}{2-\frac{1}{3}}$$

Add 4 and 8 to get 12.

$$\frac{\frac{12}{9}}{2-\frac{1}{3}}$$

Reduce the fraction \frac{12}{9} to lowest terms by extracting and canceling out 3.

$$\frac{\frac{4}{3}}{2-\frac{1}{3}}$$

Convert 2 to fraction \frac{6}{3}.

$$\frac{\frac{4}{3}}{\frac{6}{3}-\frac{1}{3}}$$

Since \frac{6}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{4}{3}}{\frac{6-1}{3}}$$

Subtract 1 from 6 to get 5.

$$\frac{\frac{4}{3}}{\frac{5}{3}}$$

Divide \frac{4}{3} by \frac{5}{3} by multiplying \frac{4}{3} by the reciprocal of \frac{5}{3}.

$$\frac{4}{3}\times \frac{3}{5}$$

Multiply \frac{4}{3} times \frac{3}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{4\times 3}{3\times 5}$$

Cancel out 3 in both numerator and denominator.

$$\frac{4}{5}$$

Factor

$$\frac{2 ^ {2}}{5} = 0.8$$