Solve 1/3+[3/9-2]

$$\frac { 1 } { 3 } + [ \frac { 3 } { 9 } - 2 ]$$

Evaluate

$$-\frac{4}{3}\approx -1.333333333$$

Solution Steps

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

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

Since \frac{1}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

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

Add 1 and 1 to get 2.

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

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

$$\frac{2}{3}-\frac{6}{3}$$

Since \frac{2}{3} and \frac{6}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2-6}{3}$$

Subtract 6 from 2 to get -4.

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

Factor

$$-\frac{4}{3} = -1\frac{1}{3} = -1.3333333333333333$$