Solve 3/2*|-2/3+1/6div3/8|=

$$\frac { 3 } { 2 } \times | - \frac { 2 } { 3 } + \frac { 1 } { 6 } \div \frac { 3 } { 8 } | =$$

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Solution Steps

Divide \frac{1}{6} by \frac{3}{8} by multiplying \frac{1}{6} by the reciprocal of \frac{3}{8}.

$$\frac{3}{2}|-\frac{2}{3}+\frac{1}{6}\times \frac{8}{3}|$$

Multiply \frac{1}{6} times \frac{8}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{2}|-\frac{2}{3}+\frac{1\times 8}{6\times 3}|$$

Do the multiplications in the fraction \frac{1\times 8}{6\times 3}.

$$\frac{3}{2}|-\frac{2}{3}+\frac{8}{18}|$$

Reduce the fraction \frac{8}{18} to lowest terms by extracting and canceling out 2.

$$\frac{3}{2}|-\frac{2}{3}+\frac{4}{9}|$$

Least common multiple of 3 and 9 is 9. Convert -\frac{2}{3} and \frac{4}{9} to fractions with denominator 9.

$$\frac{3}{2}|-\frac{6}{9}+\frac{4}{9}|$$

Since -\frac{6}{9} and \frac{4}{9} have the same denominator, add them by adding their numerators.

$$\frac{3}{2}|\frac{-6+4}{9}|$$

Add -6 and 4 to get -2.

$$\frac{3}{2}|-\frac{2}{9}|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{2}{9} is \frac{2}{9}.

$$\frac{3}{2}\times \frac{2}{9}$$

Multiply \frac{3}{2} times \frac{2}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3\times 2}{2\times 9}$$

Cancel out 2 in both numerator and denominator.

$$\frac{3}{9}$$

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

$$\frac{1}{3}$$

Factor

$$\frac{1}{3} = 0.3333333333333333$$