Solve 1/6*3/4-3/4*(1/6-frac{3}{4})=

$$\frac { 1 } { 6 } \cdot \frac { 3 } { 4 } - \frac { 3 } { 4 } \cdot ( \frac { 1 } { 6 } - \frac { 3 } { 4 } ) =$$

Evaluate

$$\frac{9}{16}=0.5625$$

Solution Steps

Multiply \frac{1}{6} times \frac{3}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 3}{6\times 4}-\frac{3}{4}\left(\frac{1}{6}-\frac{3}{4}\right)$$

Do the multiplications in the fraction \frac{1\times 3}{6\times 4}.

$$\frac{3}{24}-\frac{3}{4}\left(\frac{1}{6}-\frac{3}{4}\right)$$

Reduce the fraction \frac{3}{24} to lowest terms by extracting and canceling out 3.

$$\frac{1}{8}-\frac{3}{4}\left(\frac{1}{6}-\frac{3}{4}\right)$$

Least common multiple of 6 and 4 is 12. Convert \frac{1}{6} and \frac{3}{4} to fractions with denominator 12.

$$\frac{1}{8}-\frac{3}{4}\left(\frac{2}{12}-\frac{9}{12}\right)$$

Since \frac{2}{12} and \frac{9}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{8}-\frac{3}{4}\times \frac{2-9}{12}$$

Subtract 9 from 2 to get -7.

$$\frac{1}{8}-\frac{3}{4}\left(-\frac{7}{12}\right)$$

Multiply \frac{3}{4} times -\frac{7}{12} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{8}-\frac{3\left(-7\right)}{4\times 12}$$

Do the multiplications in the fraction \frac{3\left(-7\right)}{4\times 12}.

$$\frac{1}{8}-\frac{-21}{48}$$

Reduce the fraction \frac{-21}{48} to lowest terms by extracting and canceling out 3.

$$\frac{1}{8}-\left(-\frac{7}{16}\right)$$

The opposite of -\frac{7}{16} is \frac{7}{16}.

$$\frac{1}{8}+\frac{7}{16}$$

Least common multiple of 8 and 16 is 16. Convert \frac{1}{8} and \frac{7}{16} to fractions with denominator 16.

$$\frac{2}{16}+\frac{7}{16}$$

Since \frac{2}{16} and \frac{7}{16} have the same denominator, add them by adding their numerators.

$$\frac{2+7}{16}$$

Add 2 and 7 to get 9.

$$\frac{9}{16}$$

Factor

$$\frac{3 ^ {2}}{2 ^ {4}} = 0.5625$$