Solve 2/n+3/12-1/3/frac{2{4}-frac{1}{2}+frac{1}{8}}

$$\frac { \frac { 2 } { n } + \frac { 3 } { 12 } - \frac { 1 } { 3 } } { \frac { 2 } { 4 } - \frac { 1 } { 2 } + \frac { 1 } { 8 } }$$

Evaluate

$$-\frac{2}{3}+\frac{16}{n}$$

Short Solution Steps

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

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

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

$$\frac{\frac{2}{n}+\frac{3}{12}-\frac{4}{12}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Since \frac{3}{12} and \frac{4}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2}{n}+\frac{3-4}{12}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Subtract 4 from 3 to get -1.

$$\frac{\frac{2}{n}-\frac{1}{12}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n and 12 is 12n. Multiply \frac{2}{n} times \frac{12}{12}. Multiply \frac{1}{12} times \frac{n}{n}.

$$\frac{\frac{2\times 12}{12n}-\frac{n}{12n}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

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

$$\frac{\frac{2\times 12-n}{12n}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Do the multiplications in 2\times 12-n.

$$\frac{\frac{24-n}{12n}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{\frac{24-n}{12n}}{\frac{1}{2}-\frac{1}{2}+\frac{1}{8}}$$

Subtract \frac{1}{2} from \frac{1}{2} to get 0.

$$\frac{\frac{24-n}{12n}}{\frac{1}{8}}$$

Divide \frac{24-n}{12n} by \frac{1}{8} by multiplying \frac{24-n}{12n} by the reciprocal of \frac{1}{8}.

$$\frac{\left(24-n\right)\times 8}{12n}$$

Cancel out 4 in both numerator and denominator.

$$\frac{2\left(-n+24\right)}{3n}$$

Use the distributive property to multiply 2 by -n+24.

$$\frac{-2n+48}{3n}$$

Expand

$$-\frac{2}{3}+\frac{16}{n}$$

Short Solution Steps

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

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

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

$$\frac{\frac{2}{n}+\frac{3}{12}-\frac{4}{12}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Since \frac{3}{12} and \frac{4}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{2}{n}+\frac{3-4}{12}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Subtract 4 from 3 to get -1.

$$\frac{\frac{2}{n}-\frac{1}{12}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of n and 12 is 12n. Multiply \frac{2}{n} times \frac{12}{12}. Multiply \frac{1}{12} times \frac{n}{n}.

$$\frac{\frac{2\times 12}{12n}-\frac{n}{12n}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

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

$$\frac{\frac{2\times 12-n}{12n}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Do the multiplications in 2\times 12-n.

$$\frac{\frac{24-n}{12n}}{\frac{2}{4}-\frac{1}{2}+\frac{1}{8}}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{\frac{24-n}{12n}}{\frac{1}{2}-\frac{1}{2}+\frac{1}{8}}$$

Subtract \frac{1}{2} from \frac{1}{2} to get 0.

$$\frac{\frac{24-n}{12n}}{\frac{1}{8}}$$

Divide \frac{24-n}{12n} by \frac{1}{8} by multiplying \frac{24-n}{12n} by the reciprocal of \frac{1}{8}.

$$\frac{\left(24-n\right)\times 8}{12n}$$

Cancel out 4 in both numerator and denominator.

$$\frac{2\left(-n+24\right)}{3n}$$

Use the distributive property to multiply 2 by -n+24.

$$\frac{-2n+48}{3n}$$