Solve [n/6(4-7/3)]-1/2=

$$[ \frac { n } { 6 } ( 4 - \frac { 7 } { 3 } ) ] - \frac { 1 } { 2 } =$$

Evaluate

$$\frac{5n}{18}-\frac{1}{2}$$

Solution Steps

Convert 4 to fraction \frac{12}{3}.

$$\frac{n}{6}\left(\frac{12}{3}-\frac{7}{3}\right)-\frac{1}{2}$$

Since \frac{12}{3} and \frac{7}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{n}{6}\times \frac{12-7}{3}-\frac{1}{2}$$

Subtract 7 from 12 to get 5.

$$\frac{n}{6}\times \frac{5}{3}-\frac{1}{2}$$

Multiply \frac{n}{6} times \frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n\times 5}{6\times 3}-\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6\times 3 and 2 is 18. Multiply \frac{1}{2} times \frac{9}{9}.

$$\frac{n\times 5}{18}-\frac{9}{18}$$

Since \frac{n\times 5}{18} and \frac{9}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{n\times 5-9}{18}$$

Expand

$$\frac{5n}{18}-\frac{1}{2}$$

Solution Steps

Convert 4 to fraction \frac{12}{3}.

$$\frac{n}{6}\left(\frac{12}{3}-\frac{7}{3}\right)-\frac{1}{2}$$

Since \frac{12}{3} and \frac{7}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{n}{6}\times \frac{12-7}{3}-\frac{1}{2}$$

Subtract 7 from 12 to get 5.

$$\frac{n}{6}\times \frac{5}{3}-\frac{1}{2}$$

Multiply \frac{n}{6} times \frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n\times 5}{6\times 3}-\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6\times 3 and 2 is 18. Multiply \frac{1}{2} times \frac{9}{9}.

$$\frac{n\times 5}{18}-\frac{9}{18}$$

Since \frac{n\times 5}{18} and \frac{9}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{n\times 5-9}{18}$$