Solve (1/4+1/12)-(1/6+1/8)

$$( \frac { 1 } { 4 } + \frac { 1 } { 12 } ) - ( \frac { 1 } { 6 } + \frac { 1 } { 8 } )$$

Evaluate

$$\frac{1}{24}\approx 0.041666667$$

Solution Steps

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

$$\frac{3}{12}+\frac{1}{12}-\left(\frac{1}{6}+\frac{1}{8}\right)$$

Since \frac{3}{12} and \frac{1}{12} have the same denominator, add them by adding their numerators.

$$\frac{3+1}{12}-\left(\frac{1}{6}+\frac{1}{8}\right)$$

Add 3 and 1 to get 4.

$$\frac{4}{12}-\left(\frac{1}{6}+\frac{1}{8}\right)$$

Reduce the fraction \frac{4}{12} to lowest terms by extracting and canceling out 4.

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

Least common multiple of 6 and 8 is 24. Convert \frac{1}{6} and \frac{1}{8} to fractions with denominator 24.

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

Since \frac{4}{24} and \frac{3}{24} have the same denominator, add them by adding their numerators.

$$\frac{1}{3}-\frac{4+3}{24}$$

Add 4 and 3 to get 7.

$$\frac{1}{3}-\frac{7}{24}$$

Least common multiple of 3 and 24 is 24. Convert \frac{1}{3} and \frac{7}{24} to fractions with denominator 24.

$$\frac{8}{24}-\frac{7}{24}$$

Since \frac{8}{24} and \frac{7}{24} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8-7}{24}$$

Subtract 7 from 8 to get 1.

$$\frac{1}{24}$$

Factor

$$\frac{1}{2 ^ {3} \cdot 3} = 0.041666666666666664$$