Solve {l}{1/40+1/60}{3+2/120}{5/120}

$$\left. \begin{array} { l } { \frac { 1 } { 40 } + \frac { 1 } { 60 } } \\ { \frac { 3 + 2 } { 120 } } \\ { \frac { 5 } { 120 } } \end{array} \right.$$

Sort

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

Solution Steps

Least common multiple of 40 and 60 is 120. Convert \frac{1}{40} and \frac{1}{60} to fractions with denominator 120.

$$sort(\frac{3}{120}+\frac{2}{120},\frac{3+2}{120},\frac{5}{120})$$

Since \frac{3}{120} and \frac{2}{120} have the same denominator, add them by adding their numerators.

$$sort(\frac{3+2}{120},\frac{3+2}{120},\frac{5}{120})$$

Add 3 and 2 to get 5.

$$sort(\frac{5}{120},\frac{3+2}{120},\frac{5}{120})$$

Reduce the fraction \frac{5}{120} to lowest terms by extracting and canceling out 5.

$$sort(\frac{1}{24},\frac{3+2}{120},\frac{5}{120})$$

Add 3 and 2 to get 5.

$$sort(\frac{1}{24},\frac{5}{120},\frac{5}{120})$$

Reduce the fraction \frac{5}{120} to lowest terms by extracting and canceling out 5.

$$sort(\frac{1}{24},\frac{1}{24},\frac{5}{120})$$

Reduce the fraction \frac{5}{120} to lowest terms by extracting and canceling out 5.

$$sort(\frac{1}{24},\frac{1}{24},\frac{1}{24})$$

The list values are already in order.

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

Evaluate

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