Solve (19/24-7/12+3/8)*Delta

$$( \frac { 19 } { 24 } - \frac { 7 } { 12 } + \frac { 3 } { 8 } ) \cdot \Delta$$

Evaluate

$$\frac{7\Delta }{12}$$

Solution Steps

Least common multiple of 24 and 12 is 24. Convert \frac{19}{24} and \frac{7}{12} to fractions with denominator 24.

$$\left(\frac{19}{24}-\frac{14}{24}+\frac{3}{8}\right)\Delta$$

Since \frac{19}{24} and \frac{14}{24} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{19-14}{24}+\frac{3}{8}\right)\Delta$$

Subtract 14 from 19 to get 5.

$$\left(\frac{5}{24}+\frac{3}{8}\right)\Delta$$

Least common multiple of 24 and 8 is 24. Convert \frac{5}{24} and \frac{3}{8} to fractions with denominator 24.

$$\left(\frac{5}{24}+\frac{9}{24}\right)\Delta$$

Since \frac{5}{24} and \frac{9}{24} have the same denominator, add them by adding their numerators.

$$\frac{5+9}{24}\Delta$$

Add 5 and 9 to get 14.

$$\frac{14}{24}\Delta$$

Reduce the fraction \frac{14}{24} to lowest terms by extracting and canceling out 2.

$$\frac{7}{12}\Delta$$

Expand

$$\frac{7\Delta }{12}$$

Solution Steps

Least common multiple of 24 and 12 is 24. Convert \frac{19}{24} and \frac{7}{12} to fractions with denominator 24.

$$\left(\frac{19}{24}-\frac{14}{24}+\frac{3}{8}\right)\Delta$$

Since \frac{19}{24} and \frac{14}{24} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{19-14}{24}+\frac{3}{8}\right)\Delta$$

Subtract 14 from 19 to get 5.

$$\left(\frac{5}{24}+\frac{3}{8}\right)\Delta$$

Least common multiple of 24 and 8 is 24. Convert \frac{5}{24} and \frac{3}{8} to fractions with denominator 24.

$$\left(\frac{5}{24}+\frac{9}{24}\right)\Delta$$

Since \frac{5}{24} and \frac{9}{24} have the same denominator, add them by adding their numerators.

$$\frac{5+9}{24}\Delta$$

Add 5 and 9 to get 14.

$$\frac{14}{24}\Delta$$

Reduce the fraction \frac{14}{24} to lowest terms by extracting and canceling out 2.

$$\frac{7}{12}\Delta$$