Solve 24/13*(4/5-7/12)-7:31/2

$$2 \frac { 4 } { 13 } \cdot ( \frac { 4 } { 5 } - \frac { 7 } { 12 } ) - 7 : 3 \frac { 1 } { 2 }$$

Evaluate

$$-\frac{3}{2}=-1.5$$

Solution Steps

Multiply 2 and 13 to get 26.

$$\frac{26+4}{13}\left(\frac{4}{5}-\frac{7}{12}\right)-\frac{7}{\frac{3\times 2+1}{2}}$$

Add 26 and 4 to get 30.

$$\frac{30}{13}\left(\frac{4}{5}-\frac{7}{12}\right)-\frac{7}{\frac{3\times 2+1}{2}}$$

Least common multiple of 5 and 12 is 60. Convert \frac{4}{5} and \frac{7}{12} to fractions with denominator 60.

$$\frac{30}{13}\left(\frac{48}{60}-\frac{35}{60}\right)-\frac{7}{\frac{3\times 2+1}{2}}$$

Since \frac{48}{60} and \frac{35}{60} have the same denominator, subtract them by subtracting their numerators.

$$\frac{30}{13}\times \frac{48-35}{60}-\frac{7}{\frac{3\times 2+1}{2}}$$

Subtract 35 from 48 to get 13.

$$\frac{30}{13}\times \frac{13}{60}-\frac{7}{\frac{3\times 2+1}{2}}$$

Multiply \frac{30}{13} times \frac{13}{60} by multiplying numerator times numerator and denominator times denominator.

$$\frac{30\times 13}{13\times 60}-\frac{7}{\frac{3\times 2+1}{2}}$$

Cancel out 13 in both numerator and denominator.

$$\frac{30}{60}-\frac{7}{\frac{3\times 2+1}{2}}$$

Reduce the fraction \frac{30}{60} to lowest terms by extracting and canceling out 30.

$$\frac{1}{2}-\frac{7}{\frac{3\times 2+1}{2}}$$

Divide 7 by \frac{3\times 2+1}{2} by multiplying 7 by the reciprocal of \frac{3\times 2+1}{2}.

$$\frac{1}{2}-\frac{7\times 2}{3\times 2+1}$$

Multiply 7 and 2 to get 14.

$$\frac{1}{2}-\frac{14}{3\times 2+1}$$

Multiply 3 and 2 to get 6.

$$\frac{1}{2}-\frac{14}{6+1}$$

Add 6 and 1 to get 7.

$$\frac{1}{2}-\frac{14}{7}$$

Divide 14 by 7 to get 2.

$$\frac{1}{2}-2$$

Convert 2 to fraction \frac{4}{2}.

$$\frac{1}{2}-\frac{4}{2}$$

Since \frac{1}{2} and \frac{4}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-4}{2}$$

Subtract 4 from 1 to get -3.

$$-\frac{3}{2}$$

Factor

$$-\frac{3}{2} = -1\frac{1}{2} = -1.5$$