Solve (7/8+1/5)*(3/2-5/7)

$$( \frac { 7 } { 8 } + \frac { 1 } { 5 } ) \times ( \frac { 3 } { 2 } - \frac { 5 } { 7 } )$$

Evaluate

$$\frac{473}{560}\approx 0.844642857$$

Solution Steps

Least common multiple of 8 and 5 is 40. Convert \frac{7}{8} and \frac{1}{5} to fractions with denominator 40.

$$\left(\frac{35}{40}+\frac{8}{40}\right)\left(\frac{3}{2}-\frac{5}{7}\right)$$

Since \frac{35}{40} and \frac{8}{40} have the same denominator, add them by adding their numerators.

$$\frac{35+8}{40}\left(\frac{3}{2}-\frac{5}{7}\right)$$

Add 35 and 8 to get 43.

$$\frac{43}{40}\left(\frac{3}{2}-\frac{5}{7}\right)$$

Least common multiple of 2 and 7 is 14. Convert \frac{3}{2} and \frac{5}{7} to fractions with denominator 14.

$$\frac{43}{40}\left(\frac{21}{14}-\frac{10}{14}\right)$$

Since \frac{21}{14} and \frac{10}{14} have the same denominator, subtract them by subtracting their numerators.

$$\frac{43}{40}\times \frac{21-10}{14}$$

Subtract 10 from 21 to get 11.

$$\frac{43}{40}\times \frac{11}{14}$$

Multiply \frac{43}{40} times \frac{11}{14} by multiplying numerator times numerator and denominator times denominator.

$$\frac{43\times 11}{40\times 14}$$

Do the multiplications in the fraction \frac{43\times 11}{40\times 14}.

$$\frac{473}{560}$$

Factor

$$\frac{11 \cdot 43}{2 ^ {4} \cdot 5 \cdot 7} = 0.8446428571428571$$