Solve (7/8+21/4-1/20)*(9/7+frac{1}{16})=

$$( \frac { 7 } { 8 } + \frac { 21 } { 4 } - \frac { 1 } { 20 } ) \times ( \frac { 9 } { 7 } + \frac { 1 } { 16 } ) =$$

Evaluate

$$\frac{36693}{4480}\approx 8.190401786$$

Solution Steps

Least common multiple of 8 and 4 is 8. Convert \frac{7}{8} and \frac{21}{4} to fractions with denominator 8.

$$\left(\frac{7}{8}+\frac{42}{8}-\frac{1}{20}\right)\left(\frac{9}{7}+\frac{1}{16}\right)$$

Since \frac{7}{8} and \frac{42}{8} have the same denominator, add them by adding their numerators.

$$\left(\frac{7+42}{8}-\frac{1}{20}\right)\left(\frac{9}{7}+\frac{1}{16}\right)$$

Add 7 and 42 to get 49.

$$\left(\frac{49}{8}-\frac{1}{20}\right)\left(\frac{9}{7}+\frac{1}{16}\right)$$

Least common multiple of 8 and 20 is 40. Convert \frac{49}{8} and \frac{1}{20} to fractions with denominator 40.

$$\left(\frac{245}{40}-\frac{2}{40}\right)\left(\frac{9}{7}+\frac{1}{16}\right)$$

Since \frac{245}{40} and \frac{2}{40} have the same denominator, subtract them by subtracting their numerators.

$$\frac{245-2}{40}\left(\frac{9}{7}+\frac{1}{16}\right)$$

Subtract 2 from 245 to get 243.

$$\frac{243}{40}\left(\frac{9}{7}+\frac{1}{16}\right)$$

Least common multiple of 7 and 16 is 112. Convert \frac{9}{7} and \frac{1}{16} to fractions with denominator 112.

$$\frac{243}{40}\left(\frac{144}{112}+\frac{7}{112}\right)$$

Since \frac{144}{112} and \frac{7}{112} have the same denominator, add them by adding their numerators.

$$\frac{243}{40}\times \frac{144+7}{112}$$

Add 144 and 7 to get 151.

$$\frac{243}{40}\times \frac{151}{112}$$

Multiply \frac{243}{40} times \frac{151}{112} by multiplying numerator times numerator and denominator times denominator.

$$\frac{243\times 151}{40\times 112}$$

Do the multiplications in the fraction \frac{243\times 151}{40\times 112}.

$$\frac{36693}{4480}$$

Factor

$$\frac{3 ^ {5} \cdot 151}{2 ^ {7} \cdot 5 \cdot 7} = 8\frac{853}{4480} = 8.190401785714286$$