Solve (9/8*20/2-8/10)+3/12=frac{107}{10}

$$( \frac { 9 } { 8 } * \frac { 20 } { 2 } - \frac { 8 } { 10 } ) + \frac { 3 } { 12 } = \frac { 107 } { 10 }$$

Verify

Solution Steps

Divide 20 by 2 to get 10.

$$\frac{9}{8}\times 10-\frac{8}{10}+\frac{3}{12}=\frac{107}{10}$$

Express \frac{9}{8}\times 10 as a single fraction.

$$\frac{9\times 10}{8}-\frac{8}{10}+\frac{3}{12}=\frac{107}{10}$$

Multiply 9 and 10 to get 90.

$$\frac{90}{8}-\frac{8}{10}+\frac{3}{12}=\frac{107}{10}$$

Reduce the fraction \frac{90}{8} to lowest terms by extracting and canceling out 2.

$$\frac{45}{4}-\frac{8}{10}+\frac{3}{12}=\frac{107}{10}$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$\frac{45}{4}-\frac{4}{5}+\frac{3}{12}=\frac{107}{10}$$

Least common multiple of 4 and 5 is 20. Convert \frac{45}{4} and \frac{4}{5} to fractions with denominator 20.

$$\frac{225}{20}-\frac{16}{20}+\frac{3}{12}=\frac{107}{10}$$

Since \frac{225}{20} and \frac{16}{20} have the same denominator, subtract them by subtracting their numerators.

$$\frac{225-16}{20}+\frac{3}{12}=\frac{107}{10}$$

Subtract 16 from 225 to get 209.

$$\frac{209}{20}+\frac{3}{12}=\frac{107}{10}$$

Reduce the fraction \frac{3}{12} to lowest terms by extracting and canceling out 3.

$$\frac{209}{20}+\frac{1}{4}=\frac{107}{10}$$

Least common multiple of 20 and 4 is 20. Convert \frac{209}{20} and \frac{1}{4} to fractions with denominator 20.

$$\frac{209}{20}+\frac{5}{20}=\frac{107}{10}$$

Since \frac{209}{20} and \frac{5}{20} have the same denominator, add them by adding their numerators.

$$\frac{209+5}{20}=\frac{107}{10}$$

Add 209 and 5 to get 214.

$$\frac{214}{20}=\frac{107}{10}$$

Reduce the fraction \frac{214}{20} to lowest terms by extracting and canceling out 2.

$$\frac{107}{10}=\frac{107}{10}$$

Compare \frac{107}{10} and \frac{107}{10}.

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