Solve 180-31/5-(21/3+1/6-1/9)=

$$180 - 3 \frac { 1 } { 5 } - ( 2 \frac { 1 } { 3 } + \frac { 1 } { 6 } - \frac { 1 } { 9 } ) =$$

Evaluate

$$\frac{15697}{90}\approx 174.411111111$$

Solution Steps

Multiply 3 and 5 to get 15.

$$180-\frac{15+1}{5}-\left(\frac{2\times 3+1}{3}+\frac{1}{6}-\frac{1}{9}\right)$$

Add 15 and 1 to get 16.

$$180-\frac{16}{5}-\left(\frac{2\times 3+1}{3}+\frac{1}{6}-\frac{1}{9}\right)$$

Convert 180 to fraction \frac{900}{5}.

$$\frac{900}{5}-\frac{16}{5}-\left(\frac{2\times 3+1}{3}+\frac{1}{6}-\frac{1}{9}\right)$$

Since \frac{900}{5} and \frac{16}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{900-16}{5}-\left(\frac{2\times 3+1}{3}+\frac{1}{6}-\frac{1}{9}\right)$$

Subtract 16 from 900 to get 884.

$$\frac{884}{5}-\left(\frac{2\times 3+1}{3}+\frac{1}{6}-\frac{1}{9}\right)$$

Multiply 2 and 3 to get 6.

$$\frac{884}{5}-\left(\frac{6+1}{3}+\frac{1}{6}-\frac{1}{9}\right)$$

Add 6 and 1 to get 7.

$$\frac{884}{5}-\left(\frac{7}{3}+\frac{1}{6}-\frac{1}{9}\right)$$

Least common multiple of 3 and 6 is 6. Convert \frac{7}{3} and \frac{1}{6} to fractions with denominator 6.

$$\frac{884}{5}-\left(\frac{14}{6}+\frac{1}{6}-\frac{1}{9}\right)$$

Since \frac{14}{6} and \frac{1}{6} have the same denominator, add them by adding their numerators.

$$\frac{884}{5}-\left(\frac{14+1}{6}-\frac{1}{9}\right)$$

Add 14 and 1 to get 15.

$$\frac{884}{5}-\left(\frac{15}{6}-\frac{1}{9}\right)$$

Reduce the fraction \frac{15}{6} to lowest terms by extracting and canceling out 3.

$$\frac{884}{5}-\left(\frac{5}{2}-\frac{1}{9}\right)$$

Least common multiple of 2 and 9 is 18. Convert \frac{5}{2} and \frac{1}{9} to fractions with denominator 18.

$$\frac{884}{5}-\left(\frac{45}{18}-\frac{2}{18}\right)$$

Since \frac{45}{18} and \frac{2}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{884}{5}-\frac{45-2}{18}$$

Subtract 2 from 45 to get 43.

$$\frac{884}{5}-\frac{43}{18}$$

Least common multiple of 5 and 18 is 90. Convert \frac{884}{5} and \frac{43}{18} to fractions with denominator 90.

$$\frac{15912}{90}-\frac{215}{90}$$

Since \frac{15912}{90} and \frac{215}{90} have the same denominator, subtract them by subtracting their numerators.

$$\frac{15912-215}{90}$$

Subtract 215 from 15912 to get 15697.

$$\frac{15697}{90}$$

Factor

$$\frac{11 \cdot 1427}{2 \cdot 3 ^ {2} \cdot 5} = 174\frac{37}{90} = 174.4111111111111$$