Solve 15/6+53/4-7/8

$$1 \frac { 5 } { 6 } + 5 \frac { 3 } { 4 } - \frac { 7 } { 8 }$$

Evaluate

$$\frac{161}{24}\approx 6.708333333$$

Solution Steps

Multiply 1 and 6 to get 6.

$$\frac{6+5}{6}+\frac{5\times 4+3}{4}-\frac{7}{8}$$

Add 6 and 5 to get 11.

$$\frac{11}{6}+\frac{5\times 4+3}{4}-\frac{7}{8}$$

Multiply 5 and 4 to get 20.

$$\frac{11}{6}+\frac{20+3}{4}-\frac{7}{8}$$

Add 20 and 3 to get 23.

$$\frac{11}{6}+\frac{23}{4}-\frac{7}{8}$$

Least common multiple of 6 and 4 is 12. Convert \frac{11}{6} and \frac{23}{4} to fractions with denominator 12.

$$\frac{22}{12}+\frac{69}{12}-\frac{7}{8}$$

Since \frac{22}{12} and \frac{69}{12} have the same denominator, add them by adding their numerators.

$$\frac{22+69}{12}-\frac{7}{8}$$

Add 22 and 69 to get 91.

$$\frac{91}{12}-\frac{7}{8}$$

Least common multiple of 12 and 8 is 24. Convert \frac{91}{12} and \frac{7}{8} to fractions with denominator 24.

$$\frac{182}{24}-\frac{21}{24}$$

Since \frac{182}{24} and \frac{21}{24} have the same denominator, subtract them by subtracting their numerators.

$$\frac{182-21}{24}$$

Subtract 21 from 182 to get 161.

$$\frac{161}{24}$$

Factor

$$\frac{7 \cdot 23}{2 ^ {3} \cdot 3} = 6\frac{17}{24} = 6.708333333333333$$