Solve 13/4+21/8-31/6-41/3

$$1 \frac { 3 } { 4 } + 2 \frac { 1 } { 8 } - \frac { 31 } { 6 } - 4 \frac { 1 } { 3 }$$

Evaluate

$$-\frac{45}{8}=-5.625$$

Solution Steps

Multiply 1 and 4 to get 4.

$$\frac{4+3}{4}+\frac{2\times 8+1}{8}-\frac{31}{6}-\frac{4\times 3+1}{3}$$

Add 4 and 3 to get 7.

$$\frac{7}{4}+\frac{2\times 8+1}{8}-\frac{31}{6}-\frac{4\times 3+1}{3}$$

Multiply 2 and 8 to get 16.

$$\frac{7}{4}+\frac{16+1}{8}-\frac{31}{6}-\frac{4\times 3+1}{3}$$

Add 16 and 1 to get 17.

$$\frac{7}{4}+\frac{17}{8}-\frac{31}{6}-\frac{4\times 3+1}{3}$$

Least common multiple of 4 and 8 is 8. Convert \frac{7}{4} and \frac{17}{8} to fractions with denominator 8.

$$\frac{14}{8}+\frac{17}{8}-\frac{31}{6}-\frac{4\times 3+1}{3}$$

Since \frac{14}{8} and \frac{17}{8} have the same denominator, add them by adding their numerators.

$$\frac{14+17}{8}-\frac{31}{6}-\frac{4\times 3+1}{3}$$

Add 14 and 17 to get 31.

$$\frac{31}{8}-\frac{31}{6}-\frac{4\times 3+1}{3}$$

Least common multiple of 8 and 6 is 24. Convert \frac{31}{8} and \frac{31}{6} to fractions with denominator 24.

$$\frac{93}{24}-\frac{124}{24}-\frac{4\times 3+1}{3}$$

Since \frac{93}{24} and \frac{124}{24} have the same denominator, subtract them by subtracting their numerators.

$$\frac{93-124}{24}-\frac{4\times 3+1}{3}$$

Subtract 124 from 93 to get -31.

$$-\frac{31}{24}-\frac{4\times 3+1}{3}$$

Multiply 4 and 3 to get 12.

$$-\frac{31}{24}-\frac{12+1}{3}$$

Add 12 and 1 to get 13.

$$-\frac{31}{24}-\frac{13}{3}$$

Least common multiple of 24 and 3 is 24. Convert -\frac{31}{24} and \frac{13}{3} to fractions with denominator 24.

$$-\frac{31}{24}-\frac{104}{24}$$

Since -\frac{31}{24} and \frac{104}{24} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-31-104}{24}$$

Subtract 104 from -31 to get -135.

$$\frac{-135}{24}$$

Reduce the fraction \frac{-135}{24} to lowest terms by extracting and canceling out 3.

$$-\frac{45}{8}$$

Factor

$$-\frac{45}{8} = -5\frac{5}{8} = -5.625$$