Solve -31/5+21/4*1/2-1/3

$$- 3 \frac { 1 } { 5 } + 2 \frac { 1 } { 4 } \times \frac { 1 } { 2 } - \frac { 1 } { 3 }$$

Evaluate

$$-\frac{289}{120}\approx -2.408333333$$

Solution Steps

Multiply 3 and 5 to get 15.

$$-\frac{15+1}{5}+\frac{2\times 4+1}{4}\times \frac{1}{2}-\frac{1}{3}$$

Add 15 and 1 to get 16.

$$-\frac{16}{5}+\frac{2\times 4+1}{4}\times \frac{1}{2}-\frac{1}{3}$$

Multiply 2 and 4 to get 8.

$$-\frac{16}{5}+\frac{8+1}{4}\times \frac{1}{2}-\frac{1}{3}$$

Add 8 and 1 to get 9.

$$-\frac{16}{5}+\frac{9}{4}\times \frac{1}{2}-\frac{1}{3}$$

Multiply \frac{9}{4} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

$$-\frac{16}{5}+\frac{9\times 1}{4\times 2}-\frac{1}{3}$$

Do the multiplications in the fraction \frac{9\times 1}{4\times 2}.

$$-\frac{16}{5}+\frac{9}{8}-\frac{1}{3}$$

Least common multiple of 5 and 8 is 40. Convert -\frac{16}{5} and \frac{9}{8} to fractions with denominator 40.

$$-\frac{128}{40}+\frac{45}{40}-\frac{1}{3}$$

Since -\frac{128}{40} and \frac{45}{40} have the same denominator, add them by adding their numerators.

$$\frac{-128+45}{40}-\frac{1}{3}$$

Add -128 and 45 to get -83.

$$-\frac{83}{40}-\frac{1}{3}$$

Least common multiple of 40 and 3 is 120. Convert -\frac{83}{40} and \frac{1}{3} to fractions with denominator 120.

$$-\frac{249}{120}-\frac{40}{120}$$

Since -\frac{249}{120} and \frac{40}{120} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-249-40}{120}$$

Subtract 40 from -249 to get -289.

$$-\frac{289}{120}$$

Factor

$$-\frac{289}{120} = -2\frac{49}{120} = -2.408333333333333$$