Solve 40/3-8+81/4-27-4+8

$$\frac { 40 } { 3 } - 8 + \frac { 81 } { 4 } - 27 - 4 + 8$$

Evaluate

$$\frac{31}{12}\approx 2.583333333$$

Solution Steps

Convert 8 to fraction \frac{24}{3}.

$$\frac{40}{3}-\frac{24}{3}+\frac{81}{4}-27-4+8$$

Since \frac{40}{3} and \frac{24}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{40-24}{3}+\frac{81}{4}-27-4+8$$

Subtract 24 from 40 to get 16.

$$\frac{16}{3}+\frac{81}{4}-27-4+8$$

Least common multiple of 3 and 4 is 12. Convert \frac{16}{3} and \frac{81}{4} to fractions with denominator 12.

$$\frac{64}{12}+\frac{243}{12}-27-4+8$$

Since \frac{64}{12} and \frac{243}{12} have the same denominator, add them by adding their numerators.

$$\frac{64+243}{12}-27-4+8$$

Add 64 and 243 to get 307.

$$\frac{307}{12}-27-4+8$$

Convert 27 to fraction \frac{324}{12}.

$$\frac{307}{12}-\frac{324}{12}-4+8$$

Since \frac{307}{12} and \frac{324}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{307-324}{12}-4+8$$

Subtract 324 from 307 to get -17.

$$-\frac{17}{12}-4+8$$

Convert 4 to fraction \frac{48}{12}.

$$-\frac{17}{12}-\frac{48}{12}+8$$

Since -\frac{17}{12} and \frac{48}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-17-48}{12}+8$$

Subtract 48 from -17 to get -65.

$$-\frac{65}{12}+8$$

Convert 8 to fraction \frac{96}{12}.

$$-\frac{65}{12}+\frac{96}{12}$$

Since -\frac{65}{12} and \frac{96}{12} have the same denominator, add them by adding their numerators.

$$\frac{-65+96}{12}$$

Add -65 and 96 to get 31.

$$\frac{31}{12}$$

Factor

$$\frac{31}{2 ^ {2} \cdot 3} = 2\frac{7}{12} = 2.5833333333333335$$