Solve 5^3-4^2+5/3^2+4+2*5/5+4+3

$$\frac { 5 ^ { 3 } - 4 ^ { 2 } + 5 } { 3 ^ { 2 } } + \frac { 4 + 2 \cdot 5 } { 5 + 4 + 3 }$$

Evaluate

$$\frac{83}{6}\approx 13.833333333$$

Solution Steps

Calculate 5 to the power of 3 and get 125.

$$\frac{125-4^{2}+5}{3^{2}}+\frac{4+2\times 5}{5+4+3}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{125-16+5}{3^{2}}+\frac{4+2\times 5}{5+4+3}$$

Subtract 16 from 125 to get 109.

$$\frac{109+5}{3^{2}}+\frac{4+2\times 5}{5+4+3}$$

Add 109 and 5 to get 114.

$$\frac{114}{3^{2}}+\frac{4+2\times 5}{5+4+3}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{114}{9}+\frac{4+2\times 5}{5+4+3}$$

Reduce the fraction \frac{114}{9} to lowest terms by extracting and canceling out 3.

$$\frac{38}{3}+\frac{4+2\times 5}{5+4+3}$$

Multiply 2 and 5 to get 10.

$$\frac{38}{3}+\frac{4+10}{5+4+3}$$

Add 4 and 10 to get 14.

$$\frac{38}{3}+\frac{14}{5+4+3}$$

Add 5 and 4 to get 9.

$$\frac{38}{3}+\frac{14}{9+3}$$

Add 9 and 3 to get 12.

$$\frac{38}{3}+\frac{14}{12}$$

Reduce the fraction \frac{14}{12} to lowest terms by extracting and canceling out 2.

$$\frac{38}{3}+\frac{7}{6}$$

Least common multiple of 3 and 6 is 6. Convert \frac{38}{3} and \frac{7}{6} to fractions with denominator 6.

$$\frac{76}{6}+\frac{7}{6}$$

Since \frac{76}{6} and \frac{7}{6} have the same denominator, add them by adding their numerators.

$$\frac{76+7}{6}$$

Add 76 and 7 to get 83.

$$\frac{83}{6}$$

Factor

$$\frac{83}{2 \cdot 3} = 13\frac{5}{6} = 13.833333333333334$$