Solve 6-3/7^2-3^2-3-7/3+7

$$\frac{ 6-3 }{ { 7 }^{ 2 } - { 3 }^{ 2 } } - \frac{ 3-7 }{ 3+7 }$$

Evaluate

$$\frac{19}{40}=0.475$$

Solution Steps

Subtract 3 from 6 to get 3.

$$\frac{3}{7^{2}-3^{2}}-\frac{3-7}{3+7}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{3}{49-3^{2}}-\frac{3-7}{3+7}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{3}{49-9}-\frac{3-7}{3+7}$$

Subtract 9 from 49 to get 40.

$$\frac{3}{40}-\frac{3-7}{3+7}$$

Subtract 7 from 3 to get -4.

$$\frac{3}{40}-\frac{-4}{3+7}$$

Add 3 and 7 to get 10.

$$\frac{3}{40}-\frac{-4}{10}$$

Reduce the fraction \frac{-4}{10} to lowest terms by extracting and canceling out 2.

$$\frac{3}{40}-\left(-\frac{2}{5}\right)$$

The opposite of -\frac{2}{5} is \frac{2}{5}.

$$\frac{3}{40}+\frac{2}{5}$$

Least common multiple of 40 and 5 is 40. Convert \frac{3}{40} and \frac{2}{5} to fractions with denominator 40.

$$\frac{3}{40}+\frac{16}{40}$$

Since \frac{3}{40} and \frac{16}{40} have the same denominator, add them by adding their numerators.

$$\frac{3+16}{40}$$

Add 3 and 16 to get 19.

$$\frac{19}{40}$$

Factor

$$\frac{19}{2 ^ {3} \cdot 5} = 0.475$$