Solve (1/10+3/10)-(2^3/10+6/10)

$$( \frac{ 1 }{ 10 } + \frac{ 3 }{ 10 } )-( \frac{ { 2 }^{ 3 } }{ 10 } + \frac{ 6 }{ 10 } )$$

Evaluate

$$-1$$

Solution Steps

Since \frac{1}{10} and \frac{3}{10} have the same denominator, add them by adding their numerators.

$$\frac{1+3}{10}-\left(\frac{2^{3}}{10}+\frac{6}{10}\right)$$

Add 1 and 3 to get 4.

$$\frac{4}{10}-\left(\frac{2^{3}}{10}+\frac{6}{10}\right)$$

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

$$\frac{2}{5}-\left(\frac{2^{3}}{10}+\frac{6}{10}\right)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{2}{5}-\left(\frac{8}{10}+\frac{6}{10}\right)$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$\frac{2}{5}-\left(\frac{4}{5}+\frac{6}{10}\right)$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$\frac{2}{5}-\left(\frac{4}{5}+\frac{3}{5}\right)$$

Since \frac{4}{5} and \frac{3}{5} have the same denominator, add them by adding their numerators.

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

Add 4 and 3 to get 7.

$$\frac{2}{5}-\frac{7}{5}$$

Since \frac{2}{5} and \frac{7}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2-7}{5}$$

Subtract 7 from 2 to get -5.

$$\frac{-5}{5}$$

Divide -5 by 5 to get -1.

$$-1$$

Factor

$$-1$$