Solve (6/3+3/6+2)*1/6+3

$$( \frac{ 6 }{ 3 } + \frac{ 3 }{ 6 } +2) \times \frac{ 1 }{ 6+3 }$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

Divide 6 by 3 to get 2.

$$\left(2+\frac{3}{6}+2\right)\times \frac{1}{6+3}$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$\left(2+\frac{1}{2}+2\right)\times \frac{1}{6+3}$$

Convert 2 to fraction \frac{4}{2}.

$$\left(\frac{4}{2}+\frac{1}{2}+2\right)\times \frac{1}{6+3}$$

Since \frac{4}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

$$\left(\frac{4+1}{2}+2\right)\times \frac{1}{6+3}$$

Add 4 and 1 to get 5.

$$\left(\frac{5}{2}+2\right)\times \frac{1}{6+3}$$

Convert 2 to fraction \frac{4}{2}.

$$\left(\frac{5}{2}+\frac{4}{2}\right)\times \frac{1}{6+3}$$

Since \frac{5}{2} and \frac{4}{2} have the same denominator, add them by adding their numerators.

$$\frac{5+4}{2}\times \frac{1}{6+3}$$

Add 5 and 4 to get 9.

$$\frac{9}{2}\times \frac{1}{6+3}$$

Add 6 and 3 to get 9.

$$\frac{9}{2}\times \frac{1}{9}$$

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

$$\frac{9\times 1}{2\times 9}$$

Cancel out 9 in both numerator and denominator.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$