Solve 1/2*[2/3*3/4-(1/3+frac{1}{9})]

$$\frac { 1 } { 2 } \cdot [ \frac { 2 } { 3 } \cdot \frac { 3 } { 4 } - ( \frac { 1 } { 3 } + \frac { 1 } { 9 } ) ]$$

Evaluate

$$\frac{1}{36}\approx 0.027777778$$

Solution Steps

Multiply \frac{2}{3} times \frac{3}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}\left(\frac{2\times 3}{3\times 4}-\left(\frac{1}{3}+\frac{1}{9}\right)\right)$$

Cancel out 3 in both numerator and denominator.

$$\frac{1}{2}\left(\frac{2}{4}-\left(\frac{1}{3}+\frac{1}{9}\right)\right)$$

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

$$\frac{1}{2}\left(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{9}\right)\right)$$

Least common multiple of 3 and 9 is 9. Convert \frac{1}{3} and \frac{1}{9} to fractions with denominator 9.

$$\frac{1}{2}\left(\frac{1}{2}-\left(\frac{3}{9}+\frac{1}{9}\right)\right)$$

Since \frac{3}{9} and \frac{1}{9} have the same denominator, add them by adding their numerators.

$$\frac{1}{2}\left(\frac{1}{2}-\frac{3+1}{9}\right)$$

Add 3 and 1 to get 4.

$$\frac{1}{2}\left(\frac{1}{2}-\frac{4}{9}\right)$$

Least common multiple of 2 and 9 is 18. Convert \frac{1}{2} and \frac{4}{9} to fractions with denominator 18.

$$\frac{1}{2}\left(\frac{9}{18}-\frac{8}{18}\right)$$

Since \frac{9}{18} and \frac{8}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{2}\times \frac{9-8}{18}$$

Subtract 8 from 9 to get 1.

$$\frac{1}{2}\times \frac{1}{18}$$

Multiply \frac{1}{2} times \frac{1}{18} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 1}{2\times 18}$$

Do the multiplications in the fraction \frac{1\times 1}{2\times 18}.

$$\frac{1}{36}$$

Factor

$$\frac{1}{2 ^ {2} \cdot 3 ^ {2}} = 0.027777777777777776$$