Solve 1/6-1/3/1+(frac{1{6})(1/3)}

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

Evaluate

$$-\frac{3}{19}\approx -0.157894737$$

Solution Steps

Least common multiple of 6 and 3 is 6. Convert \frac{1}{6} and \frac{1}{3} to fractions with denominator 6.

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

Since \frac{1}{6} and \frac{2}{6} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 2 from 1 to get -1.

$$\frac{-\frac{1}{6}}{1+\frac{1}{6}\times \frac{1}{3}}$$

Multiply \frac{1}{6} times \frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\frac{1}{6}}{1+\frac{1\times 1}{6\times 3}}$$

Do the multiplications in the fraction \frac{1\times 1}{6\times 3}.

$$\frac{-\frac{1}{6}}{1+\frac{1}{18}}$$

Convert 1 to fraction \frac{18}{18}.

$$\frac{-\frac{1}{6}}{\frac{18}{18}+\frac{1}{18}}$$

Since \frac{18}{18} and \frac{1}{18} have the same denominator, add them by adding their numerators.

$$\frac{-\frac{1}{6}}{\frac{18+1}{18}}$$

Add 18 and 1 to get 19.

$$\frac{-\frac{1}{6}}{\frac{19}{18}}$$

Divide -\frac{1}{6} by \frac{19}{18} by multiplying -\frac{1}{6} by the reciprocal of \frac{19}{18}.

$$-\frac{1}{6}\times \frac{18}{19}$$

Multiply -\frac{1}{6} times \frac{18}{19} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-18}{6\times 19}$$

Do the multiplications in the fraction \frac{-18}{6\times 19}.

$$\frac{-18}{114}$$

Reduce the fraction \frac{-18}{114} to lowest terms by extracting and canceling out 6.

$$-\frac{3}{19}$$

Factor

$$-\frac{3}{19} = -0.15789473684210525$$