Solve 1/3*3/6-4/18=

$$\frac { 1 } { 3 } \cdot \frac { 3 } { 6 } - \frac { 4 } { 18 } =$$

Evaluate

$$-\frac{1}{18}\approx -0.055555556$$

Solution Steps

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

$$\frac{1}{3}\times \frac{1}{2}-\frac{4}{18}$$

Multiply \frac{1}{3} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 1}{3\times 2}-\frac{4}{18}$$

Do the multiplications in the fraction \frac{1\times 1}{3\times 2}.

$$\frac{1}{6}-\frac{4}{18}$$

Reduce the fraction \frac{4}{18} to lowest terms by extracting and canceling out 2.

$$\frac{1}{6}-\frac{2}{9}$$

Least common multiple of 6 and 9 is 18. Convert \frac{1}{6} and \frac{2}{9} to fractions with denominator 18.

$$\frac{3}{18}-\frac{4}{18}$$

Since \frac{3}{18} and \frac{4}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3-4}{18}$$

Subtract 4 from 3 to get -1.

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

Factor

$$-\frac{1}{18} = -0.05555555555555555$$