Solve |-18|div3+(1/3-5/6)*12-(-2)^2

$$| - 18 | \div 3 + ( \frac { 1 } { 3 } - \frac { 5 } { 6 } ) \times 12 - ( - 2 ) ^ { 2 }$$

Evaluate

$$-4$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -18 is 18.

$$\frac{18}{3}+\left(\frac{1}{3}-\frac{5}{6}\right)\times 12-\left(-2\right)^{2}$$

Divide 18 by 3 to get 6.

$$6+\left(\frac{1}{3}-\frac{5}{6}\right)\times 12-\left(-2\right)^{2}$$

Least common multiple of 3 and 6 is 6. Convert \frac{1}{3} and \frac{5}{6} to fractions with denominator 6.

$$6+\left(\frac{2}{6}-\frac{5}{6}\right)\times 12-\left(-2\right)^{2}$$

Since \frac{2}{6} and \frac{5}{6} have the same denominator, subtract them by subtracting their numerators.

$$6+\frac{2-5}{6}\times 12-\left(-2\right)^{2}$$

Subtract 5 from 2 to get -3.

$$6+\frac{-3}{6}\times 12-\left(-2\right)^{2}$$

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

$$6-\frac{1}{2}\times 12-\left(-2\right)^{2}$$

Express -\frac{1}{2}\times 12 as a single fraction.

$$6+\frac{-12}{2}-\left(-2\right)^{2}$$

Divide -12 by 2 to get -6.

$$6-6-\left(-2\right)^{2}$$

Subtract 6 from 6 to get 0.

$$0-\left(-2\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$0-4$$

Subtract 4 from 0 to get -4.

$$-4$$

Factor

$$-4$$