Solve -3^2*(-1/3)^2+(3/4-1/6+3/8)*(-24)

$$- 3 ^ { 2 } \times ( - \frac { 1 } { 3 } ) ^ { 2 } + ( \frac { 3 } { 4 } - \frac { 1 } { 6 } + \frac { 3 } { 8 } ) \times ( - 24 )$$

Evaluate

$$-24$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$-9\left(-\frac{1}{3}\right)^{2}+\left(\frac{3}{4}-\frac{1}{6}+\frac{3}{8}\right)\left(-24\right)$$

Calculate -\frac{1}{3} to the power of 2 and get \frac{1}{9}.

$$-9\times \frac{1}{9}+\left(\frac{3}{4}-\frac{1}{6}+\frac{3}{8}\right)\left(-24\right)$$

Multiply -9 times \frac{1}{9}.

$$-1+\left(\frac{3}{4}-\frac{1}{6}+\frac{3}{8}\right)\left(-24\right)$$

Least common multiple of 4 and 6 is 12. Convert \frac{3}{4} and \frac{1}{6} to fractions with denominator 12.

$$-1+\left(\frac{9}{12}-\frac{2}{12}+\frac{3}{8}\right)\left(-24\right)$$

Since \frac{9}{12} and \frac{2}{12} have the same denominator, subtract them by subtracting their numerators.

$$-1+\left(\frac{9-2}{12}+\frac{3}{8}\right)\left(-24\right)$$

Subtract 2 from 9 to get 7.

$$-1+\left(\frac{7}{12}+\frac{3}{8}\right)\left(-24\right)$$

Least common multiple of 12 and 8 is 24. Convert \frac{7}{12} and \frac{3}{8} to fractions with denominator 24.

$$-1+\left(\frac{14}{24}+\frac{9}{24}\right)\left(-24\right)$$

Since \frac{14}{24} and \frac{9}{24} have the same denominator, add them by adding their numerators.

$$-1+\frac{14+9}{24}\left(-24\right)$$

Add 14 and 9 to get 23.

$$-1+\frac{23}{24}\left(-24\right)$$

Express \frac{23}{24}\left(-24\right) as a single fraction.

$$-1+\frac{23\left(-24\right)}{24}$$

Multiply 23 and -24 to get -552.

$$-1+\frac{-552}{24}$$

Divide -552 by 24 to get -23.

$$-1-23$$

Subtract 23 from -1 to get -24.

$$-24$$

Factor

$$-24$$