Solve |4/5+2/3*(-12)div(-6)-(-3)^2|+|24+(-3)^3|*(-5)

$$| \frac { 4 } { 5 } + \frac { 2 } { 3 } \times ( - 12 ) \div ( - 6 ) - ( - 3 ) ^ { 2 } | + | 24 + ( - 3 ) ^ { 3 } | \times ( - 5 )$$

Evaluate

$$-\frac{122}{15}\approx -8.133333333$$

Solution Steps

Express \frac{2}{3}\left(-12\right) as a single fraction.

$$|\frac{4}{5}+\frac{\frac{2\left(-12\right)}{3}}{-6}-\left(-3\right)^{2}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Multiply 2 and -12 to get -24.

$$|\frac{4}{5}+\frac{\frac{-24}{3}}{-6}-\left(-3\right)^{2}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Divide -24 by 3 to get -8.

$$|\frac{4}{5}+\frac{-8}{-6}-\left(-3\right)^{2}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Reduce the fraction \frac{-8}{-6} to lowest terms by extracting and canceling out -2.

$$|\frac{4}{5}+\frac{4}{3}-\left(-3\right)^{2}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Least common multiple of 5 and 3 is 15. Convert \frac{4}{5} and \frac{4}{3} to fractions with denominator 15.

$$|\frac{12}{15}+\frac{20}{15}-\left(-3\right)^{2}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Since \frac{12}{15} and \frac{20}{15} have the same denominator, add them by adding their numerators.

$$|\frac{12+20}{15}-\left(-3\right)^{2}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Add 12 and 20 to get 32.

$$|\frac{32}{15}-\left(-3\right)^{2}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Calculate -3 to the power of 2 and get 9.

$$|\frac{32}{15}-9|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Convert 9 to fraction \frac{135}{15}.

$$|\frac{32}{15}-\frac{135}{15}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Since \frac{32}{15} and \frac{135}{15} have the same denominator, subtract them by subtracting their numerators.

$$|\frac{32-135}{15}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

Subtract 135 from 32 to get -103.

$$|-\frac{103}{15}|+|24+\left(-3\right)^{3}|\left(-5\right)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{103}{15} is \frac{103}{15}.

$$\frac{103}{15}+|24+\left(-3\right)^{3}|\left(-5\right)$$

Calculate -3 to the power of 3 and get -27.

$$\frac{103}{15}+|24-27|\left(-5\right)$$

Subtract 27 from 24 to get -3.

$$\frac{103}{15}+|-3|\left(-5\right)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -3 is 3.

$$\frac{103}{15}+3\left(-5\right)$$

Multiply 3 and -5 to get -15.

$$\frac{103}{15}-15$$

Convert 15 to fraction \frac{225}{15}.

$$\frac{103}{15}-\frac{225}{15}$$

Since \frac{103}{15} and \frac{225}{15} have the same denominator, subtract them by subtracting their numerators.

$$\frac{103-225}{15}$$

Subtract 225 from 103 to get -122.

$$-\frac{122}{15}$$

Factor

$$-\frac{122}{15} = -8\frac{2}{15} = -8.133333333333333$$