Solve 2/3-8*[4-6(2/5-8)]

$$\frac { 2 } { 3 } - 8 \cdot [ 4 - 6 ( \frac { 2 } { 5 } - 8 ) ]$$

Evaluate

$$-\frac{5942}{15}\approx -396.133333333$$

Solution Steps

Convert 8 to fraction \frac{40}{5}.

$$\frac{2}{3}-8\left(4-6\left(\frac{2}{5}-\frac{40}{5}\right)\right)$$

Since \frac{2}{5} and \frac{40}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2}{3}-8\left(4-6\times \frac{2-40}{5}\right)$$

Subtract 40 from 2 to get -38.

$$\frac{2}{3}-8\left(4-6\left(-\frac{38}{5}\right)\right)$$

Express 6\left(-\frac{38}{5}\right) as a single fraction.

$$\frac{2}{3}-8\left(4-\frac{6\left(-38\right)}{5}\right)$$

Multiply 6 and -38 to get -228.

$$\frac{2}{3}-8\left(4-\frac{-228}{5}\right)$$

Fraction \frac{-228}{5} can be rewritten as -\frac{228}{5} by extracting the negative sign.

$$\frac{2}{3}-8\left(4-\left(-\frac{228}{5}\right)\right)$$

The opposite of -\frac{228}{5} is \frac{228}{5}.

$$\frac{2}{3}-8\left(4+\frac{228}{5}\right)$$

Convert 4 to fraction \frac{20}{5}.

$$\frac{2}{3}-8\left(\frac{20}{5}+\frac{228}{5}\right)$$

Since \frac{20}{5} and \frac{228}{5} have the same denominator, add them by adding their numerators.

$$\frac{2}{3}-8\times \frac{20+228}{5}$$

Add 20 and 228 to get 248.

$$\frac{2}{3}-8\times \frac{248}{5}$$

Express 8\times \frac{248}{5} as a single fraction.

$$\frac{2}{3}-\frac{8\times 248}{5}$$

Multiply 8 and 248 to get 1984.

$$\frac{2}{3}-\frac{1984}{5}$$

Least common multiple of 3 and 5 is 15. Convert \frac{2}{3} and \frac{1984}{5} to fractions with denominator 15.

$$\frac{10}{15}-\frac{5952}{15}$$

Since \frac{10}{15} and \frac{5952}{15} have the same denominator, subtract them by subtracting their numerators.

$$\frac{10-5952}{15}$$

Subtract 5952 from 10 to get -5942.

$$-\frac{5942}{15}$$

Factor

$$-\frac{5942}{15} = -396\frac{2}{15} = -396.1333333333333$$