Solve 1/18[64/3+16)-(-8/3+4)

$$\frac { 1 } { 18 } [ \frac { 64 } { 3 } + 16 ) - ( \frac { - 8 } { 3 } + 4 )$$

Evaluate

$$\frac{20}{27}\approx 0.740740741$$

Solution Steps

Convert 16 to fraction \frac{48}{3}.

$$\frac{1}{18}\left(\frac{64}{3}+\frac{48}{3}\right)-\left(\frac{-8}{3}+4\right)$$

Since \frac{64}{3} and \frac{48}{3} have the same denominator, add them by adding their numerators.

$$\frac{1}{18}\times \frac{64+48}{3}-\left(\frac{-8}{3}+4\right)$$

Add 64 and 48 to get 112.

$$\frac{1}{18}\times \frac{112}{3}-\left(\frac{-8}{3}+4\right)$$

Multiply \frac{1}{18} times \frac{112}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1\times 112}{18\times 3}-\left(\frac{-8}{3}+4\right)$$

Do the multiplications in the fraction \frac{1\times 112}{18\times 3}.

$$\frac{112}{54}-\left(\frac{-8}{3}+4\right)$$

Reduce the fraction \frac{112}{54} to lowest terms by extracting and canceling out 2.

$$\frac{56}{27}-\left(\frac{-8}{3}+4\right)$$

Fraction \frac{-8}{3} can be rewritten as -\frac{8}{3} by extracting the negative sign.

$$\frac{56}{27}-\left(-\frac{8}{3}+4\right)$$

Convert 4 to fraction \frac{12}{3}.

$$\frac{56}{27}-\left(-\frac{8}{3}+\frac{12}{3}\right)$$

Since -\frac{8}{3} and \frac{12}{3} have the same denominator, add them by adding their numerators.

$$\frac{56}{27}-\frac{-8+12}{3}$$

Add -8 and 12 to get 4.

$$\frac{56}{27}-\frac{4}{3}$$

Least common multiple of 27 and 3 is 27. Convert \frac{56}{27} and \frac{4}{3} to fractions with denominator 27.

$$\frac{56}{27}-\frac{36}{27}$$

Since \frac{56}{27} and \frac{36}{27} have the same denominator, subtract them by subtracting their numerators.

$$\frac{56-36}{27}$$

Subtract 36 from 56 to get 20.

$$\frac{20}{27}$$

Factor

$$\frac{2 ^ {2} \cdot 5}{3 ^ {3}} = 0.7407407407407407$$