Solve (62/3-1/3)*2,5+6

$$( 6 \frac { 2 } { 3 } - \frac { 1 } { 3 } ) \cdot 2,5 + 6$$

Evaluate

$$\frac{131}{6}\approx 21,833333333$$

Solution Steps

Multiply 6 and 3 to get 18.

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

Add 18 and 2 to get 20.

$$\left(\frac{20}{3}-\frac{1}{3}\right)\times 2,5+6$$

Since \frac{20}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{20-1}{3}\times 2,5+6$$

Subtract 1 from 20 to get 19.

$$\frac{19}{3}\times 2,5+6$$

Convert decimal number 2,5 to fraction \frac{25}{10}. Reduce the fraction \frac{25}{10} to lowest terms by extracting and canceling out 5.

$$\frac{19}{3}\times \frac{5}{2}+6$$

Multiply \frac{19}{3} times \frac{5}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{19\times 5}{3\times 2}+6$$

Do the multiplications in the fraction \frac{19\times 5}{3\times 2}.

$$\frac{95}{6}+6$$

Convert 6 to fraction \frac{36}{6}.

$$\frac{95}{6}+\frac{36}{6}$$

Since \frac{95}{6} and \frac{36}{6} have the same denominator, add them by adding their numerators.

$$\frac{95+36}{6}$$

Add 95 and 36 to get 131.

$$\frac{131}{6}$$

Factor

$$\frac{131}{2 \cdot 3} = 21\frac{5}{6} = 21.833333333333332$$