Solve (5-11/3*11/6)*27/31

$$( 5 - 1 \frac { 1 } { 3 } \cdot 1 \frac { 1 } { 6 } ) \cdot \frac { 27 } { 31 }$$

Evaluate

$$3$$

Solution Steps

Multiply 1 and 3 to get 3.

$$\left(5-\frac{3+1}{3}\times \frac{1\times 6+1}{6}\right)\times \frac{27}{31}$$

Add 3 and 1 to get 4.

$$\left(5-\frac{4}{3}\times \frac{1\times 6+1}{6}\right)\times \frac{27}{31}$$

Multiply 1 and 6 to get 6.

$$\left(5-\frac{4}{3}\times \frac{6+1}{6}\right)\times \frac{27}{31}$$

Add 6 and 1 to get 7.

$$\left(5-\frac{4}{3}\times \frac{7}{6}\right)\times \frac{27}{31}$$

Multiply \frac{4}{3} times \frac{7}{6} by multiplying numerator times numerator and denominator times denominator.

$$\left(5-\frac{4\times 7}{3\times 6}\right)\times \frac{27}{31}$$

Do the multiplications in the fraction \frac{4\times 7}{3\times 6}.

$$\left(5-\frac{28}{18}\right)\times \frac{27}{31}$$

Reduce the fraction \frac{28}{18} to lowest terms by extracting and canceling out 2.

$$\left(5-\frac{14}{9}\right)\times \frac{27}{31}$$

Convert 5 to fraction \frac{45}{9}.

$$\left(\frac{45}{9}-\frac{14}{9}\right)\times \frac{27}{31}$$

Since \frac{45}{9} and \frac{14}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{45-14}{9}\times \frac{27}{31}$$

Subtract 14 from 45 to get 31.

$$\frac{31}{9}\times \frac{27}{31}$$

Multiply \frac{31}{9} times \frac{27}{31} by multiplying numerator times numerator and denominator times denominator.

$$\frac{31\times 27}{9\times 31}$$

Cancel out 31 in both numerator and denominator.

$$\frac{27}{9}$$

Divide 27 by 9 to get 3.

$$3$$

Factor

$$3$$