Solve 1441/10*5/7-31/3

$$14 \frac { 41 } { 10 } \times \frac { 5 } { 7 } - 3 \frac { 1 } { 3 }$$

Evaluate

$$\frac{403}{42}\approx 9.595238095$$

Solution Steps

Multiply 14 and 10 to get 140.

$$\frac{140+41}{10}\times \frac{5}{7}-\frac{3\times 3+1}{3}$$

Add 140 and 41 to get 181.

$$\frac{181}{10}\times \frac{5}{7}-\frac{3\times 3+1}{3}$$

Multiply \frac{181}{10} times \frac{5}{7} by multiplying numerator times numerator and denominator times denominator.

$$\frac{181\times 5}{10\times 7}-\frac{3\times 3+1}{3}$$

Do the multiplications in the fraction \frac{181\times 5}{10\times 7}.

$$\frac{905}{70}-\frac{3\times 3+1}{3}$$

Reduce the fraction \frac{905}{70} to lowest terms by extracting and canceling out 5.

$$\frac{181}{14}-\frac{3\times 3+1}{3}$$

Multiply 3 and 3 to get 9.

$$\frac{181}{14}-\frac{9+1}{3}$$

Add 9 and 1 to get 10.

$$\frac{181}{14}-\frac{10}{3}$$

Least common multiple of 14 and 3 is 42. Convert \frac{181}{14} and \frac{10}{3} to fractions with denominator 42.

$$\frac{543}{42}-\frac{140}{42}$$

Since \frac{543}{42} and \frac{140}{42} have the same denominator, subtract them by subtracting their numerators.

$$\frac{543-140}{42}$$

Subtract 140 from 543 to get 403.

$$\frac{403}{42}$$

Factor

$$\frac{13 \cdot 31}{2 \cdot 3 \cdot 7} = 9\frac{25}{42} = 9.595238095238095$$