Solve (3/7*49/9-40/6)+02

$$( \frac { 3 } { 7 } \times \frac { 49 } { 9 } - 40 / 6 ) + 02$$

Evaluate

$$-\frac{13}{3}\approx -4.333333333$$

Solution Steps

Multiply \frac{3}{7} times \frac{49}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3\times 49}{7\times 9}-\frac{40}{6}+0\times 2$$

Do the multiplications in the fraction \frac{3\times 49}{7\times 9}.

$$\frac{147}{63}-\frac{40}{6}+0\times 2$$

Reduce the fraction \frac{147}{63} to lowest terms by extracting and canceling out 21.

$$\frac{7}{3}-\frac{40}{6}+0\times 2$$

Reduce the fraction \frac{40}{6} to lowest terms by extracting and canceling out 2.

$$\frac{7}{3}-\frac{20}{3}+0\times 2$$

Since \frac{7}{3} and \frac{20}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7-20}{3}+0\times 2$$

Subtract 20 from 7 to get -13.

$$-\frac{13}{3}+0\times 2$$

Multiply 0 and 2 to get 0.

$$-\frac{13}{3}+0$$

Add -\frac{13}{3} and 0 to get -\frac{13}{3}.

$$-\frac{13}{3}$$

Factor

$$-\frac{13}{3} = -4\frac{1}{3} = -4.333333333333333$$