Solve 7/3+21/8(-41/9+5/7*7)

$$\frac{ 7 }{ 3 } + \frac{ 21 }{ 8 } ( \frac{ -41 }{ 9 } + \frac{ 5 }{ 7 } \times 7)$$

Evaluate

$$\frac{7}{2}=3.5$$

Solution Steps

Fraction \frac{-41}{9} can be rewritten as -\frac{41}{9} by extracting the negative sign.

$$\frac{7}{3}+\frac{21}{8}\left(-\frac{41}{9}+\frac{5}{7}\times 7\right)$$

Cancel out 7 and 7.

$$\frac{7}{3}+\frac{21}{8}\left(-\frac{41}{9}+5\right)$$

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

$$\frac{7}{3}+\frac{21}{8}\left(-\frac{41}{9}+\frac{45}{9}\right)$$

Since -\frac{41}{9} and \frac{45}{9} have the same denominator, add them by adding their numerators.

$$\frac{7}{3}+\frac{21}{8}\times \frac{-41+45}{9}$$

Add -41 and 45 to get 4.

$$\frac{7}{3}+\frac{21}{8}\times \frac{4}{9}$$

Multiply \frac{21}{8} times \frac{4}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7}{3}+\frac{21\times 4}{8\times 9}$$

Do the multiplications in the fraction \frac{21\times 4}{8\times 9}.

$$\frac{7}{3}+\frac{84}{72}$$

Reduce the fraction \frac{84}{72} to lowest terms by extracting and canceling out 12.

$$\frac{7}{3}+\frac{7}{6}$$

Least common multiple of 3 and 6 is 6. Convert \frac{7}{3} and \frac{7}{6} to fractions with denominator 6.

$$\frac{14}{6}+\frac{7}{6}$$

Since \frac{14}{6} and \frac{7}{6} have the same denominator, add them by adding their numerators.

$$\frac{14+7}{6}$$

Add 14 and 7 to get 21.

$$\frac{21}{6}$$

Reduce the fraction \frac{21}{6} to lowest terms by extracting and canceling out 3.

$$\frac{7}{2}$$

Factor

$$\frac{7}{2} = 3\frac{1}{2} = 3.5$$