Solve 4/3+7/9(1/7+4/9)-frac{1}{5}=

$$\frac { 4 } { 3 } + \frac { 7 } { 9 } ( \frac { 1 } { 7 } + \frac { 4 } { 9 } ) - \frac { 1 } { 5 } =$$

Evaluate

$$\frac{644}{405}\approx 1.590123457$$

Solution Steps

Least common multiple of 7 and 9 is 63. Convert \frac{1}{7} and \frac{4}{9} to fractions with denominator 63.

$$\frac{4}{3}+\frac{7}{9}\left(\frac{9}{63}+\frac{28}{63}\right)-\frac{1}{5}$$

Since \frac{9}{63} and \frac{28}{63} have the same denominator, add them by adding their numerators.

$$\frac{4}{3}+\frac{7}{9}\times \frac{9+28}{63}-\frac{1}{5}$$

Add 9 and 28 to get 37.

$$\frac{4}{3}+\frac{7}{9}\times \frac{37}{63}-\frac{1}{5}$$

Multiply \frac{7}{9} times \frac{37}{63} by multiplying numerator times numerator and denominator times denominator.

$$\frac{4}{3}+\frac{7\times 37}{9\times 63}-\frac{1}{5}$$

Do the multiplications in the fraction \frac{7\times 37}{9\times 63}.

$$\frac{4}{3}+\frac{259}{567}-\frac{1}{5}$$

Reduce the fraction \frac{259}{567} to lowest terms by extracting and canceling out 7.

$$\frac{4}{3}+\frac{37}{81}-\frac{1}{5}$$

Least common multiple of 3 and 81 is 81. Convert \frac{4}{3} and \frac{37}{81} to fractions with denominator 81.

$$\frac{108}{81}+\frac{37}{81}-\frac{1}{5}$$

Since \frac{108}{81} and \frac{37}{81} have the same denominator, add them by adding their numerators.

$$\frac{108+37}{81}-\frac{1}{5}$$

Add 108 and 37 to get 145.

$$\frac{145}{81}-\frac{1}{5}$$

Least common multiple of 81 and 5 is 405. Convert \frac{145}{81} and \frac{1}{5} to fractions with denominator 405.

$$\frac{725}{405}-\frac{81}{405}$$

Since \frac{725}{405} and \frac{81}{405} have the same denominator, subtract them by subtracting their numerators.

$$\frac{725-81}{405}$$

Subtract 81 from 725 to get 644.

$$\frac{644}{405}$$

Factor

$$\frac{2 ^ {2} \cdot 7 \cdot 23}{3 ^ {4} \cdot 5} = 1\frac{239}{405} = 1.5901234567901235$$