Solve 7/9*3/5+7/9*4/5

$$\frac { 7 } { 9 } \cdot \frac { 3 } { 5 } + \frac { 7 } { 9 } \cdot \frac { 4 } { 5 }$$

Evaluate

$$\frac{49}{45}\approx 1.088888889$$

Solution Steps

Multiply \frac{7}{9} times \frac{3}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7\times 3}{9\times 5}+\frac{7}{9}\times \frac{4}{5}$$

Do the multiplications in the fraction \frac{7\times 3}{9\times 5}.

$$\frac{21}{45}+\frac{7}{9}\times \frac{4}{5}$$

Reduce the fraction \frac{21}{45} to lowest terms by extracting and canceling out 3.

$$\frac{7}{15}+\frac{7}{9}\times \frac{4}{5}$$

Multiply \frac{7}{9} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator.

$$\frac{7}{15}+\frac{7\times 4}{9\times 5}$$

Do the multiplications in the fraction \frac{7\times 4}{9\times 5}.

$$\frac{7}{15}+\frac{28}{45}$$

Least common multiple of 15 and 45 is 45. Convert \frac{7}{15} and \frac{28}{45} to fractions with denominator 45.

$$\frac{21}{45}+\frac{28}{45}$$

Since \frac{21}{45} and \frac{28}{45} have the same denominator, add them by adding their numerators.

$$\frac{21+28}{45}$$

Add 21 and 28 to get 49.

$$\frac{49}{45}$$

Factor

$$\frac{7 ^ {2}}{3 ^ {2} \cdot 5} = 1\frac{4}{45} = 1.0888888888888888$$