Find the value of $\frac{5x+9y}{45xy}$, given $x = \frac{3}{5}$ and $y = \frac{7}{9}$.
To evaluate the fraction, I need to substitute the given values of x and y into the numerator and denominator. That means I have to replace x with $\frac{3}{5}$ and y with $\frac{7}{9}$ wherever they appear in the fraction. So I get $\frac{5(\frac{3}{5})+9(\frac{7}{9})}{45(\frac{3}{5})(\frac{7}{9})}$. Now I can simplify the fraction by multiplying the coefficients and the fractions separately. For the numerator, I get $5(\frac{3}{5})+9(\frac{7}{9}) = \frac{15}{5} + \frac{63}{9} = 3 + 7 = 10$. For the denominator, I get $45(\frac{3}{5})(\frac{7}{9}) = \frac{135}{5} \cdot \frac{7}{9} = 27 \cdot \frac{7}{9} = 21$. So the fraction becomes $\frac{10}{21}$. This is the simplest form of the fraction, since 10 and 21 have no common factors other than 1. So the value of the fraction is $\frac{10}{21}$. 