smc

We can rewrite the left-hand side as $\frac{a^{2}b+a}{b+a{b}^2}$ by multiplying the numerator and denominator by $ab$. Now we can multiply the top part of the fraction by the denominator to give $a^{2}b+a=13b+13a{b}^2⟹ab(a-13b)=13b-a=-(a-13b)⟹(ab+1)(a-13b)=0$. Now, as $a$ and $b$ are positive, we cannot have $ab+1=0$, so we must have $a-13b=0⟹a=13b$, so the solutions are $(13,1),(26,2),(39,3),(52,4),(65,5),(78,6),(91,7)$, which is 7 ordered pairs. $\boxed{C}$ # Solution 2 The equation is equivalent to $\frac{a+\frac{1}{b}}{\frac{1}{a}+b}=13$ or $\frac{\frac{ab+1}{b}}{\frac{ab+1}{a}}=13$. The $ab+1$s cancel out leaving us with $\frac{a}{b}=13$. This is asking us for the positive multiples of $13$ up to $100$, and we know that $13\cdot 7=91$, so our answer is $\boxed{7}$.