To find relatively prime ordered pairs of positive integers $(a,b)$ such that $ \dfrac ab +\dfrac {14b}{9a}$ is an integer How many ordered pairs $(a,b)$ of positive integers are there such that g.c.d.$(a,b)=1$ , and 
$ \dfrac ab +\dfrac {14b}{9a}$ is an integer ? 

$$\frac ab+\frac{14b}{9a}=\frac{9a^2+14b^2}{9ab}$$
If this expression is an integer, since $a$ and $b$ are coprime, $a$ divides $14$ and $b$ divides $9$.
If $b=1$, we get
$$a+\frac{14}{9a}$$
which is not an integer.
If $b=9$,
$$\frac{a}9+\frac{14}a=\frac{a^2+126}{9a}$$
which is neither an integer, since $a$ can't be a multiple of $9$.
If $b=3$,
$$\frac a3+\frac{14}{3a}=\frac{a^2+14}{3a}$$
which is an integer for every possible value of $a$ (namely, $1$, $2$, $7$ and $14$).
So there is $4$ ordered pairs of positive integers.
