<s>[INST] 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 ? 
 [/INST] If $(a,b)$ is an ordered pair of positive integers such that $\gcd(a,b)=1$ and
$$\frac{a}{b}+\frac{14b}{9a}=\frac{9a^2+14b^2}{9ab},$$
is an integer, then $9ab$ divides $9a^2+14b^2$. In particular $a$ and $b$ both divide $9a^2+14b^2$, and so $a$ divides $14b^2$ and $b$ divides $9a^2$. Because $\gcd(a,b)=1$ it follows that $a$ divides $14$ and $b$ divides $9$. If $b=9$ then $81a$ divides $9a^2+14\cdot 81$, so $81$ divides $9a^2$ and hence $3$ divides $a$, contradicting the fact that $\gcd(a,b)=1$. Of course $b\neq1$ because $9$ does not divide 14. Hence $b=3$, and we conclude that there are precisely four such pairs.
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