$$\frac{a}{x}+\frac{b}{y}=\frac{a}{2}+\frac{b}{3} \;\;\; \ldots(i)$$ $$x+1=y \;\; \;\ldots(ii)$$
We have to solve for $x$ and $y$, only this time using the method of elimination.
From equation $(ii)$, we get, $$\frac{1}{x+1}=\frac{1}{y} \Rightarrow \frac{b}{x+1}=\frac{b}{y} \;\;\; \ldots(iii)$$
Subtracting $(iii)$ from $(i)$, we get, $$\frac{a}{x}+\frac{b}{y}-\frac{b}{y}=\frac{a}{2} + \frac{b}{3}- \frac{b}{x+1}$$ $$\Rightarrow \frac{a}{x}=\frac{a}{2} + \frac{b}{3} - \frac{b}{x+1}$$
After that,I really cannot find anything to do.I have taken quite a few other routes, but have hit nothing but dead ends. At this a point a little hint will be appreciated.
You've already done the "elimination" when you subtract (iii) from (i), and it looks like you have a typo, $y$ should be completely eliminated when you do the subtraction, and you should get
$$a/x = a/2 + b/3 - b/(x+1)$$
If you then multiply this equation by $x(x+1)$ on both sides (keeping in mind then that $x$ cannot be $0$ or $-1$), and cancel the $x$ and $(x+1)$ in denominators, and then expand and bring all terms to one side then you will get a quadratic in $x$ that is set to $0$, which you can solve for $x$. Then you can substitute into (ii) to easily get $y$.
Note that when you solve the quadratic, you are assuming either $a \neq 0$ or $b \neq 0$. If $a = b = 0$ then any $x$ gives a solution except $x = 0 $ and $x = -1$.