smc

Let $m=\frac{x-a}{b}$ and $n=\frac{x-b}{a}$ then we have $$m+n=\frac{1}{m}+\frac{1}{n}$$ $$m+n=\frac{m+n}{mn}$$ Notice that the equation is possible iff $m+n=0$ or $mn=1$. If $m+n=0$ then $$\frac{x-a}{b}+\frac{x-b}{a}=0$$ $$\frac{x-a}{b}=\frac{b-x}{a}$$ $$x=\frac{a^2+b^2}{a+b}$$ Which yields $1$ solution for $x$. If $mn=0$ then $$(\frac{x-a}{b})(\frac{x-b}{a})=1$$ $$x^2-(a+b)x=0$$ Solving the quadratic gets another $2$ solutions for $x$. Thus there are $\boxed{\text{three}}$ solutions in total.