For $c \in \mathbb{R}$, the equation $\dfrac{1}{x-a_1} + \dfrac{1}{x-a_2} + \ldots + \dfrac{1}{x-a_n} = c$ has $n-1$ real solutions 
For $c \in \mathbb{R}$, the equation $\dfrac{1}{x-a_1} + \dfrac{1}{x-a_2} + \ldots + \dfrac{1}{x-a_n} = c$ has $n-1$ real solutions

The follow up to this is that prove that the equation has $n$ solutions if $c \neq 0$.  Here $a_1 < a_2 < \ldots < a_n$ and $c \in \mathbb{R}$. How does this equation even have solutions, I'm really confused with this!
 A: The function 
$$ f(x)=\sum_{k=1}^{n}\frac{1}{x-a_k} $$
is a continuous and differentiable function over its domain, $\mathbb{R}\setminus\{a_1,\ldots,a_n\}$. Since
$$ f'(x)=-\sum_{k=1}^{n}\frac{1}{(x-a_k)^2} \leq 0 $$
$f(x)$ is a decreasing function on the connected components of the domain $I_0=(-\infty,a_1), I_1=(a_1,a_2),\ldots,I_n=(a_n,+\infty)$. Since for any $k\in[1,n]$ we have
$$ \lim_{x\to a_k^+}f(x)=+\infty,\qquad \lim_{x\to a_k^-}f(x)=-\infty $$
and $$\lim_{x\to \pm\infty}f(x)=0,$$
by the mean value property of continuous functions we have that there is exactly a solution of $f(x)=c$ in $I_1,I_2,\ldots,I_{n-1}$ and, assuming $c\neq 0$, a solution of $f(x)=c$ either in $I_0$ or in $I_n$ according to the sign of $c$.
A: Hint let $f 1(x)=\frac {1}{(x-a_1) }$  differentiating we have $\frac {-1}{(x-a_1)^2} $ thus function is decreasing hence $f1 (x) $ cuts x axis at least one time then the given function is summation of all such functions thus given function has atleast n solutions .
