Expanding on the comment by player100:
The function $f(x) = x-\min(bx,c) $ has two regions:
$f(x) = (1-b)x \quad \text{if } bx \le c$
$f(x) = x-c \quad \text{if } bx \gt c$
Setting $f(x) = a$ then we have
$x = \frac{a}{1-b} \quad \text{if } \frac{ab}{1-b} \le c$
$x = c+a \quad \text{if } b(c+a) \gt c$
Note that we can re-arrange the condition $b(c+a) \gt b$ as $ab \gt (1-b)c$, so the two conditions are mutually exclusive if $1-b > 0$ i.e. $b < 1$. So if $b< 1$ then the equation $f(x)=a$ has exactly one solution.
However, if $b > 1$ then the equation $f(x)=a$ will have two solutions if $ab > (1-b)c$, one solution if $ab = (1-b)c$ and no solutions if $ab < (1-b)c$.
*mean the product? – Mikasa Sep 04 '18 at 12:33