1. Show that, if \(a\) and \(b\) are either both positive or both negative, then the equation \[\frac{x}{x-a}+\frac{x}{x-b}=1\] has two distinct real solutions.

How can we handle equations which involve fractions? Does it make a difference if \(x\) is in the denominator?

  1. Show that the equation \[\frac{x}{x-a}+\frac{x}{x-b}=1+c\] has exactly one real solution if \(c^2=-\dfrac{4ab}{(a-b)^2}\). Show that this equation can be written \(c^2=1-\left(\dfrac{a+b}{a-b}\right)^2\) and deduce that it can only hold if \(0 < c^2 \leq 1\).

This interactive applet shows the graph of \(y = \dfrac{x}{x-a}+\dfrac{x}{x-b}-1\). You can change the values of \(a\) and \(b\).

The blue lines are the lines \(y=c\) which intersect the red curve exactly once.