Sketch the graph of $\displaystyle y=\frac{(2(x+a))}{(x-1)}$, where $\displaystyle a>2$. Hence find the set of values of x such that $\displaystyle \frac{2(x+a)}{x-1}\leqslant |x+a|$
They intersect twice. Good. Now look at the inequality again. You are looking for all points on the 2(x + a)/(x - 1) curve that are lower on the graph (or equal to) the graph of |x + a|. So what parts of the function 2(x + a)/(x - 1) are below the lines?
-Dan