gmatbull wrote:

|x-1| + |x+2| <= 5

what is the shortest approach to solving this question?

\(|x-1|+|x+2|\leq{5}\)

Two key points: \(x=-2\) and \(x=1\) (key points are the values of x when absolute values equal to zero), thus three ranges to check:

---------{-2}--------{1}---------1. \(x<-2\) (blue range) --> \(|x-1|+|x+2|\leq{5}\) becomes: \(-x+1-x-2\leq{5}\) --> \(-3\leq{x}\) --> \(-3\leq{x}<-2\);

2. \(-2\leq{x}\leq{1}\) (green range) --> \(|x-1|+|x+2|\leq{5}\) becomes: \(-x+1+x+2\leq{5}\) --> \(3\leq{5}\), which is true. This means that inequality \(|x-1|+|x+2|\leq{5}\) is ALWAYS true when \(x\) is in the range \(-2\leq{x}\leq{1}\);

3. \(x>1\) (red range)--> \(|x-1|+|x+2|\leq{5}\) becomes: \(x-1+x+2\leq{5}\) --> \(x\leq{2}\) --> \(1<x\leq{2}\).

The ranges above give the following final range for which given inequality is true: \(-3\leq{x}\leq{2}\).

Hope it's clear.

Am grateful, it's very clear now. +1 Kudos to you.