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.