# Absolute value problem $|x-5|<|x+1|$

So I have this basic absolute value problem: $|x-5|<|x+1$|. From what I understand, I need to consider what happens in every case. There are four cases, right? One where both sides are positive, one where both sides are negative and two where one side is negative and the other is positive.

The case were both are negative gives me something absurd ($5<-1$). Since the solution set is the intersection of the solutions for every case, this would mean that it has no solution. But it has.

Where is the error in my reasoning?

• Since the solution set is the intersection of the solutions for every case... This is correct but the case where both sides(in the absolute value) are negative does not provide a solution so it can not be considered in the intersection... – Fermat Mar 3 '15 at 20:42

The left hand side is the distance from $x$ to $5$ and the right hand side is the distance from $x$ to $-1$. When is $x$ closer to $5$? The midpoint is $2$. So when $x>2$
Start solving: $$x-5>0 \rightarrow x>5 \quad \land \quad x+1>0 \rightarrow x>-1$$ so $\mathbb{R}$ splits in three intervals $(-\infty,-1) \bigcup [-1, 5) \bigcup [5,+\infty)$, and your inequality split in three systems: $$\begin{cases} x<-1\\ 5-x<-x-1 \end {cases} \quad \lor \quad \begin{cases} -1\le x<5\\ 5-x<x+1 \end {cases} \quad \lor \quad \begin{cases} 5\le x\\ x-5<x+1 \end {cases}$$ that you can easely solve (and the final solution is the union of the solutions).
Square: $$(x-5)^2<(x+1)^2\\ x^2-10x+25<x^2+2x+1\\ 12x>24\\ x>2$$