# Absolute inequality question.

I'm stuck with this kind of absolute inequality:

$$|x+1|>x+2. \tag1$$

Firstly, when I solve this one:

$$|x+1|=x+2, \tag2$$

I make sure the right side of the equation is greater than zero; Condition:

$$x+2\ge0;\quad x\ge-2.$$

When I solve this by separating it in two cases and finding a solution, I get, in this example: $x=-1.5$, which belongs to the interval of $[-2,+\infty)$, so it's the right solution.

However, when I solve the first inequality $(1)$, I get the solution:

$$x<-1.5$$.

$-1.5$ is in the interval of the condition for the equation $(1)$ (which is the same for the second one -- the inequality): $[-2,+\infty )$, therefore, the WHOLE interval of $(-∞, -1.5)$ is the solution. Is this the right way of solving this kind of inequality?

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You have some good ideas.

You don't have a problem with the expression on the right-hand side. You need to know though whether $$|x+1|=x+1 \text{ or }-x-1$$ and this depends on whether $x+1\ge 0$ or $x\ge-1$.

If this is true then the equation becomes $$x+1\gt x+2$$ which is always false.

If $x\lt -1$ the equation becomes $$-x-1\gt x+2 \text{ or, as you computed }x\lt -\frac 32$$

Here there are two conditions, as you observed, and only the values of $x$ which satisfy both are admissible as solutions, and this is true when $x\lt -\frac 32$

It doesn't actually matter whether the right-hand side is positive - you aren't multiplying or dividing anything by it, or, for example squaring both sides of the equation (when you do have to be careful, so your instinct is a good one).

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So basically, I don't need any checks or conditions when dealing with these inequalities. I asked this question, because when I have another problem like this one: |x+2|>|x+1|, I have to make 3 intervals, then check the solutions and make a cross-section \cap (opposite of union, I don't know the word) within intervals. That's why I thought I should make the same with the original inequality I presented here. For everything to be clear: it obviously differs from the way of solving |x+2|>|x+1|, because there's a cross-section, and here's the whole interval. – Guest Mar 9 '14 at 13:43
@Guest When you have an absolute value on each side you can consider (i) where both are positive (ii) where both are negative (iii) where one is positive and the other negative [which may be two cases in more complex situations]. When you take away the absolute value sign you need to know whether you replace it with $+$ or $-$. It is very useful to sketch a graph, and being very clear and organised in your work, because you need to keep track of cases and signs. – Mark Bennet Mar 9 '14 at 13:53

Yes, you arrived at the correct solution.

One way to approach this is to note that given $|x + 1| > x + 2$, there are two cases to consider: $$x + 1 > x+2 \tag 1$$ $$-(x + 1) > x + 2\tag 2$$

Case $(1)$ never holds for any $x$, since $$x + 1 > x + 2 \iff 1 \gt 2$$ which clearly is false.

Case $(2)$ gives us $$-(x + 1) > x + 2 \iff -x - 1 > x + 2 \iff 0 > 2x + 3 \iff x \lt -1.5$$

Hence, indeed, we have $$x\in (-\infty, -1.5)$$

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Well, I think the best method you can used to be clear is separate cases:

$|x+1|=x+1$ if $x\geq -1\quad (1)$

$|x+1|=-x-1$ if $x< -1\quad (2)$

So, in the first case you have $x>x+1$ thats it's always false.

For the rest of the solution you we look at the second case: $-x-1<x+1 \Rightarrow x<-1.5$ so the another part is $(-\infty,-1.5)$

The complete solution is $$(-\infty,-1.5)$$

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