Inequalities with two absolute values with greater than symbol. Please tell me the proper way of doing $|4x-1|>|3x+2|$

Question

I have finals in a couple of days. I need help with inequalities. I have spent around 15 hours trying my hand at inequalities and I am still trying to figure things out. In this case I need help with an inequality with two absolute values with greater than symbol.

The following is a picture of the inequality ($|4x-1|>|3x+2|$) and the procedure I tried, which is something I applied from an answer by Isaac to another question in this site, although I think I did something wrong because it looks like I got the answer wrong or incomplete.

Please tell me what I did wrong and how to do it the right way.

Thanks in advance.

PS: Does this method also apply if there is a less than sign?

Answer 1

You're doing almost everything right; you're just misinterpreting the results at the end.

In particular, the idea is to combine intervals where the inequality holds. In the first region, $\left(-\infty, -\frac23\right)$, you arrived at $x<3$. It seems that you rejected this result, judging by the "$\times$" beside your work. This is where you went wrong.

Instead, think of it like this:

Within the region $\left(-\infty, -\frac23\right)$, which $x$ satisfy $x<3$?

All $x$ do within this region. Another way to see this is that you are effectively taking the intersection of the two: $$\left(-\infty, -\frac23\right) \cap (-\infty,3) = \left(-\infty, -\frac23\right)$$

The other two regions are handled the same way.

Answer 2

Here’s another way to look at it: $$|4x-1|>|3x+2| \iff (4x-1)^2-(3x+2)^2>0 \iff (7x+1)(x-3)>0$$

Now LHS has only two obvious zeros, and outside the interval with those end points, it has to be positive.

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P.S. When you have something like $f(x)>0$, where $f$ is continuous, it can change sign only when there are roots. So its roots effectively partition the real line into intervals which are either solutions or not. In the above case, two roots are $-\frac17, 3$, so we need to consider only the three intervals $(-\infty, -\frac17), (-\frac17, 3)$ and $(3, \infty)$. Obviously the LHS is positive for $x\to \pm\infty$ and negative for $x=0$, so the first and last intervals are solutions, the middle one isn’t.

BTW, whether the intervals to consider are open / closed depends on whether the inequality is strict or $\geqslant$.

Answer 3

I tried again, following the answer of Théophile, and I was able to reach the correct solution. Below is the pic of the procedure I used. Thanks!