I would like to highlight that I am aware that there are different approaches to solving absolute value inequalities. However, I am interested in why this particular method, that I often use for single absolute value problems, does not work for those with two absolute values.
So for solving the most simple inequality problems like this one:
|x|≥3
I would divide it into two cases:
Case 1:
The initial domain are all positive numbers and 0.
x≥3
Case 2:
The initial domain are all negative numbers and 0
x≤-3
Then, I would find the union of the two solutions, to obtain the solution below: x∈ ]-∞,-3] ∪[3,∞[
On the other hand, when applying this method for a problem with two absolute values it does not work well.
Let's consider this problem:
|2x+1|≥|x-2|
First, I would calculate the domains:
If we consider 2x+1>0, then x>-0.5
If we consider 2x+1<0, then x<-0.5
If we consider x-2>0, then x>2
If we consider x-2<0, then X<2
Thus, next I will take into account all the cases:
Case 1 - both sides positive - domain of x is x>-0.5;
then I perform the calculations;
2x+1≥x-2;
x>-3;
The part in the domain is: [-0.5, ∞[
Case 2 - left side negative, right side positive - domain ]-∞,-0.5] ∪[2,∞[;
Now the left side is negative, so I put a - sign before the first function.
-2x-1>x-2;
x<0.333;
The part in the domain is: ]-∞,-0.5]
Case 3 - left side positive, right side negative - domain [-0.5, 2];
Now I consider the negative side for the other value:
2x+1>-x+2;
3x>1;
x>0.333;
The part in the domain: [0.333, 2]
Case 4 - both sides negative - domain ]-∞,2];
-2x-1>-x+2;
x<-3;
The part in the domain: ]-∞,-3]
Then analogically I take the union of all the solutions that fit in their domains. I obtain the following result: x∈ ]-∞,2]
In reality the result should be the following: x∈ ]-∞,-3] ∪[0.333,∞[
I really do not know where I have done a mistake, as the approach is analogous to the one for the simpler example. If anyone would show me any mistakes or false assumptions that I have made, I would be very thankful. Btw, I know the approaches with squaring both sides and dividing the inequality into 3 sections according to their x-intercepts.
This is my first post, so I would also be up for any feedback that you guys could give me.
This method does work, but you made some mistakes:
That's not really the case.
$$ 2x + 1 > 0 \land x - 2 > 0 \iff x > 2 $$
So for Case 1 the domain is $ x > 2 $
For Case 2 domain is empty:
$$ 2x + 1 < 0 \land x - 2 > 0 \iff x \in \emptyset $$
You should continue your calculations for the domain of each case similarly. In fact, you should notice that the cases cover the whole $ \mathbb{R} $ completely and their intersections are empty (each point is covered only by one case). (Edit: this paragraph is true for at least as long, as each term to which absolute value is applied is a linear.)
For each case you should calculate the solution and take it's intersection with the domain, as you have noted.
Use TeX to format your posts, they would be much nicer to read. It's also a useful skill for any mathematician or STEM student. To use TeX formatting on StackExchange type something like
$ x > 2 $which will render as $ x > 2 $. For separate, centered formulas use$$ x \in \emptyset $$which will render as: $$ x \in \emptyset $$I hope you find this answer useful. Just redo your calculations being a little bit more careful and good luck :)