If |X-2| = |2X-3| , what are the possible values of x?

Because there are two absolute value expressions, each of which yields two algebraic cases, it seems that we need to test four cases overall.

1. Positive/Positive: (x-2) = (2x-3)
2. Positive/Negative: (x-2) = -(2x-3)
3. Negative/Positive: -(x-2) = (2x-3)
4. Negative/Negative -(x-2) = -(2x-3)

The book states that case 1 and 4 yield the same equation. And 2 and 3 yield the same equation....therefore we only need to test the positive/positive case, and a case where the sign of one equation changes. (1 and 3 for example)

Also, the two cases tested are if 1) neither expression changes signs or 2) one expression changes signs. I am just trying to figure out the step by step approach for the "complex" absolute value equations. Are you supposed to test all four cases first, then see which are different equations?

The example in the problem set (#6) states that we test same same, and both have opposite sign.

In terms of a general strategy: what are we looking for when we decide to take this approach? (is it simply, an absolute value equation with one variable and several constraints?). And lastly, is the only way to figure out which cases to test -- to write out all four and see which are distinct?

@rockrock: Your method is absolutely flawless, but not the simplest of all

We can square terms (within the modulus signs) whenever they are equal in nature.

ie; The equation, |X-2| = |2X-3|
can be rewritten as (x-2)^2=(2x-3)^2

This reduces to 3x^2-8x+5=0 (After solving)

Thus x=1 or x=5/3.....

which is the same as you will get according to your method.



@MGMAT Tutors: Please advise whether this technique is appropriate for this case and similar ones.

Yeah I'd really like to know. It's way easier! Also think that another option is to split it up and test regions...but not sure how efficient that method would be.

Also - quick question for an inquality like x-2. I realize the critical point is 2.

How does it work for the other equation - 2X-3, when graphing? Is the critical point then 3/2? If so, how would we then solve by graphing?

The easiest way out:


|x|=|y|

=>x=y
or
=>x=-y....This is applicable for all cases where 2 quantities within the modulus signs are equal.

Coming to the problem,

Main Equation: |X-2| = |2X-3|


x-2=2x-3 =>x=1
x-2=-2x+3 =>x=5/3

So the solution has to be either x=1 or x=5/3 or both

Check both solutions by substituting them in the given equation:

Substituting x=1 in the main equation, we get,
|1|=|1|...Hence satisfied

Substituting x=5/3 in the main equation, we get,
|-1/3|=|1/3|...=>1/3=1/3...Hence Satisfied.

Thus both solutions satisfy the condition.
You can forget about range , critical points, graphs , number lines, squaring the inequalities so on and so forth for such kind of problems. The only thing you must do is, check whether the solutions satisfy the main equation.

PS:
If |x|=|y|...Always apply the aforementioned method to save time but do not forget to check.
If |x|=y...Then split into cases.
1. x=y
2. x=-y...Solve, substitute in the main equation and confirm.

Thanks. So I'm assuming the critical regions, number line etc etc is only relevant for the inequalities, not for absolute value equations. I think I've already decided that if I get an absolute value inequality problem I'm guessing ! The equation methodology is very easy to understand.

your best bet for a generic strategy is just to try all the possibilities and then ALSO plug your answers back into the original equation to check. absolute value problems are tricky, so unless you have a REALLY solid handle on the theory behind what's going on, be careful about using any slick shortcuts. not to say they won't work, but a clever question writer can throw in a lot of traps that could ensnare you..

_________________
Tim Sanders
Manhattan GMAT Instructor