Checking solutions of absolute value equations

[yo4561]

I was reading All the Quant companion and on page 16, the following example is given:
"If |x-2| = |2x-3|, what are the possible values for x?"
Through case 1, MP shows that you get x=1 and through case 2 you get x=5/3. Then MP says you need to go back and test the cases in which when you test x=1 into the original, you get x-1. But, for x=5/3, the equation spits out 1/3. MP says both solutions are valid...

For x=5/3 to be a solution, when you test to see if it works, wouldn't the equation need to spit back 5/3? If that's not the case, what is an example of an "invalid" answer that MP talks about in the book?

Also, do you need to test your answers for simple absolute value equations or just complex ones?

Many thanks

[esledge]

Hi yo, for me personally, starting absolute value problems by drawing a number line helps a lot. The meaning of the absolute value sign depends on whether the stuff inside is positive or negative: if the stuff inside is already positive, the absolute value signs do nothing and can just be dropped, but if the stuff inside is negative, the absolute value signs "flip the sign."

So I can't draw it here, but my number line has vertical dotted lines at x = 1.5 and at x = 2, as these are where 2x - 3 and x -2, respectively, change sign. Then I label the "zones" with + and - signs for each of the expressions. There are actually 3 cases I would test.

Case 1 is for the x<1.5 zone where both expressions are negative and the absolute value signs "flip the sign" of both expressions.
Solve for Case 1:
- (x - 2) = - (2x - 3)
2 - x = 3 - 2x
x = 1
Check that x=1 is actually in the x<1.5 zone. Since it is, x = 1 is a real solution.

Case 2 is for the 1.5<x<2 zone where (x-2) is negative and the absolute value signs "flip the sign," but where 2x-3 is already positive and the absolute value sign does nothing.
Solve for Case 2:
- (x - 2) = 2x - 3
2 - x = 2x - 3
5 = 3x
5/3 = x
Check that x=5/3 = 1.666... is actually in the 1.5<x<2 zone. Since it is, x = 5/3 is a real solution.

Case 3 is for the 2<x zone where both expressions are positive and the absolute value sign does nothing and can just be dropped.
Solve for Case 3:
x - 2 = 2x - 3
1 = x
Check that x=1 is actually in the 2<x zone. It is not, and this would be how a false/phantom solution might appear. But x = 1 is still a real solution to this problem since it was also found and confirmed for Case 1.

And I generally do a quick double check of any answers, just because we know that phantom solutions could be an issue on these:
Check x = 1: |1-2| = |2(1)-3| becomes |-1| = |-1|, which is true.
Check x = 5/3: |(5/3)-2| = |2(5/3)-3| becomes |-1/3| = |1/3|, which is true.

How would a false solution arise?

Try |x - 2| + | x - 3| = 6 using the method above. There are two legitimate solutions, but the middle "zone" where 2<x<3 produces a non-solution result (slightly different from a false/phantom solution).

Try |x - 5| + |2 - 3x| = 4 using the method above. There are 3 false solutions and no real solutions! This would be too hard or "gotcha" for the GMAT, in my opinion.