feedrom wrote:GMATGuruNY wrote:
Statement 2: |x|=4x-15
Case 1: No signs are changed
x = 4x-15
15 = 3x
x=5.
To confirm that x=5 is a valid solution, plug x=5 into the original equation:
|5| = 4*5 - 15
5 = 5.
Since the two sides of the equation are equal, x=5 is a valid solution.
Case 2: Flip the signs for one side of the equation
-x = 4x-15
15 = 5x
x=3.
To confirm that x=3 is a valid solution, plug x=3 into the original equation:
|3| = 4*3 - 15
3 = -3.
Since the two sides of the equation are NOT equal, x=3 is NOT a valid solution.
Thus, x=5.
SUFFICIENT.
The correct answer is B.
This is an interesting case! So if I don't check the two values of x again, I'll automatically think that (2) is insufficient.
My question is: when I see |a| = b, how do I know a has only 1 value without testing all values again? Thank you.
|x|=4x-15.
As Ceilidh has pointed out, whenever we have the same variable on both sides but absolute value on only one side, we need to confirm that both solutions are valid.
The easiest approach is to plug the two solutions back into the equation, as I did in my post above.
Alternate approach:
The absolute value notation on the lefthand side implies that the lefthand side represents a NONNEGATIVE VALUE.
Thus, the righthand side -- 4x-15 -- must also represent a nonnegative value.
This restricts the range of x as follows:
4x-15 ≥ 0
4x ≥ 15
x ≥ 3.75.
Thus, for the righthand side to be nonnegative, x must be greater than or equal to 3.75.
Implication:
A solution is valid only if it is greater than or equal to 3.75.
In my post above, algebra yielded two possible values for x:
3 and 5.
But x=3 is invalid because it is less than 3.75.
Thus, the only valid solution is x=5.
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.At the outset these seem very simple so our feeling is we got it right, but alas....the GMAT algorithm is working in the background 
