Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0

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1. check the answers
the maximum range provided in the answer is -6, -2, 2, 0 and 6
check for this range

-6, gives us that |-12-|-6||<6
|-12-6| <6
18<6 not true
Eliminate E

next value given in the answer series is -2
Holds good, so left value in the number line should be 2

Now check for right value by putting 6
Holds good so the range should be -2 to 6

So clearly answer should be C

=>

We should split real number of x into two cases x ≥ 0 and x < 0.

Case 1: x ≥ 0
|2x-|x||<6 ⬄ | 2x – x | < 6 ⬄ |x| < 6 ⬄ -6<x<6.
From the assumption of the case, x ≥ 0, we have 0 ≤ x < 6.

Case 2: x < 0
|2x-|x||<6 ⬄ | 2x + x | < 6 ⬄ |3x| < 6 ⬄ |x| < 2 ⬄ -2<x<2.
From the assumption of the case, x < 0, we have -2<x<0.

Thus, we have -2 < x < 6.

Answer: C

Bunuel - Can you help me understand where did I go wrong?

hey there MathRevolution, i dont get ...when we have expression |x|<a doesnt it mean only ONE case which is -a<x<a


as far as i know if |x|<a then -a<x<a


also if Case 2: x < 0 shouldnt it be - (2x - x ) < 6

and one more question when we have nested absolute value should we work from outside to inside please show how to do it in this case

got so confused with this question !!!!

ibra10 maybe you can shed some absolute light on my questions

Hi,
can you tell
why 2x is positive when x <0.?

Hi,
can you tell
why 2x is positive when x <0.?

IanStewart Bunuel can you help?

Posted from my mobile device

IanStewart thanks a lot! Your explanation was amazing

Thanks Ian for that great explanation.
What would you do in instances when you have mods with multiple variables in the inequality or equation?
e.g. |x + 3 | + |y| < ...
How would you deal with those types?

First, bear in mind that the question in this thread is not an official question. Most official absolute value questions can be solved without using cases at all -- they can instead be solved much more easily by thinking about distances, something I've explained in other solutions on this forum, and that I go over in detail in my Algebra book.

So the method I used to solve this problem is one you only rarely need on the real test, and more complicated variants on this problem are even less likely to matter. If an official question did contain an inequality like |x + 3| + |y| < 1, I'd really want to see the actual question, because there'd almost always be some shortcut built into an official question containing an expression like that, and you'd be able to bypass a detailed analysis.

But if you did want to know how to think about an inequality like |x + 3| + |y| < 1 in general, it makes sense to first start with this equation:

|x| + |y| = 1

You might notice that would look exactly like the equation of a line, in coordinate geometry, if we erased the absolute values. In general, if you start throwing absolute values into an equation of a line, you end up with a coordinate geometry picture that looks like pieces of lines stuck together. So in this case, you can instantly see that (1, 0), (0, 1), (-1, 0) and (0, -1) will all be points that satisfy |x| + |y| = 1. It turns out if you just connect those four dots with straight lines, making a diamond shape, then you'll have the correct graph of that equation. You can see why that is true: if you imagine first that x and y are both positive, so we're in the first quadrant, then the absolute values don't change anything, and we get the equation x + y = 1, or y = -x + 1. So you get the line of slope -1 with y-intercept at 1, which is the line connecting (0, 1) and (1, 0). But because absolute value makes negatives into positives, the same points will work if we make one or both of the coordinates negative, which is how we get all four lines of the diamond.

Now if you change this (or any other coordinate geometry equation) into an inequality:

|x| + |y| < 1

then the picture we get when we make this an equation |x| + |y| = 1 becomes the borderline separating the points where |x| + |y| < 1 and the points where |x| + |y| > 1. So the inequality |x| + |y| < 1 is just all of the points strictly inside the diamond we drew above.

If you want to get even more complicated, by replacing the "x" with "x + 3", then you're dealing with a 'translation'. I've never seen a real GMAT question where you'd need to understand translations like this, but if you take a coordinate geometry equation, and replace every "x" with "x + 3", then you're translating the picture 3 units to the left. Similarly if you replace every "y" with "y + 3", you're translating the picture 3 units down. So the inequality |x + 3| + |y| < 1 is just a diamond identical to the one we drew above in size, but which is 'centred' at (-3, 0) instead of at (0, 0). In other words, its four corners will be at (-4, 0), (-3, 1), (-2, 0) and (-3, -1); you can see by plugging those four points into the left side that they each make the left side exactly equal to 1, so they are points on the boundary.

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