Which of the following inequalities is equal to |2x-|x||

Question

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

fmik7894 · Accepted Answer

You can write the inequality as: -6 < |2x-|x|| < 6 Case 1: If x >= 0 ; |x| = x |2x-|x|| becomes 2x - x = x -6 < x < 6 But as x>=0 we only take the values as 0 < x < 6 Case 2: If x < 0 ; |x| = -x |2x-|x|| becomes 2x + x = 3x -6 < 3x < 6 -2 < x < 2 But as x < 0 we only take the values as -2 < x < 0 Combining the results of Case 1 and 2 -2 < x < 6 Answer : C

IanStewart · Answer

If you were stuck on this question (and two inequalities cannot be "equal" to each other, so the wording should be different), you could always just plug in numbers. If you plug in x = 4, you discover the inequality works, so x < 2 is false, and B, C or E is correct. If you plug in x = -4, the inequality does not work, so the minimum value of x is greater than -4, and E is wrong. Clearly the inequality will work for negative values close to zero, so B is wrong and C must be right (or you can plug in x = -1 to establish that).

If you wanted to do the problem algebraically, you can use the fact that:

|k| = k when k > 0
|k| = -k when k < 0

It's much simpler to apply this to the "|x|" first, rather than to the "|2x - |x||" -- someone asked above whether we work from the inside outwards or from the outside inwards in situations like this, and technically either approach is fine, but it's going to be much easier to apply absolute value cases with the simplest possible expressions first, so with "|x|" in this question.

So when x > 0:

|2x - |x|| = |2x - x| = |x|, and since we're assuming x > 0, |x| = x. So when x > 0, the inequality becomes x < 6. We haven't even considered negative values yet, so all we've proved is that the inequality certainly works whenever 0 < x < 6. But it might also work for some negative values of x.

When x < 0, |x| = -x, so:

|2x - |x|| = |2x - (-x)| = |2x + x| = |3x| = 3|x|, and since we're assuming x < 0, this is equal to -3x. So when x < 0, the inequality says -3x < 6, or, dividing by -3 on both sides and flipping the inequality (because we're dividing by a negative), x > -2. So when x is a negative number, x > -2 must be true, and the inequality also works when -2 < x < 0.

Either case includes the possibility that x = 0, so combining the two cases, -2 < x < 6.

Bunuel · Answer

Similar question: https://gmatclub.com/forum/which-of-the ... 47913.html

nkmungila · Answer

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

MathRevolution · Answer

=>

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

pikolo2510 · Answer

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

dave13 · Answer

hey there

MathRevolution , i dont get ...when we have expression |x|

please show how to do it in this case

got so confused with this question !!!!

dave13 · Answer

ibra10 maybe you can shed some absolute light on my questions

Ahmed9955 · Answer

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

Ahmed9955 · Answer

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

Ahmed9955 · Answer

IanStewart   Bunuel  can you help?

Posted from my mobile device

siriusly · Answer

IanStewart  thanks a lot! Your explanation was amazing

CEdward · Answer

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?

IanStewart · Answer

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.

bumpbot · Answer

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

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