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wsypinero
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Absolute Range

by wsypinero Wed Sep 09, 2009 6:53 pm

The explanation for (1) confused me. My understanding with absolute value is that the value inside the absolute value lines can be positive or negative. Therefore, if positive then x = 3 and if negative then x = 3 again. I get this part.

What I don't get is the explanation in red text below. This is completely new approach to me.

Please advise.


Is |x| < 1 ?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0
We can rephrase the question by opening up the absolute value sign. In other words, we must solve all possible scenarios for the inequality, remembering that the absolute value is always a positive value. The two scenarios for the inequality are as follows:

If x > 0, the question becomes "Is x < 1?"
If x < 0, the question becomes: "Is x > -1?"
We can also combine the questions: "Is -1 < x < 1?"

Since Statement 2 is less complex than Statement 1, begin with Statement 2 and a BD/ACE grid.

(1) INSUFFICIENT: There are three possible equations here if we open up the absolute value signs:

1. If x < -1, the values inside the absolute value symbols on both sides of the equation are negative, so we must multiply each through by -1 (to find its opposite, or positive, value):

|x + 1| = 2|x -1| -(x + 1) = 2(1 - x) x = 3
(However, this is invalid since in this scenario, x < -1.)

2. If -1 < x < 1, the value inside the absolute value symbols on the left side of the equation is positive, but the value on the right side of the equation is negative. Thus, only the value on the right side of the equation must be multiplied by -1:

|x + 1| = 2|x -1| x + 1 = 2(1 - x) x = 1/3


3. If x > 1, the values inside the absolute value symbols on both sides of the equation are positive. Thus, we can simply remove the absolute value symbols:

|x + 1| = 2|x -1| x + 1 = 2(x - 1) x = 3

Thus x = 1/3 or 3. While 1/3 is between -1 and 1, 3 is not. Thus, we cannot answer the question.

(2) INSUFFICIENT: There are two possible equations here if we open up the absolute value sign:

1. If x > 3, the value inside the absolute value symbols is greater than zero. Thus, we can simply remove the absolute value symbols:

|x - 3| > 0 x - 3 > 0 x > 3

2. If x < 3, the value inside the absolute value symbols is negative, so we must multiply through by -1 (to find its opposite, or positive, value).

|x - 3| > 0 3 - x > 0 x < 3

If x is either greater than 3 or less than 3, then x is anything but 3. This does not answer the question as to whether x is between -1 and 1.

(1) AND (2) SUFFICIENT: According to statement (1), x can be 3 or 1/3. According to statement (2), x cannot be 3. Thus using both statements, we know that x = 1/3 which IS between -1 and 1.

The correct answer is C.
anoo.anand
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Re: Absolute Range

by anoo.anand Sat Sep 19, 2009 3:31 pm

can u please post all the questions individually ..so that we can have a better discussion....

Thanks
RonPurewal
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Re: Absolute Range

by RonPurewal Sat Sep 26, 2009 3:06 am

yeah, that's a needlessly cumbersome explanation. ick. we should fix this.

this should be better:
if you have
| QUANTITY 1 | = | QUANTITY 2 |
(with NOTHING ADDED to, or SUBTRACTED from, the abs. values)
then
just SOLVE TWO EQUATIONS:
* QUANTITY 1 = QUANTITY 2
* QUANTITY 1 = -(QUANTITY 2)

this is all you need. (there are also the possibilities with a negative sign in front of quantity 1, but those will just be equivalent the two already written here.)

wsypinero Wrote:the value inside the absolute value lines can be positive or negative.


yeah, but in this case there are TWO DIFFERENT quantities in absolute value. hence, you can potentially have any combination of +/- in front of these quantities.
marc.gagnon
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Re: Absolute Range

by marc.gagnon Tue May 17, 2011 9:38 pm

As a follow up on this question stem:
The answer discussion for statement one is as follows:
1. If x < -1, the values inside the absolute value symbols on both sides of the equation are negative, so we must multiply each through by -1 (to find its opposite, or positive, value):

|x + 1| = 2|x -1| --->> -(x + 1) = 2(1 - x) --->> x = 3
(However, this is invalid since in this scenario, x < -1.)


When statement one is solved if -1 < x < 1, the result is
|x + 1| = 2|x -1| --->> x + 1 = 2(1 - x) --->> x = 1/3

I understand that based on this we determined that the only valid solution for statement one is x = 1/3; as we determined that x <> 3. With this, can you help me understand why statement 1 alone isn't sufficient.
RonPurewal
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Re: Absolute Range

by RonPurewal Sat May 21, 2011 4:29 am

marc.gagnon Wrote:As a follow up on this question stem:
The answer discussion for statement one is as follows:
1. If x < -1, the values inside the absolute value symbols on both sides of the equation are negative, so we must multiply each through by -1 (to find its opposite, or positive, value):

|x + 1| = 2|x -1| --->> -(x + 1) = 2(1 - x) --->> x = 3
(However, this is invalid since in this scenario, x < -1.)


When statement one is solved if -1 < x < 1, the result is
|x + 1| = 2|x -1| --->> x + 1 = 2(1 - x) --->> x = 1/3

I understand that based on this we determined that the only valid solution for statement one is x = 1/3; as we determined that x <> 3. With this, can you help me understand why statement 1 alone isn't sufficient.


well, again, this whole "breaking down into different parts of the number line" approach is needlessly bulky and labor-intensive -- as i wrote in the post above yours (which you should read, if you haven't already), you can just solve the following two equations
QUANTITY1 = QUANTITY2
QUANTITY1 = -QUANTITY2
... and just check the answers to see whether they work.
"QUANTITY1 = QUANTITY2" is "x + 1 = 2x - 2", which gives x = 3. if you plug this back in --> |4| = 2|2| -- it works.
"QUANTITY1 = -QUANTITY2" is "x + 1 = -2x + 2", which gives x = 1/3. if you plug this back in --> |4/3| = 2|-2/3| -- it also works.
therefore, 3 and 1/3 are both solutions; no need to worry about this number line range stuff. as i said, we'll edit the explanation so that it doesn't go into such unnecessary things.

--

if you do go into the "breaking down the number line" thing, then, when you solve for x > 1, you'll get x = 3 again -- and this time it will count (because 3 > 1).
try it; you'll see.
(note that x > 1 is currently missing from your analysis; you have only x < -1 and -1 < x < 1. this is the problem.)

--

what does "<>" mean?
marc.gagnon
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Re: Absolute Range

by marc.gagnon Wed May 25, 2011 11:14 pm

<> is used in some programming languages to indicate Not Equal to.
RonPurewal
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Re: Absolute Range

by RonPurewal Thu May 26, 2011 6:09 pm

marc.gagnon Wrote:<> is used in some programming languages to indicate Not Equal to.


ah, i get it, it's like "not =, so the other 2 remaining options are < and >". gotcha

does the explanation for the problem make sense?
ali_sha12j
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Re: Absolute Range

by ali_sha12j Fri Jul 29, 2011 12:23 pm

The question is if -1<x<1.agreed!

Cant we solve the statement 1 as -x-1=-2x+2 and x+1=2x-2?
And similarly statement 2? But this ends up really confusing. Can someone help in solving question from beginning?
RonPurewal
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Re: Absolute Range

by RonPurewal Sat Jul 30, 2011 3:31 am

ali_sha12j Wrote:The question is if -1<x<1.agreed!

Cant we solve the statement 1 as -x-1=-2x+2 and x+1=2x-2?

no; that's the same equation twice.
if you don't see why, consider the equations "a = b" and "-a = -b"; these two are obviously the same. the same problem exists here.

see this post (from this thread)
post31369.html#p31369

Can someone help in solving question from beginning?


have you read the thread? the problem is solved from start to finish already.
ali_sha12j
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Re: Absolute Range

by ali_sha12j Sun Jul 31, 2011 9:06 pm

RonPurewal Wrote:yeah, that's a needlessly cumbersome explanation. ick. we should fix this.

this should be better:
if you have
| QUANTITY 1 | = | QUANTITY 2 |
(with NOTHING ADDED to, or SUBTRACTED from, the abs. values)
then
just SOLVE TWO EQUATIONS:
* QUANTITY 1 = QUANTITY 2
* QUANTITY 1 = -(QUANTITY 2)

this is all you need. (there are also the possibilities with a negative sign in front of quantity 1, but those will just be equivalent the two already written here.)

wsypinero Wrote:the value inside the absolute value lines can be positive or negative.


yeah, but in this case there are TWO DIFFERENT quantities in absolute value. hence, you can potentially have any combination of +/- in front of these quantities.


i got the point that abosolute quantity 1=absolute quantity 2 equation. but, in the statement one there is 2 outside the absolute equation. doesnt a constant make a difference? and also when there is a < or > sign, do we need to solve it one equation with - and other with + just like in statement 2?
ali_sha12j
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Re: Absolute Range

by ali_sha12j Mon Aug 01, 2011 9:04 am

i slept and when got up tried the question all over again. i got my answers.. thanks!
RonPurewal
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Re: Absolute Range

by RonPurewal Mon Aug 08, 2011 1:06 am

ali_sha12j Wrote:i slept and when got up tried the question all over again. i got my answers.. thanks!


all good.