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Need Help - Inequalities

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Need Help - Inequalities [#permalink] New post   25 Sep 2012, 09:21
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Hi Experts,

I need help with inequalities.

I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.

I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.

Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.


How many solutions does \(|x+3| - |4-x| = |8+x|\) have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), which is not a valid solution since we are considering \(x < -8\) range and -1 is out of it;

If \(-8\leq{x}\leq{-3}\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;

If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;

If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.

So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).

Thanks in anticipation.
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Re: Need Help - Inequalities [#permalink] New post   25 Sep 2012, 11:39
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GetThisDone wrote:
Hi Experts,

I need help with inequalities.

I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.

I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.

Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.


How many solutions does \(|x+3| - |4-x| = |8+x|\) have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), which is not a valid solution since we are considering \(x < -8\) range and -1 is out of it;

If \(-8\leq{x}\leq{-3}\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;

If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;

If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.

So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).

Thanks in anticipation.


Use the definition of the absolute value: \(|x| = x\) if \(x\geq0\) and \(|x|=-x\)if \(x<0.\)
Obviously, instead of \(x\) you can have any algebraic expression. So, for example \(|x+8|=x+8\) if \(x+8\geq0\) or \(x\geq-8\) and \(|x+8| = -(x+8)\) if \(x+8<0\) or \(x<-8\), etc.

Absolute value expresses a distance between two numbers and it is always non-negative.
\(|x|\) is the distance on the number line between \(x\) and \(0\). The distance between \(x\) and \(0\) is the same as the distance between \(-x\) and \(0.\)
\(|7|=7\) and \(|-7|=7\). How can we connect \(-7\) and \(7\)? There is no mathematical operation "just drop the minus sign..." But obviously \(7=(-1)*(-7)\).
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Re: Need Help - Inequalities [#permalink] New post   25 Sep 2012, 12:11
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|x+3| - |4-x| = |8+x|
Hey
I would be happy to help though I did not get your question very clearly. Do you want to know how did the solution find out the ranges as given? Then I will try to help you out.
I assume you know the basic of the integer function which is:
|x|=x when x>=0
|x|=-x when x<0
So, you can see the checkpoints here are -8, -3 and 4
I hope till now you have followed.
Now lets look at each check points and see how the integer function behaves
When \(x<-8,\)
x+3 is negative, so \(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is negative, so \(|8+x|=-(8+x)\)
So the equation, in this case, becomes,
\(-(x+3)-(4-x)=-(8+x)\)
\(=> x=-1\) which violates the range we have considered

Similarly when\(-8<x<-3\)
x+3 is negative, so\(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so\(|8+x|=(8+x)\)
So equation becomes
\(-(x+3)-(4-x)=(8+x)\)
=>\(x=-15\) which violates the range we have considered

Again, when \(-3<x<4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
So the equation becomes,
\((x+3)-(4-x)=(8+x)\)
 \(2x-1=8+x\)
 \(X=9\) which violates the range we have considered

Last, when \(x>4\)
x+3 is positive, so \(|x+3|=(x+3)\)
4-x is negative, so \(|4-x|=-(4-x)\)
8+x is positive, so \(|8+x|=(8+x)\)
\((x+3)+(4-x)=(8+x)\)
\(=>x=-1\) which again violates the range

SO no solution
Did this help?
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Re: Need Help - Inequalities [#permalink] New post   25 Sep 2012, 18:56
Thank you souvik101990 & EvaJager.

The section that souvik101990 explained is where I had confusion. Got a new question after going through the explanation :). If you could help me with the below, it will help me put this topic to rest.

The range chosen by souvik below is different from the one in the original post \(-8\leq{x}\leq{-3}\). My understanding is that the possible values in the range help you decide the +ve/-ve sign for that term. If we were to consider the original range, what happens to 8+x in the below expansion when x tends to -8.

Or is it that we choose ranges in such a way that we can draw a +ve/-ve sign.

Similarly when\(-8<x<-3\)
x+3 is negative, so\(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so\(|8+x|=(8+x)\)
So equation becomes
\(-(x+3)-(4-x)=(8+x)\)
=>\(x=-15\) which violates the range we have considered
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Re: Need Help - Inequalities [#permalink] New post   25 Sep 2012, 21:28
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GetThisDone wrote:
Thank you souvik101990 & EvaJager.

The section that souvik101990 explained is where I had confusion. Got a new question after going through the explanation :). If you could help me with the below, it will help me put this topic to rest.

The range chosen by souvik below is different from the one in the original post \(-8\leq{x}\leq{-3}\). My understanding is that the possible values in the range help you decide the +ve/-ve sign for that term. If we were to consider the original range, what happens to 8+x in the below expansion when x tends to -8.

Or is it that we choose ranges in such a way that we can draw a +ve/-ve sign.

Similarly when\(-8<x<-3\)
x+3 is negative, so\(|x+3|=-(x+3)\)
4-x is positive, so \(|4-x|=(4-x)\)
8+x is positive, so\(|8+x|=(8+x)\)
So equation becomes
\(-(x+3)-(4-x)=(8+x)\)
=>\(x=-15\) which violates the range we have considered


here \(-8<=x<=-3\) serves essentially the same function because:

When \(x=-8\) the both \((8+x)\) and\(-(8+x)=0\), so it does not change your equation in any way.
Same for (x+3)

Think basic
|\(x|=x when x>=0; \\
|x|=-x when x<0\)

We could very well write this (and not change the meaning) as:
\(|x|=x when x>0; \\
|x|=-x when x<=0\)

It does not matter where you put the \("="\) sign because both the funtions would result in \(0 when x=0.\)
Hope this helps.
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Need Help - Inequalities [#permalink] New post   Updated on: 11 Oct 2022, 02:16
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GetThisDone wrote:
Hi Experts,

I need help with inequalities.

I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.

I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.

Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.


How many solutions does \(|x+3| - |4-x| = |8+x|\) have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), which is not a valid solution since we are considering \(x < -8\) range and -1 is out of it;

If \(-8\leq{x}\leq{-3}\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;

If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;

If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.

So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).

Thanks in anticipation.


You can also try working with a graphical approach (number line). Souvik has already explained the algebraic approach so I will not take it. But let me demonstrate how you can solve this question using a number line.

\(|x+3| - |4-x| = |8+x|\)
\(|x+3| = |8+x| + |x - 4|\)

(|4-x| = |x-4| since there is a mod. After all, |x| = |-x|)

What does this imply? It implies 'x is a point whose sum of distance from -8 and 4 is equal to its distance from -3'

i.e. x is a point where the sum of red and green arrows is equal to the blue arrow.
Look at all the 4 ranges - is it possible anywhere?

x < -8
Attachment:
Ques3.jpg
Ques3.jpg [ 3.99 KiB | Viewed 3300 times ]

Can the sum of red and green arrows is equal to the blue arrow? No.

-8 < x< -3
Attachment:
Ques4.jpg
Ques4.jpg [ 3.37 KiB | Viewed 3315 times ]

Can the sum of red and green arrows is equal to the blue arrow? No.

Same thing for the other two ranges too.

Originally posted by KarishmaB on 02 Oct 2012, 20:49.
Last edited by KarishmaB on 11 Oct 2022, 02:16, edited 1 time in total.
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Re: Need Help - Inequalities [#permalink] New post   02 Oct 2012, 22:12
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The graphical approach is instrumental in solving problems quickly!
+1 to Karishma
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Re: Need Help - Inequalities [#permalink] New post   04 Oct 2012, 02:28
VeritasPrepKarishma wrote:
GetThisDone wrote:
Hi Experts,

I need help with inequalities.

I am finding difficulty in finding solutions to inequalities in complex problems. I understood the concept of finding the solution in simpler problems (to the extent of finding the range and plotting it on the number line). But in the below example, I understand till identifying the check points. Also understood the point of using the check points to create ranges for verification. But could not understand how in each of the scenarios the equation is expanded. On what grounds -ve sings are applied to the equations while expanding.

I have gone through MGMAT material on Inequalities. But could not relate this approach anything there. Also checked the GMAT club math book example. Guess even that is on the same lines and could not understand it for the same reasons.

Could someone please explain the strategy/concept here. What basics am I missing here. Are there any links or material I can refer to better understand the basics I am missing. Help on this will be greatly appreciated.


How many solutions does \(|x+3| - |4-x| = |8+x|\) have?

Basically the same here: we have three check points -8, -3 and 4 and thus four ranges to check.

If \(x < -8\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=-(8+x)\) --> \(x = -1\), which is not a valid solution since we are considering \(x < -8\) range and -1 is out of it;

If \(-8\leq{x}\leq{-3}\) then \(|x+3| - |4-x| = |8+x|\) expands as \(-(x+3)-(4-x)=(8+x)\) --> \(x=-15\), which is also not a valid solution since we are considering \(-8\leq{x}\leq{-3}\) range;

If \(-3<x<4\) then \(|x+3| - |4-x| = |8+x|\) expands as \((x+3)-(4-x)=(8+x)\) --> \(x = 9\), which is also not a valid solution since we are considering \(-3<x<4\) range;

If \(x\geq{4}\) then \(|x+3|-|4-x|=|8+x|\) expands as \((x+3)+(4-x)=(8+x)\) --> \(x = -1\), which is also not a valid solution since we are considering \(x>4\) range.

So we have that no solution exist for \(|x+3|-|4-x|=|8+x|\) (no x can satisfy this equation).

Thanks in anticipation.


You can also try working with a graphical approach (number line). Souvik has already explained the algebraic approach so I will not take it. But let me demonstrate how you can solve this question using a number line.

\(|x+3| - |4-x| = |8+x|\)
\(|x+3| = |8+x| + |x - 4|\)

(|4-x| = |x-4| since there is a mod. After all, |x| = |-x|)

What does this imply? It implies 'x is a point whose sum of distance from -8 and 4 is equal to its distance from -3'

i.e. x is a point where the sum of red and green arrows is equal to the blue arrow.
Look at all the 4 ranges - is it possible anywhere?

x < -8
Attachment:
Ques3.jpg

Can the sum of red and green arrows is equal to the blue arrow? No.

-8 < x< -3
Attachment:
Ques4.jpg

Can the sum of red and green arrows is equal to the blue arrow? No.

Same thing for the other two ranges too.

Check out these posts on solving inequalities and mods using number line:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/01 ... edore-did/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/01 ... s-part-ii/


Huh I never know Inequalities with modulus was so easy !. Graphical concept makes sense and of course so easy.
Thank you Karishma .


+1 Kudos to you.
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