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# Absolute modulus : A better understanding

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Math Expert
Joined: 02 Sep 2009
Posts: 59561
Re: Absolute modulus : A better understanding  [#permalink]

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08 Aug 2017, 05:11
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Re: Absolute modulus : A better understanding  [#permalink]

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15 Aug 2017, 05:43
1
chetan2u wrote:
periyar22 wrote:
This is a very confusing method of solving. What is rear? I cant believe this person who spells rare as rear is an expert

Mr periyar,
Firstly, these are the methods used to solve modulus. So, if you find them confusing, you will have to learn them or leave the topic or invent a new method to solve it.
Secondly this post is related to quant, and I hope you are not mistaking this to be a verbal post, as you seem to be more interested in spellings. Thankfully, you found only one mistake in the big post above. But it is surprising, you could not avoid a mistake in even a single line you have written in last two years. It is not cant but can't.
Anyway, experts badge does not mean anyone should follow what he writes and in the same way, it is not necessary if an intelligent person but without a expert badge, like you, writes people will not follow. Please use this forum and assist others and surely people will follow you more than they follow any expert.
By the way, I will not edit rare and rear as it is a mark of your keen observation.

chetan2u
bro

What is more, it is not correct to say, I can't believe this person....bla bla bla.... is an expert, the correct way, however, to say so is, I can't believe this person ...bla bla bla... to be an expert...

So, bro, just forget such negative comments, and keep posting your valuables ...

thanks
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Re: Absolute modulus : A better understanding  [#permalink]

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22 Sep 2017, 07:28
gmatexam439 wrote:
Thank you Chetan for this. I would like to share an alternate method for solving absolute equations which can be used for as many variables: [I used this while persuing my engg.]

While solving the below equation:
|x+3|-|4-x|=|8+x|

=> |x+3|-|4-x|-|8+x|=0
Therefore the critical points will be x=-8,-3 and 4

Case 1. Now when x<-8 on the number line
|x+3| will open with a -ve sign, |4-x| will open with a +ve sign and |8+x| will open with a -ve sign
Therefore, equation will be
=> -x-3-4+x+8+x=0
=> x=-1; Now if we notice this value of x doesn't match with the case value i.e. x<-8 therefore this is not a correct answer.

Case 2. Now when -8<x<-3 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3-4+x-8-x=0
=> x=9; Now if we notice this value of x doesn't match with the case value i.e. -8<x<-3 therefore this is not a correct answer.

Case 3. Now when -3<x<4 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3-4+x-8-x=0
=> x=9; Now if we notice this value of x doesn't match with the case value i.e. -3<x<4 therefore this is not a correct answer.

Case 4. Now when x>4 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a -ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3+4-x-8-x=0
=> x=-1; Now if we notice this value of x doesn't match with the case value i.e. x>4 therefore this is not a correct answer.

Therefore there is no solution to this equation.

I think in case 2: X should be equal to -15 not 9.
Since -8<x<-3,
(x+3) will be NEGATIVE not positive,
Therefore, equation will be
=> -(x+3)-(4-x)-8-x=0
=> -x-3-4+x-8-x=0
=> -15=x
not that it really matters coz the answer is still invalid but thought I'd still point it out.
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Re: Absolute modulus : A better understanding  [#permalink]

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26 Oct 2017, 06:15
chetan2u & Bunuel Thanks a lot for being in this forum. God Bless you!
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Re: Absolute modulus : A better understanding  [#permalink]

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14 Nov 2017, 04:24
how to apply such method in the actual test?
Or, is this theory that helps to understand the concept of the absolute modulus.
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Re: Absolute modulus : A better understanding  [#permalink]

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28 Jan 2018, 22:12
chetan2u wrote:
AR15J wrote:
Hi Chetan,

First of all, thanks for sharing this information. In graphical method, I could not get how did you draw the graph for two modes at one side (x+3|-|4-x|) . My question might be silly but I could not get it. Please explain. Thanks once again.

Hi,

The LHS is|x+3|-|4-x|...
Take this to be equal to y, so y=|x+3|-|4-x|•••••
Now find y for different values of x..
When x=0, y=|0+3|-|4-0|=-1...
When x=1, y=|1+3|-|4-1|=1...
For all values of x as 7 & above 7, y will be a constant 7... and below -7, y will be -7...
You can draw the graph with these points..

So can I say the method is more like plug and test method? Or is there more genuine way to find the graph?
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Re: Absolute modulus : A better understanding  [#permalink]

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01 Apr 2018, 23:29
Bunuel wrote:
udbhav2412 wrote:
nice post and exactly what i was looking for

In addition to Chetan's excellent post on absolute values, you can find the following topics useful:

For other topics please refer to ALL YOU NEED FOR QUANT ! ! !.

Hope it helps.

Thank you Bunuel. Exactly what I was looking for.
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Re: Absolute modulus : A better understanding  [#permalink]

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21 Jul 2018, 07:50
chetan2u wrote:
periyar22 wrote:
This is a very confusing method of solving. What is rear? I cant believe this person who spells rare as rear is an expert

Mr periyar,
Firstly, these are the methods used to solve modulus. So, if you find them confusing, you will have to learn them or leave the topic or invent a new method to solve it.
Secondly this post is related to quant, and I hope you are not mistaking this to be a verbal post, as you seem to be more interested in spellings. Thankfully, you found only one mistake in the big post above. But it is surprising, you could not avoid a mistake in even a single line you have written in last two years. It is not cant but can't.
Anyway, experts badge does not mean anyone should follow what he writes and in the same way, it is not necessary if an intelligent person but without a expert badge, like you, writes people will not follow. Please use this forum and assist others and surely people will follow you more than they follow any expert.
By the way, I will not edit rare and rear as it is a mark of your keen observation.

I was practicing some quants and almost get bored of it. But now just read this post and got fresh now. Anyway earning an Expert Badge at GMAT Club is not much easy. Thanks to chetan2u for this topic. Kudos!
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Re: Absolute modulus : A better understanding  [#permalink]

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12 Aug 2018, 22:54
chetan2u wrote:
Attachment:
docu1.png

I had a PM and a profile comment asking about the absolute modulus, its concept and in particular a Question discussed on various occassion " How many solutions does the equation |x+3|-|4-x|=|8+x| have?....
Just thought to write down few concepts I have gathered. I have not gone through various Topics on Absolute Modulus in this Forum, so maybe few points are repetition.

Although difficult for a topic like this, I'll try to follow KISS- Keep It Short and Simple. So, let me touch the concepts now..

what is absolute modulus?

Absolute modulus is the numeric value of any number without any sign or in other words ' the distance from the origin'. It will always be positive.

What kind of Qs can one see in GMAT?

The Q will ask either the values of x or how many values can x take?..
most often what one can encounter is a linear Equation with...
a) only one Mod
eg.. |x+2| + 2x= 3..
b) two mods..
|x+2|=|x-3|+1..
c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..

What are the methods

..
three methods..
1) As the property suggests, Open each modulus in both +ive and -ive ....
2) Critical value
3) Graphical method..

Opening each modulus

It is a time consuming process where you open each mod in both positive and negative and the number of Equations thus formed will increase as we increase the no of mods..

a) only one Mod
eg.. |x+2| + 2x= 3..

i) (x+2) + 2x=3.. 3x+2=3 x=1/3.. valid value
ii) -(x+2)+2x=3.. x-2=3..x=5...
but if we substitute x=5 in |x+2| + 2x= 3..... |x+2| will turn out to be a negative value so discard
so one value of x..

b) two mods..
|x+2|=|x-3|+1..
here you will have four equations..
i)(x+2)=(x-3)+1.. both positive

ii)-(x+2)=-(x-3)+1.. both negative

iii)-(x+2)=(x-3)+1..one positive and other negative

iv)(x+2)=-(x-3)+1.. opposite of the one on top

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
it will further increase the number of equations..

Suggestion.. time consuming and susceptible to errors in opening of brackets and at times requires more to negate the values found as in first example.

Critical method

lets find what happens in this before trying the Qs as this was the main query..
Step 1 :- for each mod, there is a value of x that will make that mod to 0..
Step 2 :- the minimum value of a mod will be 0 and at this value of x, the mod has the lowest value...
Once we know this critical value, we work on the mod for values lesser than(<) that or more than(>)that and including the critical value in either of them,
we assign a sign, + or -, depending on what will happen to the value inside the mod in each scenario(in one scenario it will be positive and in other, negative)..
Step 3 :- after assigning the sign, we solve for x and see if the value of x that we get is possible depending on which side of critical value we are working on..

So what are we doing here
We are assuming a certain region for value of x and then solving for x.. If the value found matches the initial assumption, we take that as a solution or discard that value, which would mean that there is no value of x in that assumed region

lets see the three examples
a) only one Mod
eg.. |x+2| + 2x= 3..
here x+2 will be 0 at x=-2..
so Critical value =-2..
so two regions are <-2 and >= -2

i)when x<-2, |x+2|will be given negative sign.. for this assign any value in that region say -3 in this case x+2 will become -3+2=-1 hence a negative sign..
-(x+2)+2x=3..
x-2=3.. x=5, which is not in the region <-2.. so not a valid value..

ii)when x>=-2, |x+2|will be given positive sign.. for this assign any value in that region say 3 in this case x+2 will become 3+2= 5 hence a positive sign..
(x+2)+2x=3..
3x+2=3.. x=1/3, which is in the region >=-2.. so a valid value..

b) two mods..
|x+2|=|x-3|+1..
critical values -2 and 3...
so regions are <-2, -2<=x<3, x>=3..

i) x<-2...
x+2 will be -ive and x-3 will be negative ..
eq becomes -(x+2)=-(x-3)+1.. both negative
-x-2=-x+3+1..... no values..

ii) $$-2<=x<3$$..
x+2 will be positive and x-3 will be negative ..
eq becomes (x+2)=-(x-3)+1..
x+2=-x+3+1..
x=1.. valid value

iii)x>=3..
x+2 will be positive and x-3 will be positive ..
eq becomes (x+2)=(x-3)+1..
x+2=x-3+1..
no valid value..
so the solution is x=1

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
its time consuming and can be solved similarly..

Graphical method

for graphical method we will have to again use the critical point..
at critical point, it is the lowest value of mod and on either side it increases with a negative slope on one side and positive slope on other side
so it forms a 'V' shape in linear equation and a 'U ' curve for Quadratic Equation..
If the mod has a negative sign in front, -|x+3|, it will have an "inverted V" shape with max value at critical value..

lets see the three examples..

a) only one Mod
eg.. |x+2| + 2x= 3..
critical value at -2 and equation can be written as
|x+2| = 3-2x..
we take y=|x+2| and draw a graph and then take y=3-2x and again draw graph..
the point of intersection is our value..

b) two mods..
|x+2|=|x-3|+1..
here we will take y=|x+2| and y=|x-3|+1
again the point of intersection of two sides will give us the value of x..

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
Here we have three critical values, but the graph will still be only two, one for LHS and one for RHS..
It will not be three for three mod as someone has drawn it in one of the discussions on this Q..
again we see the intersection of two graph..

there are no points of intersection , so no solution

THE FINER POINT

1) Opening modulus is time consuming, susceptible to error, and the answer found can still be wrong and has to checked by putting the values in mod again..
should be least priority and should be used by someone has not been able to grasp finer points of other two methods..

2) "Critical method" should be the one used in most circumstances although it requires a good understanding of signs given to the mod when opened within a region.
It has to be the method, when you are looking for values of X..

3) "Graphical method" is useful in finding the number of values of x, as getting accurate values of x may be difficult while estimating from free hand graphs..
but if understood much faster and easier to find sol for Q like How many solutions does the equation |x+3|-|4-x|=|8+x| have?....

Hope it helps atleast a few of you..

Hi Chetan,

First of all, thank you for the detailed explanation given!
I have a question on the graphical method used. In graphical method for one mod - you have mentioned about the critical point. What is significance of that there because i could not see the critical value being used anywhere...
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Re: Absolute modulus : A better understanding  [#permalink]

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15 Oct 2018, 00:23
Hi Chetan,

Could you please explain why 'x+2 will be positive' and 'x-3 will be negative'?

−2<=x<3−2<=x<3 ..
x+2 will be positive and x-3 will be negative ..
eq becomes (x+2)=-(x-3)+1..
x+2=-x+3+1..
x=1.. valid value

Thanks,
Megha

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Re: Absolute modulus : A better understanding  [#permalink]

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29 Oct 2018, 21:38
gmatexam439 wrote:
Thank you Chetan for this. I would like to share an alternate method for solving absolute equations which can be used for as many variables: [I used this while persuing my engg.]

While solving the below equation:
|x+3|-|4-x|=|8+x|

=> |x+3|-|4-x|-|8+x|=0
Therefore the critical points will be x=-8,-3 and 4

Case 1. Now when x<-8 on the number line
|x+3| will open with a -ve sign, |4-x| will open with a +ve sign and |8+x| will open with a -ve sign
Therefore, equation will be
=> -x-3-4+x+8+x=0
=> x=-1; Now if we notice this value of x doesn't match with the case value i.e. x<-8 therefore this is not a correct answer.

Case 2. Now when -8<x<-3 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3-4+x-8-x=0
=> x=9; Now if we notice this value of x doesn't match with the case value i.e. -8<x<-3 therefore this is not a correct answer.

Case 3. Now when -3<x<4 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3-4+x-8-x=0
=> x=9; Now if we notice this value of x doesn't match with the case value i.e. -3<x<4 therefore this is not a correct answer.

Case 4. Now when x>4 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a -ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3+4-x-8-x=0
=> x=-1; Now if we notice this value of x doesn't match with the case value i.e. x>4 therefore this is not a correct answer.

Therefore there is no solution to this equation.

HI gmatexam439 , wanted to understand the logic behind the negative/positive signs you have given while opening the Mods such as " |x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign"?
Anyone else who would like to give a shot at this explanation? Bunuel ?
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Joined: 02 Sep 2009
Posts: 59561
Re: Absolute modulus : A better understanding  [#permalink]

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29 Oct 2018, 21:52
Arpitkumar wrote:
gmatexam439 wrote:
Thank you Chetan for this. I would like to share an alternate method for solving absolute equations which can be used for as many variables: [I used this while persuing my engg.]

While solving the below equation:
|x+3|-|4-x|=|8+x|

=> |x+3|-|4-x|-|8+x|=0
Therefore the critical points will be x=-8,-3 and 4

Case 1. Now when x<-8 on the number line
|x+3| will open with a -ve sign, |4-x| will open with a +ve sign and |8+x| will open with a -ve sign
Therefore, equation will be
=> -x-3-4+x+8+x=0
=> x=-1; Now if we notice this value of x doesn't match with the case value i.e. x<-8 therefore this is not a correct answer.

Case 2. Now when -8<x<-3 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3-4+x-8-x=0
=> x=9; Now if we notice this value of x doesn't match with the case value i.e. -8<x<-3 therefore this is not a correct answer.

Case 3. Now when -3<x<4 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3-4+x-8-x=0
=> x=9; Now if we notice this value of x doesn't match with the case value i.e. -3<x<4 therefore this is not a correct answer.

Case 4. Now when x>4 on the number line
|x+3| will open with a +ve sign, |4-x| will open with a -ve sign and |8+x| will open with a +ve sign
Therefore, equation will be
=> x+3+4-x-8-x=0
=> x=-1; Now if we notice this value of x doesn't match with the case value i.e. x>4 therefore this is not a correct answer.

Therefore there is no solution to this equation.

HI gmatexam439 , wanted to understand the logic behind the negative/positive signs you have given while opening the Mods such as " |x+3| will open with a +ve sign, |4-x| will open with a +ve sign and |8+x| will open with a +ve sign"?
Anyone else who would like to give a shot at this explanation? Bunuel ?

This question is discussed in great detail here: https://gmatclub.com/forum/x-3-4-x-8-x- ... 48996.html

Hope it helps.
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Posts: 8284
Re: Absolute modulus : A better understanding  [#permalink]

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29 Oct 2018, 22:57
1
Megha1119 wrote:
Hi Chetan,

Could you please explain why 'x+2 will be positive' and 'x-3 will be negative'?

−2<=x<3−2<=x<3 ..
x+2 will be positive and x-3 will be negative ..
eq becomes (x+2)=-(x-3)+1..
x+2=-x+3+1..
x=1.. valid value

Thanks,
Megha

Posted from my mobile device

HI Megha1119,

the equation is |x+2|=|x-3|+1
now we take the range .. $$-2\leq{x}<3$$
when you substitute any value in the given range say 0..
1) x+2 becomes 0+2=2 so POSITIVE, hence |x+2|$$\geq$$0 in this range
2) x-3 becomes 0-3=-3 so NEGATIVE, hence |x-3|<0 in this range
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Re: Absolute modulus : A better understanding  [#permalink]

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05 Dec 2018, 04:33
chetan2u wrote:
Attachment:
The attachment docu1.png is no longer available

I had a PM and a profile comment asking about the absolute modulus, its concept and in particular a Question discussed on various occassion " How many solutions does the equation |x+3|-|4-x|=|8+x| have?....
Just thought to write down few concepts I have gathered. I have not gone through various Topics on Absolute Modulus in this Forum, so maybe few points are repetition.

Although difficult for a topic like this, I'll try to follow KISS- Keep It Short and Simple. So, let me touch the concepts now..

what is absolute modulus?

Absolute modulus is the numeric value of any number without any sign or in other words ' the distance from the origin'. It will always be positive.

What kind of Qs can one see in GMAT?

The Q will ask either the values of x or how many values can x take?..
most often what one can encounter is a linear Equation with...
a) only one Mod
eg.. |x+2| + 2x= 3..
b) two mods..
|x+2|=|x-3|+1..
c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..

What are the methods

..
three methods..
1) As the property suggests, Open each modulus in both +ive and -ive ....
2) Critical value
3) Graphical method..

Opening each modulus

It is a time consuming process where you open each mod in both positive and negative and the number of Equations thus formed will increase as we increase the no of mods..

a) only one Mod
eg.. |x+2| + 2x= 3..

i) (x+2) + 2x=3.. 3x+2=3 x=1/3.. valid value
ii) -(x+2)+2x=3.. x-2=3..x=5...
but if we substitute x=5 in |x+2| + 2x= 3..... |x+2| will turn out to be a negative value so discard
so one value of x..

b) two mods..
|x+2|=|x-3|+1..
here you will have four equations..
i)(x+2)=(x-3)+1.. both positive

ii)-(x+2)=-(x-3)+1.. both negative

iii)-(x+2)=(x-3)+1..one positive and other negative

iv)(x+2)=-(x-3)+1.. opposite of the one on top

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
it will further increase the number of equations..

Suggestion.. time consuming and susceptible to errors in opening of brackets and at times requires more to negate the values found as in first example.

Critical method

lets find what happens in this before trying the Qs as this was the main query..
Step 1 :- for each mod, there is a value of x that will make that mod to 0..
Step 2 :- the minimum value of a mod will be 0 and at this value of x, the mod has the lowest value...
Once we know this critical value, we work on the mod for values lesser than(<) that or more than(>)that and including the critical value in either of them,
we assign a sign, + or -, depending on what will happen to the value inside the mod in each scenario(in one scenario it will be positive and in other, negative)..
Step 3 :- after assigning the sign, we solve for x and see if the value of x that we get is possible depending on which side of critical value we are working on..

So what are we doing here
We are assuming a certain region for value of x and then solving for x.. If the value found matches the initial assumption, we take that as a solution or discard that value, which would mean that there is no value of x in that assumed region

lets see the three examples
a) only one Mod
eg.. |x+2| + 2x= 3..
here x+2 will be 0 at x=-2..
so Critical value =-2..
so two regions are <-2 and >= -2

i)when x<-2, |x+2|will be given negative sign.. for this assign any value in that region say -3 in this case x+2 will become -3+2=-1 hence a negative sign..
-(x+2)+2x=3..
x-2=3.. x=5, which is not in the region <-2.. so not a valid value..

ii)when x>=-2, |x+2|will be given positive sign.. for this assign any value in that region say 3 in this case x+2 will become 3+2= 5 hence a positive sign..
(x+2)+2x=3..
3x+2=3.. x=1/3, which is in the region >=-2.. so a valid value..

b) two mods..
|x+2|=|x-3|+1..
critical values -2 and 3...
so regions are <-2, -2<=x<3, x>=3..

i) x<-2...
x+2 will be -ive and x-3 will be negative ..
eq becomes -(x+2)=-(x-3)+1.. both negative
-x-2=-x+3+1..... no values..

ii) $$-2<=x<3$$..
x+2 will be positive and x-3 will be negative ..
eq becomes (x+2)=-(x-3)+1..
x+2=-x+3+1..
x=1.. valid value

iii)x>=3..
x+2 will be positive and x-3 will be positive ..
eq becomes (x+2)=(x-3)+1..
x+2=x-3+1..
no valid value..
so the solution is x=1

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
its time consuming and can be solved similarly..

Graphical method

for graphical method we will have to again use the critical point..
at critical point, it is the lowest value of mod and on either side it increases with a negative slope on one side and positive slope on other side
so it forms a 'V' shape in linear equation and a 'U ' curve for Quadratic Equation..
If the mod has a negative sign in front, -|x+3|, it will have an "inverted V" shape with max value at critical value..

lets see the three examples..

a) only one Mod
eg.. |x+2| + 2x= 3..
critical value at -2 and equation can be written as
|x+2| = 3-2x..
we take y=|x+2| and draw a graph and then take y=3-2x and again draw graph..
the point of intersection is our value..

b) two mods..
|x+2|=|x-3|+1..
here we will take y=|x+2| and y=|x-3|+1
again the point of intersection of two sides will give us the value of x..

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
Here we have three critical values, but the graph will still be only two, one for LHS and one for RHS..
It will not be three for three mod as someone has drawn it in one of the discussions on this Q..
again we see the intersection of two graph..

there are no points of intersection , so no solution

THE FINER POINT

1) Opening modulus is time consuming, susceptible to error, and the answer found can still be wrong and has to checked by putting the values in mod again..
should be least priority and should be used by someone has not been able to grasp finer points of other two methods..

2) "Critical method" should be the one used in most circumstances although it requires a good understanding of signs given to the mod when opened within a region.
It has to be the method, when you are looking for values of X..

3) "Graphical method" is useful in finding the number of values of x, as getting accurate values of x may be difficult while estimating from free hand graphs..
but if understood much faster and easier to find sol for Q like How many solutions does the equation |x+3|-|4-x|=|8+x| have?....

Hope it helps atleast a few of you..

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Re: Absolute modulus : A better understanding  [#permalink]

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05 Dec 2018, 04:34
deddex wrote:
chetan2u wrote:
Attachment:
The attachment docu1.png is no longer available

I had a PM and a profile comment asking about the absolute modulus, its concept and in particular a Question discussed on various occassion " How many solutions does the equation |x+3|-|4-x|=|8+x| have?....
Just thought to write down few concepts I have gathered. I have not gone through various Topics on Absolute Modulus in this Forum, so maybe few points are repetition.

Although difficult for a topic like this, I'll try to follow KISS- Keep It Short and Simple. So, let me touch the concepts now..

what is absolute modulus?

Absolute modulus is the numeric value of any number without any sign or in other words ' the distance from the origin'. It will always be positive.

What kind of Qs can one see in GMAT?

The Q will ask either the values of x or how many values can x take?..
most often what one can encounter is a linear Equation with...
a) only one Mod
eg.. |x+2| + 2x= 3..
b) two mods..
|x+2|=|x-3|+1..
c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..

What are the methods

..
three methods..
1) As the property suggests, Open each modulus in both +ive and -ive ....
2) Critical value
3) Graphical method..

Opening each modulus

It is a time consuming process where you open each mod in both positive and negative and the number of Equations thus formed will increase as we increase the no of mods..

a) only one Mod
eg.. |x+2| + 2x= 3..

i) (x+2) + 2x=3.. 3x+2=3 x=1/3.. valid value
ii) -(x+2)+2x=3.. x-2=3..x=5...
but if we substitute x=5 in |x+2| + 2x= 3..... |x+2| will turn out to be a negative value so discard
so one value of x..

b) two mods..
|x+2|=|x-3|+1..
here you will have four equations..
i)(x+2)=(x-3)+1.. both positive

ii)-(x+2)=-(x-3)+1.. both negative

iii)-(x+2)=(x-3)+1..one positive and other negative

iv)(x+2)=-(x-3)+1.. opposite of the one on top

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
it will further increase the number of equations..

Suggestion.. time consuming and susceptible to errors in opening of brackets and at times requires more to negate the values found as in first example.

Critical method

lets find what happens in this before trying the Qs as this was the main query..
Step 1 :- for each mod, there is a value of x that will make that mod to 0..
Step 2 :- the minimum value of a mod will be 0 and at this value of x, the mod has the lowest value...
Once we know this critical value, we work on the mod for values lesser than(<) that or more than(>)that and including the critical value in either of them,
we assign a sign, + or -, depending on what will happen to the value inside the mod in each scenario(in one scenario it will be positive and in other, negative)..
Step 3 :- after assigning the sign, we solve for x and see if the value of x that we get is possible depending on which side of critical value we are working on..

So what are we doing here
We are assuming a certain region for value of x and then solving for x.. If the value found matches the initial assumption, we take that as a solution or discard that value, which would mean that there is no value of x in that assumed region

lets see the three examples
a) only one Mod
eg.. |x+2| + 2x= 3..
here x+2 will be 0 at x=-2..
so Critical value =-2..
so two regions are <-2 and >= -2

i)when x<-2, |x+2|will be given negative sign.. for this assign any value in that region say -3 in this case x+2 will become -3+2=-1 hence a negative sign..
-(x+2)+2x=3..
x-2=3.. x=5, which is not in the region <-2.. so not a valid value..

ii)when x>=-2, |x+2|will be given positive sign.. for this assign any value in that region say 3 in this case x+2 will become 3+2= 5 hence a positive sign..
(x+2)+2x=3..
3x+2=3.. x=1/3, which is in the region >=-2.. so a valid value..

b) two mods..
|x+2|=|x-3|+1..
critical values -2 and 3...
so regions are <-2, -2<=x<3, x>=3..

i) x<-2...
x+2 will be -ive and x-3 will be negative ..
eq becomes -(x+2)=-(x-3)+1.. both negative
-x-2=-x+3+1..... no values..

ii) $$-2<=x<3$$..
x+2 will be positive and x-3 will be negative ..
eq becomes (x+2)=-(x-3)+1..
x+2=-x+3+1..
x=1.. valid value

iii)x>=3..
x+2 will be positive and x-3 will be positive ..
eq becomes (x+2)=(x-3)+1..
x+2=x-3+1..
no valid value..
so the solution is x=1

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
its time consuming and can be solved similarly..

Graphical method

for graphical method we will have to again use the critical point..
at critical point, it is the lowest value of mod and on either side it increases with a negative slope on one side and positive slope on other side
so it forms a 'V' shape in linear equation and a 'U ' curve for Quadratic Equation..
If the mod has a negative sign in front, -|x+3|, it will have an "inverted V" shape with max value at critical value..

lets see the three examples..

a) only one Mod
eg.. |x+2| + 2x= 3..
critical value at -2 and equation can be written as
|x+2| = 3-2x..
we take y=|x+2| and draw a graph and then take y=3-2x and again draw graph..
the point of intersection is our value..

b) two mods..
|x+2|=|x-3|+1..
here we will take y=|x+2| and y=|x-3|+1
again the point of intersection of two sides will give us the value of x..

c) three mods.. very rare
|x+3|-|4-x|=|8+x| ..
Here we have three critical values, but the graph will still be only two, one for LHS and one for RHS..
It will not be three for three mod as someone has drawn it in one of the discussions on this Q..
again we see the intersection of two graph..

there are no points of intersection , so no solution

THE FINER POINT

1) Opening modulus is time consuming, susceptible to error, and the answer found can still be wrong and has to checked by putting the values in mod again..
should be least priority and should be used by someone has not been able to grasp finer points of other two methods..

2) "Critical method" should be the one used in most circumstances although it requires a good understanding of signs given to the mod when opened within a region.
It has to be the method, when you are looking for values of X..

3) "Graphical method" is useful in finding the number of values of x, as getting accurate values of x may be difficult while estimating from free hand graphs..
but if understood much faster and easier to find sol for Q like How many solutions does the equation |x+3|-|4-x|=|8+x| have?....

Hope it helps atleast a few of you..

Attachments

File comment: solution of |x+3|-|4-x|=|8+x| through critical method as explained by Bunel (Part 2)

WhatsApp Image 2018-12-05 at 10.35.43 AM.jpeg [ 83.3 KiB | Viewed 991 times ]

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Re: Absolute modulus : A better understanding  [#permalink]

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28 Jul 2019, 23:37
Thanks for sharing and writing this in detail. It has definitely helps me.
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Re: Absolute modulus : A better understanding  [#permalink]

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30 Jul 2019, 23:16
Hi, can you explain how to solve two mods equations having inequality? Ex. Is |x-10| > |x-30|?

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Re: Absolute modulus : A better understanding  [#permalink]

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03 Aug 2019, 02:39
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Re: Absolute modulus : A better understanding  [#permalink]

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11 Sep 2019, 23:52
chetan2u

i dont know how you drew that 3 modulus graphical method..

especially the LHS, as RHS side is easy as |8+x|

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Re: Absolute modulus : A better understanding  [#permalink]

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05 Nov 2019, 15:25
chetan2u

Great post, thanks!

Could you elaborate how come:

ii) -(x+2)+2x=3.. x-2=3..x=5...
but if we substitute x=5 in |x+2| + 2x= 3..... |x+2| will turn out to be a negative value so discard.

How will it turn negative if we substitute with 5? And what will become neg. exactly?

Re: Absolute modulus : A better understanding   [#permalink] 05 Nov 2019, 15:25

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