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Inequalities

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Intern
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Joined: 11 Aug 2013
Posts: 32
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Kudos [?]: 6 [0], given: 9

Inequalities [#permalink] New post 16 Aug 2013, 16:19
4|x+1/2|=18

I'm not following this, please explain.
4 KUDOS received
Manager
Manager
Joined: 21 Aug 2012
Posts: 150
Followers: 2

Kudos [?]: 27 [4] , given: 41

Re: Inequalities [#permalink] New post 16 Aug 2013, 20:44
4
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selfishmofo wrote:
4|x+1/2|=18

I'm not following this, please explain.


Hi,

First lets discuss about MODULUS ( | | ) .
Modulus is represented by ||

MODULUS basically refers to distance. It is never negative, because distance can never be in negative.
We never say that the distance from Chicago to New York is -234 miles...

Now, the beauty of modulus is that it converts the value of any number to positive, irrespective of the original value.

Example: suppose x = 5 and y = -2

Modulus of x i .e |X| = 5
Modulus of y i .e |Y| = 2

So, you see what happened here... when a number comes in MODULUS, the modulus of that number is always positive.

Here, x was already positive, so the modulus had no impact on it, but for y the modulus of Y is positive.

Now, basically talking in terms of mathematical theory.

MODULUS refers to the distance on the number line..!!!! [ Very important ]

if i say |X|=5, then it means that the distance of X from origin is 5.

Lets see how this happens.

|X| = 5
|X - 0 | = 5 ... the derived form is | X - a |

That is, the distance of X from 0 (origin) is 5...
Remember it is the distance that is 5, not the value of X.

See the number line below,

<----- 5 -----------><------- 5 ------>
------------|--- ---------------|-------------------|-------
<-............ -5 ................. 0 .................. 5 ......->


I hope it gets clear from this number line.
The distance of X from origin is 5, i.e |X|=5, the value of X could be 5 or -5. X cannot be both 5 and -5.

Since, we are here concerned with MODULUS, we will talk about distance.

if |X| = 5,
it means x could be 5 or -5.

We are not sure about the exact value of X..

Now coming to the question:

4|x+1/2|=18

So, |x+1/2| = 18/4 = 9/2

Now, |x+1/2| = 9/2

That means, the distance of X from -1/2 is 9/2.

| x - (-1/2) | = 9/2 ... NOTE: always convert the number to the form | X-a | .... X - a means distance of X from a .

As states earlier, if |X| =5, then x could be 5 or -5.

In the same way,

x + 1/2 = 9/2 or -9/2

Solving,
x+1/2 = 9/2 or x+1/2 = -9/2

x = 9/2-1/2 or x = -9/2 -1/2
x= 4 or -5

Now, lets cross verify our answer on the number line.

-----------|------------|-------------|--------
<-...... -5 ............-1/2 ........ 4 .......->

| x - (-1/2) | = 9/2
the distance of X from -1/2.

Lets calculate the distance on both the ends.

For -5: -1/2 - (-5) =>-1/2 +5 => 9/2 (Remember we always measure distance by subtracting left terms from right terms)
For 4: 4 - ( -1/2) => 4 + 1/2 => 9/2


Hence, we have got the answer correctly.

Thanks,
Jai

KUDOS if it helped..!!! :)
_________________

MODULUS Concept ---> inequalities-158054.html#p1257636
HEXAGON Theory ---> hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308

Intern
Intern
Joined: 11 Aug 2013
Posts: 32
Followers: 0

Kudos [?]: 6 [0], given: 9

Re: Inequalities [#permalink] New post 19 Aug 2013, 14:01
Jai, thank you for your response. It was extremely clear how you explained it.
Manager
Manager
Joined: 21 Aug 2012
Posts: 150
Followers: 2

Kudos [?]: 27 [0], given: 41

Re: Inequalities [#permalink] New post 19 Aug 2013, 23:05
Narenn wrote:
Good Job, Jai. +1


Thanks Narenn..!!! :)
_________________

MODULUS Concept ---> inequalities-158054.html#p1257636
HEXAGON Theory ---> hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308

Manager
Manager
Joined: 21 Aug 2012
Posts: 150
Followers: 2

Kudos [?]: 27 [0], given: 41

Re: Inequalities [#permalink] New post 19 Aug 2013, 23:06
selfishmofo wrote:
Jai, thank you for your response. It was extremely clear how you explained it.


Your welcome selfishmofo..!!! :)
_________________

MODULUS Concept ---> inequalities-158054.html#p1257636
HEXAGON Theory ---> hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308

Re: Inequalities   [#permalink] 19 Aug 2013, 23:06
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