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Inequality - question - concept

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Inequality - question - concept [#permalink] New post 29 Dec 2006, 13:17
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|2x+3| < x+6

We know how to solve this

2x+3 < x+6 or -(2x+3) < x+6

and then test the goodness of regions on the number lines

now here is a variation

|2x+3| > x+6

I think we need to consider 3 equations (although) the x values divide the number line in to 3 regions (2 equations yield same x value with < or > relationship) . Please comment if you think only 2 equations are SUFFICIENT.

This is how I solve it

Equ 1 ........2x+3>x+6 ----> x>3
equ 2.........-(2x+3)>x+6 ----->x<-3
eqn3 .........-(2x+3)<x+6 ----->x>-3

equation 2 and 3

yeild the same point on number line and that is what we are concerned with but will eqn 3 yeild a third point on the number line (if I form a different question)? i am begining to think no. can someone please confirm?

thank you
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 [#permalink] New post 29 Dec 2006, 14:19
Actually, the way to solve this kind of inequation is to determine the crucial points at which the sign of every absolute flips from negative to positive.

Following this rule, we have only 1 crucial point : x = -3/2.

So, we have 2 domains to study : x < -3/2 and x >= 3/2. Each domain gives 1 simplified inequation.

In your case, the inequation 1 is affected to x < -3/2. As well, the inequation 2 is related to x >= 3/2.

But, your last inequation is not linked to any domain. Indeed, it never represents the original inequation (flip of sign + minus). :)
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Re: Inequality - question - concept [#permalink] New post 13 Oct 2010, 23:52
Financier wrote:
hemantS wrote:

|2x+3| > x+6



Really, what is the proper way to fix it? Crucial point does not work here :shock: ...


Not sure what the confusion is, you just need to take two cases for the sign of (2x+3).
If 2x+3>0 (x>-3/2) then 2x+3>x+6 OR x>3
If 2x+3<0 (x<-3/2) then -2x-3>x+6 OR 3x<-9 OR x<-3
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Re: Inequality - question - concept [#permalink] New post 14 Oct 2010, 00:00
Thanks. I got mental block and made a huge mistake. Now I'm going through Brutal inequality questions and need you advice:
Does this compilation represent all gmat-like tricks or you advise to look at some extra stuff?

absolute-mpractices-list-of-gmat-questions-39533.html (11th post)
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Re: [#permalink] New post 24 Feb 2012, 21:52
Fig wrote:
Actually, the way to solve this kind of inequation is to determine the crucial points at which the sign of every absolute flips from negative to positive.

Following this rule, we have only 1 crucial point : x = -3/2.

So, we have 2 domains to study : x < -3/2 and x >= 3/2. Each domain gives 1 simplified inequation.

In your case, the inequation 1 is affected to x < -3/2. As well, the inequation 2 is related to x >= 3/2.

But, your last inequation is not linked to any domain. Indeed, it never represents the original inequation (flip of sign + minus). :)


I've seen quite a few references to crucial point in the context of inequalities. What exactly is it? By, at which the sign of every absolute flips from negative to positive are we implying, the value of x for which the expression equates to 0?
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Re: Re: [#permalink] New post 24 Feb 2012, 22:10
Expert's post
fortsill wrote:
Fig wrote:
Actually, the way to solve this kind of inequation is to determine the crucial points at which the sign of every absolute flips from negative to positive.

Following this rule, we have only 1 crucial point : x = -3/2.

So, we have 2 domains to study : x < -3/2 and x >= 3/2. Each domain gives 1 simplified inequation.

In your case, the inequation 1 is affected to x < -3/2. As well, the inequation 2 is related to x >= 3/2.

But, your last inequation is not linked to any domain. Indeed, it never represents the original inequation (flip of sign + minus). :)


I've seen quite a few references to crucial point in the context of inequalities. What exactly is it? By, at which the sign of every absolute flips from negative to positive are we implying, the value of x for which the expression equates to 0?


"Crucial point", aka "check point", aka "key point" is the value of x for which the value of an expression in modulus equals to zero.

For example for |x-3| the check point is 3. Notice that if x<3 then x-3<0 and |x-3|=-(x-3) and if x>=3 then x-3>=0 and |x-3|=(x-3).

Check Absolute Value chapter of Math Book for more: math-absolute-value-modulus-86462.html

DS questions on absolute value to practice: search.php?search_id=tag&tag_id=37
PS questions on absolute value to practice: search.php?search_id=tag&tag_id=58

Tough absolute value and inequity questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Re: Re:   [#permalink] 24 Feb 2012, 22:10
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