3 Inequations

[arjmen]

Q3: I recommend solving it geometrically. Much easier for linear absolute values.

Yep, I would recommend this too! In fact I'd initially solved this problem using the graphing approach.
I explained it differently here because I was too lazy to draw a graph.

[Synergy79]

Guys, I find this number line approach excellent for solving such problems.

Thanks arjmen & dimas

[daipayan_b]

Guys i tried to use the graph....
Can you just plot a simple example and demonstrate how to plot it on the graph....

I might sound silly...but just do this once..

Thanks

[DoingMyPart]

Can some one elaborate a little more about this simple step..

(x-4)(x-3) < 0
Therefore, 3 < x < 4

Logicaly its fine ... I understand that that 1 should be +ve and other -ve.. but is there a more mechanical way..to find x .... as it would be if for (x-4)(x-3) = 0.. x =3,4 ??

[garbage100]

For question number 3: I have a confusion

|x+8| < |x-3|

Here are the four equations:

Eq 1 => x+8 < x-3 (Cannot infer anything)

Eq 2 =>
x+8 < -(x-3)
x+8 < -x+3
2x < -5
x < -5/2 (everyones answer)

Eq 3 => -(x+8) < -(x+3) (Cannot infer anything)

Eq 4 =>
-(x+8) < x+3
-x-8 < x+3
-5<2x
-5/2<x
x>-5/2
whats wrong with equation 4? Why is x>-5/2 not a solution?

Update:
I saw a similar problem discussed. I found from Graycellz explanation(below link) that for the second set of equations (in my case eqn 3,4) we should change sign. But I dont understand why.
http://www.urch.com/forums/gmat-math...nequality.html (absolute value inequality)

[manish8109]

G100,
The best way of solving this problem, is t square both sides, then

(x+8)^2<(x-3)^2
=> x^2+64+16x<x^2+9-6x
=> 22x<-55
=> x<-5/2(ANS)

[arjmen]

Eq 4 =>
-(x+8) < x+3
-x-8 < x+3
-5<2x
-5/2<x
x>-5/2
whats wrong with equation 4? Why is x>-5/2 not a solution?

The reason you can't write, -(x+8) < x-3 is that this is akin to saying that, x+8 < 0, but x-3 >= 0. This is not possible!

If |x+8| < |x-3|, what are the boundary conditions you can scrutinise?

1) x+8 > 0
2) x+8 <= 0
3) x-3 >= 0
4) x-3 < 0

1 and 2, I've looked at in my previous explanatory post. What I'll try to prove here is that we don't need to go any further.

Looking at the 3rd BC: this is not posible given the absolute value inequation. For x-3 >=0 , we need x>0. But then, |x+8| > |x-3|

Now, the 4th BC: If x-3 < 0, then for the original inequation to hold true, -(x-3) > x+8 OR -(x-3) > -(x+8)
We've already looked at both of these before!

[garbage100]

Thanks Arjmen. Thanks Manish.

So it looks like we have to keenly watch the boundary conditions which we select while removing ||.

Manish, your method looks simple....would it always work...(means i wouldnt have to worry about these boundary conditions then).

[manish8109]

G100,
if the Equation is in this form |x+8| < |x-3| means mod sign on both sides then u can use it.

thanx

[jjaacc]

The squaring thing is a neat solution. I got caught by the boundary problems too.

Thanks for the clarification guys.