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solving absolute values..!

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solving absolute values..!

by vzzai » Tue Nov 08, 2011 3:22 am
|x-2|+|x-3|<2
How do I solve this?

I solved it by squaring both sides?
|x-2| < 2-|x-3|
(x-2)^2 < (2-(x-3))^2
I got x<21/6

Is this a right approach? are is there a quicker way?... Please explain.
Thank you,
Vj
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by Anurag@Gurome » Tue Nov 08, 2011 3:35 am
No. This is not right.
Always remember that whenever you square both sides of a equation, you are incorporating some extra solutions.

In this case you got x < 21/6.
Check for x = 0, |x - 2| + |x - 3| = 2 + 3 = 5 > 2
Hence, your answer is wrong.

Probably the most easiest way to solve this problem is to visualize 2, 3, and x on the number line. Now, |x - 2| and |x - 3| are nothing but the distance of x from 2 and 3 on the number line. Hence, according to the given equation the sum of the distance of x from 2 and 3 on the number line must be less than 2.

From simple observation we can notice that if x lies between 2 and 3, the sum will be equal to 1, i.e. less than 2.

Now what about the points less than 2 or greater than 3?
Draw a number line as follows,

<--------------0----1----2----3----4------------>

Now, we can see that for any x less than 2 or greater than 3, the sum of the distances is (1 + twice the distance of x from 2 or 3 respectively).

Say, x is either d less than 2 or d greater than 3.
Then, the sum of distances = 1 + 2d < 2 ----> d < 0.5

So, (2 -0.5) < x < (3 + 0.5)
--> 1.5 < x < 3.5

This is the final solution.

Note: This approach may appear lengthy, but once someone grabs the basic idea behind it most of it can be done mentally and real fast.
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by rijul007 » Tue Nov 08, 2011 3:39 am
In such type of ques.. you need to take case and use a number line


|x-2|+|x-3|<2

Case1: x>3

x-2 + x-3 < 2
2x < 7
x < 7/2

Case 2: 2 < x < 3

x-2 + 3 - x < 2
1 < 2

Case 3: x < 2

2 - x + 3 - x < 2
5 -2x < 2
x > 3/2



Image

3/2 < x < 7/2
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by shankar.ashwin » Tue Nov 08, 2011 3:48 am
Just my way of doing absolute value based sums,

whenever you have a term like |x-2|, try removing the mod, you get 2 possibilities

|x-2| = x-2 (When x-2 is +ve) and
|x-2| = -x+2 (When x-2 is -ve)

Now solve the sum for both cases;

For positive;

x - 2 + x -3 < 2

2x -5 < 2
2x < 7 (or) x < 7/2

Case 2:

When negative

-x + 2 -x + 3 < 2

-2x + 5 < 2

-2x < -3

2x > 3

x > 3/2.

Combining we have 3/2 > x > 7/2
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by Anurag@Gurome » Tue Nov 08, 2011 3:58 am
shankar.ashwin wrote:|x-2| = x-2 (When x-2 is +ve) and
|x-2| = -x+2 (When x-2 is -ve)

...

x - 2 + x -3 < 2

I'm not sure if your understanding is wrong or you just typed it in short, but I'd suggest anyone following this method to be careful.

When (x - 2) is positive, then |x - 2| = (x - 2)
But that doesn't mean |x - 3| = (x - 3)

When x is greater than both 2 and 3, i.e. x > 3, then |x - 2| = (x - 2) and |x - 3| = (x - 3)

But when 2 < x < 3, then |x - 2| = (x - 2) and |x - 3| = -(x - 3) = (3 - x)

Luckily for this question such a logical mistake didn't create any problem.

However, for the other case, i.e. when (x - 2) is negative, i.e. x < 2, x is also less than 3 and (x - 3) is negative. Hence, |x - 2| = -(x - 2) and |x - 3| = -(x - 3)

The method followed by rijul007 is the proper algebraic way to tackle absolute value problems.
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by pemdas » Tue Nov 08, 2011 4:26 am
if you could just sneak into the 4th edition of MGMAT VICs, inequalities, pages 96-97 you would be familiar with the concept I am referring to >>>
the center points of two graphs are 2 and 3 (also they are called critical points)
now id you plot on the number line ------2-----3------
our inequality must be less than 2, hence all the negative will only add to the result will be greater than 2. We consider only intervals (some positive number-2), (2-3) and (3-positive some number). We note that the interval (2-3) is the certain solution area here. Also, it's possible only for 1.5<x and x<3.5. Thus 1.5<x<3.5. We also should account for the possibility of mode to be set as 0.
vzzai wrote:|x-2|+|x-3|<2
How do I solve this?

I solved it by squaring both sides?
|x-2| < 2-|x-3|
(x-2)^2 < (2-(x-3))^2
I got x<21/6

Is this a right approach? are is there a quicker way?... Please explain.
Success doesn't come overnight!
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by shankar.ashwin » Tue Nov 08, 2011 4:46 am
Thanks!! Always miss to consider non integer cases..!!
Anurag@Gurome wrote:
shankar.ashwin wrote:|x-2| = x-2 (When x-2 is +ve) and
|x-2| = -x+2 (When x-2 is -ve)

...

x - 2 + x -3 < 2

I'm not sure if your understanding is wrong or you just typed it in short, but I'd suggest anyone following this method to be careful.

When (x - 2) is positive, then |x - 2| = (x - 2)
But that doesn't mean |x - 3| = (x - 3)

When x is greater than both 2 and 3, i.e. x > 3, then |x - 2| = (x - 2) and |x - 3| = (x - 3)

But when 2 < x < 3, then |x - 2| = (x - 2) and |x - 3| = -(x - 3) = (3 - x)

Luckily for this question such a logical mistake didn't create any problem.

However, for the other case, i.e. when (x - 2) is negative, i.e. x < 2, x is also less than 3 and (x - 3) is negative. Hence, |x - 2| = -(x - 2) and |x - 3| = -(x - 3)

The method followed by rijul007 is the proper algebraic way to tackle absolute value problems.
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by pemdas » Tue Nov 08, 2011 5:01 am
all approaches are interesting, i.e. Anurag's and rijul007, but there's not universal approach here
the cases can be different, then we must account for mod=0 too
the modes can be more than 2, 3,4 ... then we cannot make simple observation and apply d-formula
the intervals may not be easily joined as 2<x<3 and we have to consider each case separately
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