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General question with Absolute values

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General question with Absolute values

by beater » Wed Oct 08, 2008 6:28 am
Guys - I'm a little confused with Absolute values. Could someone please solve the equation listed below. I'd like to see your workings and understand your approach.

Solve | x – 3 | = | 3x + 2 | – 1

Thanks for your help!
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by gupta.amit3 » Wed Oct 08, 2008 9:36 am
| x – 3 | = | 3x + 2 | – 1

The modulus mentioned in equations will change signs at x=3 and x=-2/3

First consider x<-2/3

-(x-3) = -(3x+2) - 1
Solving, x=-3

Putting it back in the modulus equ you will find that both sides are equal to 6, so x=-3 is one solution

consider -2/3<x<3
-(x-3) = (3x+2) - 1
Solving, x=1/2

Putting it back in the modulus equ you will find that both sides are equal to 5/2, so x=1/2 is another solution

consider x>3
x-3=3x+2-1

Solving x=-2

Putting it back in the modulus equ you will find that RHS is 5 while LHS is 3, which are not equal. So x=-2 is not a solution.

I hope this explanation helps
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by beater » Mon Oct 20, 2008 7:03 pm
Could someone please explain the as to how x = 1/2 for numbers between -2/3 and 3.

Why do you put negative sign in front of (x-3)?
That is, -(x-3) = (3x+2) - 1
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by cramya » Mon Oct 20, 2008 7:49 pm
If x is just on one side of the equation then by definition of modulus

|x| = -x if x is negative

|x| = x if x is positive or 0

For example |X| = 2

Case1: x=2
Case2: -x = 2
x = -2

Since |2| = 2 and | - 2 | = - ( - 2) = 2

By definition this holds good


PART 2

If x is on both sides of the equation

then we have 4 cases +/+,-/-,+/- and -/+


Case 1: +/+

| x – 3 | = | 3x + 2 | – 1

x-3 = 3x+2-1

Case 2: -/-

- (x-3) = - (3x+2) - 1

Case 3: +/-

x-3 = - (3x+2) - 1

Case 4: -/+

- (x-3) = 3x+2-1


You could get different values of x but you need to subsitute it back in the equation to see which value of x works i.e satisfies the equation


Hope this helps!

P.S I was just as confused as you in the beginning when I saw absolute value equations; now i feel slighlty better :) I learnt PART 2 from one of Ron's(GMAT expert on this site) posts in a different forum.
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