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absolute value change sign on what side?

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absolute value change sign on what side? [#permalink] New post 10 Jan 2012, 20:33
Why the value of x is different in the two scenarios? Is there any rule related to what side expression sign should be change?

Please advise.

|x+3| < 4

positive case:
1st approach: changing sign on the absolute expression side i.e. LHS
x+3 < 4
x < 1

negative case:
-x-3 < 4
-x < 7
x > -7

so -7 < x < 1


2nd approach: changing sign on the RHS

positive case:
x+3 < 4
x < 1

negative case:
x+3 < -4
x < -7

so x < -7

Last edited by agarwalmanoj2000 on 10 Jan 2012, 23:34, edited 1 time in total.
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Re: absolute value change sign on what side? [#permalink] New post 10 Jan 2012, 22:32
|x+3| < 4

ur 1st solution is good. the second one is not good :)

u stated that x<-7. please check it out. let x=-8 ,then u have got
|-8+3| < 4 |-5| < 4 or 5<4 it is obviously wrong
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Re: absolute value change sign on what side? [#permalink] New post 11 Jan 2012, 00:34
In general, |a| = a when a is positive (or zero). When a is negative, then |a| = -a (because a is negative, -a will be positive). So if you see an expression like |x + 3|, there are two possibilities:

x + 3 > 0, so x > -3, in which case |x + 3| = x + 3

x + 3 < 0, so x < -3, in which case |x + 3| = -(x + 3)

When you break your problem into cases, you seem to be ignoring the assumptions you're making - that is, in the positive case, you're assuming in advance that x > -3, so you shouldn't be including any solutions less than -3 when you look only at that case.

Now, when you solve an equation like |x + 3| = 5, you can certainly break the problem into two cases. The 'positive case' is straightforward. In the negative case (where x < -3), we replace |x + 3| with -(x + 3). We then have

-(x + 3) = 5

Notice that since we can multiply an equation safely on both sides by -1, we can rewrite this as follows:

x + 3 = -5

We can *only* do this with an equation. If we instead had an inequality like |x + 3| < 5, then when we consider the 'negative case' where x < -3, we can still replace |x + 3| with -(x + 3) to get

-(x + 3) < 5

Notice however that we are *not* free to instead put the negative sign on the right side of the inequality. To do that, we'd need to multiply both sides by -1, and when you multiply on both sides of an inequality by a negative number, you need to *reverse* the inequality. That was the problem with your second solution.

All of that said, I don't like case-by-case approaches to absolute value. If you understand that |a - b| is the distance between a and b on the number line, then you can interpret the expression |x + 3| = |x - (-3)| as the distance between x and -3. So if |x - (-3)| is less than 4, then x is less than 4 away from -3, so must be between -7 and 1.
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Re: absolute value change sign on what side? [#permalink] New post 11 Jan 2012, 01:33
Hi Ian,

Thanks a lot for clear response !!!

I takeaway from your response :
1) If absolute value expression is involved in an equation then we can change sign on any of two sides (LHS or RHS) of the equation.

|x+3| = 4

-(x+3) = 4
-x -3 = 4
-x=7
x=-7

OR

x+3 = -4
x=-7

Result is same.

2) If absolute value expression is involved in an inequality then:
2a) We can change sign on absolute expression side without changing the inequality sign.

|x+3| < 4
- (x+3) < 4 (inequality sign is not changed)
-x-3 < 4
-x < 7
x > -7

2b) We can change sign on other side of absolute expression side but we have to change the inequality sign.
|x+3| < 4
(x+3) > -4 (inequality sign is changed)
x+3 > -4
x > -7


For consistency and simplicity sake, I will change the sign on the absolute expression side, so that I need not change the inequality sign and I can use same approach for equation and inequality with absolute value expression.
Re: absolute value change sign on what side?   [#permalink] 11 Jan 2012, 01:33
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