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# x+|x–2|–|x+3|=5

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30 May 2003, 03:19
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

x+|x–2|–|x+3|=5
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30 May 2003, 04:39
Looks fine with them
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30 May 2003, 22:15
ABNY2002 wrote:
How did you get -6?

You have to correctly open up moduls by using intervals—and not simply terminate them.
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02 Jun 2003, 07:46
When I plug in -6 back into the equation, I dont get 5.

I think only 10 works.

Can someone put -6 back into the equation and see what they get.
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02 Jun 2003, 15:17
true, I don't get 5 either when x=-6
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02 Jun 2003, 21:24
Sorry, guys.
Tzolkin made an error and I didn't check it. The equation has the only solution 10. When we open up moduls, -6 is here but is not within an appropriate interval.
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04 Jul 2005, 05:46
sparky wrote:
Recycling ..

I got 10 as the only soln. But in a very round about way , breaking x into intervals of x>=2 , x betwenn -3 and 2 , and x < -3.

Is there any other simpler way of doing it.

HMTG.
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04 Jul 2005, 12:19
No,
with absolute values
first you need to break everything in intervals
then solve for each interval
if the solution falls within an interval it's valid.
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04 Jul 2005, 13:00
IMO its 0 as well. x+(-(x-2))-(-(x+3))=5 => x-x+2+x+3=5 => x+5=5 => x=0 OR x+x-2-x-3=5 => x=10
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Re: modul equation [#permalink]

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04 Jul 2005, 14:00
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stolyar wrote:
Try this one:

x+|xâ€“2|â€“|x+3|=5

let's see

(x-2) > 0 => x>2
(x+3)>0 => x>-3

so

x<-3, x - (x-2) + (x+3) = 5 => x = 0 > -3, so no solution
-3<x<2, x -(x-2) - (x+3) = 5 => x=-6 , no solution
x>2, x + (x-2) - (x+3) = 5 => x = 10, x>2 so 10 is a solution here

so the answer is x=10
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Re: modul equation [#permalink]

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30 Jan 2009, 14:01
stolyar wrote:
Try this one:

x+|x–2|–|x+3|=5

Agree with x = 10.

X = 0 and -6 do not work.
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Re: modul equation   [#permalink] 30 Jan 2009, 14:01
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# x+|x–2|–|x+3|=5

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