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# x+|x–2|–|x+3|=5

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SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [1] , given: 0

x+|x–2|–|x+3|=5 [#permalink]  30 May 2003, 03:19
1
KUDOS
x+|x–2|–|x+3|=5
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

Looks fine with them
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

ABNY2002 wrote:
How did you get -6?

You have to correctly open up moduls by using intervals—and not simply terminate them.
Intern
Joined: 15 Apr 2003
Posts: 23
Followers: 0

Kudos [?]: 0 [0], given: 0

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.
Manager
Joined: 25 May 2003
Posts: 54
Followers: 1

Kudos [?]: 3 [0], given: 0

true, I don't get 5 either when x=-6
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

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.
Senior Manager
Joined: 17 Apr 2005
Posts: 375
Location: India
Followers: 1

Kudos [?]: 17 [0], given: 0

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.
Director
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

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.
VP
Joined: 30 Sep 2004
Posts: 1490
Location: Germany
Followers: 4

Kudos [?]: 91 [0], given: 0

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
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Director
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [1] , given: 0

Re: modul equation [#permalink]  04 Jul 2005, 14:00
1
KUDOS
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

SVP
Joined: 29 Aug 2007
Posts: 2497
Followers: 57

Kudos [?]: 556 [0], given: 19

Re: modul equation [#permalink]  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.
_________________
Re: modul equation   [#permalink] 30 Jan 2009, 14:01
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