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modules

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B
yezz
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modules [#permalink] New post   22 Oct 2006, 14:38
solve for x

/6+x/ = 2x+1



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Re: modules [#permalink] New post   22 Oct 2006, 15:15
|6+x| = 2x+1

o If x>-6, |6+x|=6+x
=> 6+x = 2x+1
<=> x = 5

o If x<-6, |6+x|=-x-6
=> -6-x = 2x+1
<=> 3*x = -7
<=> x= -7/3
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Re: modules [#permalink] New post   22 Oct 2006, 15:52
Fig wrote:
|6+x| = 2x+1

o If x>-6, |6+x|=6+x
=> 6+x = 2x+1
<=> x = 5

o If x<-6, |6+x|=-x-6
=> -6-x = 2x+1
<=> 3*x = -7
<=> x= -7/3


i believe this equation has only one solution x = 5

|6+x| >= 0
2x + 1 >= 0 or x >= -1/2
hence x cannot be -7/3

and if you substitute x = -7/3 in the original equation, it won't hold.
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Re: modules [#permalink] New post   22 Oct 2006, 16:41
Can anyone explain how did you guys figure out from this equation |6+x|=2x+1 that |6+x| >=0 and carried on with the solution????
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Re: modules [#permalink] New post   22 Oct 2006, 23:12
iced_tea wrote:
Fig wrote:
|6+x| = 2x+1

o If x>-6, |6+x|=6+x
=> 6+x = 2x+1
<=> x = 5

o If x<-6, |6+x|=-x-6
=> -6-x = 2x+1
<=> 3*x = -7
<=> x= -7/3


i believe this equation has only one solution x = 5

|6+x| >= 0
2x + 1 >= 0 or x >= -1/2
hence x cannot be -7/3

and if you substitute x = -7/3 in the original equation, it won't hold.


Yes :)... The second one is out :) We have to check back in the original equation as u did :)
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Re: modules [#permalink] New post   23 Oct 2006, 17:01
[quote="ishtmeet"]Can anyone explain how did you guys figure out from this equation |6+x|=2x+1 that |6+x| >=0 and carried on with the solution????[/quot

an absolute value can never be negative.



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Re: modules [#permalink]
23 Oct 2006, 17:01
Moderator:
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

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