anonymous
  • anonymous
Will medal! Solve: 6/(x-3)=3/x for x and determine if the solution is extraneous.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
x = -2, extraneous x = -2, non-extraneous x = -3, non-extraneous x = -3, extraneous
anonymous
  • anonymous
x= -3 I'm not sure but I think it would non-extraneous
anonymous
  • anonymous
When you plug that in you would get -1=-1 so hence the solution is non-extranoeus right?

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anonymous
  • anonymous
Could you help with this one too?
anonymous
  • anonymous
x = 3, non-extraneous x = 3, extraneous x = -3, extraneous x = -3, non-extraneous
anonymous
  • anonymous
Wait, what was the question? :o
anonymous
  • anonymous
Okay whoops
anonymous
  • anonymous
3/(x-3)=(x/x-3)-(3/2)
anonymous
  • anonymous
3/(x-3)=(x/x-3)-(3/2)
anonymous
  • anonymous
3/(x-3) - x/(x-3) = - (3/2) (3 - x) / (x - 3) = - (3/2) 3 - x = -3/2 (x-3) 3 - x = -(3/2)x + 9/2 6 - 2x = -3x + 9 x = 3
anonymous
  • anonymous
I think it would x = 3 would be extraneous? I'm not quite sure.
anonymous
  • anonymous
yup thats what i got
anonymous
  • anonymous
Thanks so much :D
anonymous
  • anonymous
You're welcome! :)

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