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Mathematics 14 Online
OpenStudy (anonymous):

Will medal! Solve: 6/(x-3)=3/x for x and determine if the solution is extraneous.

OpenStudy (anonymous):

x = -2, extraneous x = -2, non-extraneous x = -3, non-extraneous x = -3, extraneous

OpenStudy (anonymous):

x= -3 I'm not sure but I think it would non-extraneous

OpenStudy (anonymous):

When you plug that in you would get -1=-1 so hence the solution is non-extranoeus right?

OpenStudy (anonymous):

Could you help with this one too?

OpenStudy (anonymous):

x = 3, non-extraneous x = 3, extraneous x = -3, extraneous x = -3, non-extraneous

OpenStudy (anonymous):

Wait, what was the question? :o

OpenStudy (anonymous):

Okay whoops

OpenStudy (anonymous):

3/(x-3)=(x/x-3)-(3/2)

OpenStudy (anonymous):

3/(x-3)=(x/x-3)-(3/2)

OpenStudy (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

OpenStudy (anonymous):

I think it would x = 3 would be extraneous? I'm not quite sure.

OpenStudy (anonymous):

yup thats what i got

OpenStudy (anonymous):

Thanks so much :D

OpenStudy (anonymous):

You're welcome! :)

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