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