SOLUTION: please help me: {{{2y/y-1 = 5/y + 10-8y/y^2-y}}}

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Question 253925: please help me:
2y%2Fy-1+=+5%2Fy+%2B+10-8y%2Fy%5E2-y
Found 3 solutions by stanbon, richwmiller, dabanfield:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2y/(y-1) = 5/y + 10-8y/(y^2-y)
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Multiply thru by y(y-1) to get:
2y(y) = 5(y-1) + 10y(y-1) - 8y
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2y^2 = 5y-1 + 10y^2 - 10y - 8y
2y^2 = -13y + 10y^2
8y^2 - 13y = 0
y(8y - 13) = 0
y = 0 or y = 13/8
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Cheers,
Stan H.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!

2y%2Fy-1+=+5%2Fy+%2B+10-8y%2Fy%5E2-y
8y/y^2=8/y
eventually you will have to multiply both sides by y-1 and also by y
There will be three solutions.
what does that tell you about exponents?

Answer by dabanfield(803) About Me  (Show Source):
You can put this solution on YOUR website!
please help me: 2y%2Fy-1+=+5%2Fy+%2B+10-8y%2Fy%5E2-y
First we simplify:
Remember 2y/y = 2 and y/y^2 = 8/y.
So we have:
2 - 1 = 5/y + 10 - 8/y - y
Now multiply both sides by y:
2y - y = 5 + 10y - 8 - y^2
So we have:
y^2 - 9y + 3 = 0
Solve for y.