SOLUTION: I am having a lot of trouble with this problem, and cannot get it to work out.
{{{((6)/(y-3))-((2)/(y))=((7y+3)/(y^2-9))}}}
After going through several different methods I co
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Question 321707: I am having a lot of trouble with this problem, and cannot get it to work out.
After going through several different methods I could think of, and each time getting a different and wrong answer, I still don't see how the answers of -1 and 6 are calculated.
From what I remember, the first thing is that (y-9)(y-9)(y) is the common denominator, and so all is multiplied by a factor of one in that form, canceling out all denominators, which would leave us with 6(y-3)(y) - 2(y-3)(y-3) = 7y+3(y)
This becomes -3y^2 - 9y - 18 = 0
And then -3(y^2 + 3y + 6) = 0
I cannot factor this, and so in using the quadratic formula I end up with a crazy number that is not the solution given on the practice test.
Any help in figuring out where I'm going wrong would be much appreciated.
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
No, the common denominator is because .
.
.
.
Make that change and try it again.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You are sooooooo close...
\left(\frac{y(y\,-\,3)(y\,+\,3)}{y(y\,-\,3)(y\,+\,3)}\right)\ =\ 0)
\ -\ 2(y\,-\,3)(y\,+\,3)\ -\ (7y\ +\ 3)y}{y(y\,-\,3)(y\,+\,3)} =\ 0)
And right there is your mistake. Check the sign on the binomial factor in the first term of the numerator.
\ =\ 0)
And the quadratic factors...
John
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