Question 78011
As I understand the equation you are to solve, it is:
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{{{y = 3y^2/(3y+1)}}}
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One thing to note in passing is that y cannot be {{{-1/3}}} because that would give you
a denominator of zero and division by 0 is not allowed.
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Let's get rid of the denominator by multiplying both sides by (3y + 1).  When you do that
the denominator on the right side cancels with this multiplier and the equation becomes:
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{{{y*(3y + 1) = 3y^2}}}
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Multiply out the left side and the equation becomes:
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{{{3y^2 + y = 3y^2}}}
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Subtract 3y^2 from both sides and you are left with:
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{{{y = 0}}}
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That's the solution to the problem.  You can check by going back to the original problem
and substituting 0 for y and you will find that the equation reduces to 0 = 0/1 which is 0 = 0.
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Hope this helps.