SOLUTION: y=3y(squared) over 3y+1

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Question 78011This question is from textbook GLENCOE
: y=3y(squared) over 3y+1 This question is from textbook GLENCOE

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
As I understand the equation you are to solve, it is:
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y+=+3y%5E2%2F%283y%2B1%29
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One thing to note in passing is that y cannot be -1%2F3 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%2A%283y+%2B+1%29+=+3y%5E2
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Multiply out the left side and the equation becomes:
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3y%5E2+%2B+y+=+3y%5E2
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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.