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Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
1.1 Text Problems
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Can you help me with this question out o text 1.1. I keep getting a different answer than the book has
#61.
(2/y+3)+(3/y-4)=(5/y+6)
(2/y+3)*(y-4)(y+6) = 2y-8(y+6) = 2y^2+12y-8y-48 = 2y^2-4y-48
(3/y-4)*(y+3)(y+6 = 3y+9(y+6) = 3y^2+18y+9y+54 = 3y^2+27y+54
(5/y+6)*(y+3)(y-4) = 5y+15 = 5y^2-20y+15y-60 = 5y^2+23y+6
2y^2-4y-48 + 3y^2+27y+54 = 5y^2+23y+6
5y^2+23y+6 = 5y^2-5y-60
-5y^2 -5y^2
23y+6 = -5y-60
-6 -6
23y = -5y-66
+5y = +5y
28y = -66
y = -66/28
The answer in the book is -11/6
Where did I go wrong???
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Good question.
I suspect that the equation is
2/(y+3)+3/(y-4)=5/(y+6).
If this is the case, then the least common denominator would be
(y+3) (y - 4) (y + 6).
Multiplying all terms by this denominator the equation becomes
2/(y+3) * (y+3) (y - 4) (y + 6) +3/(y-4) * (y+3) (y - 4) (y + 6) = 5/(y+6) * (y+3) (y - 4) (y + 6)
For each term the denominator matches up with one of the factors of the numerator and the two divide out. For example on the first term
2/(y+3) * (y+3) (y - 4) (y + 6)
the first (y + 3) is in the denominator, the second in the numerator, so the term can be written
2 * ((y+3) / (y+3) ) * (y - 4) (y + 6)
and since (y+3) / (y + 3) = 1, this term becomes
2 * 1 * (y - 4) (y + 6) = (y - 4) ( y + 6).
Something very similar happens with each of the other terms, and your equation ends up being
2 (y - 4) ( y + 6) + 3 (y + 3) ( y + 6) = 5 ( y + 3) ( y - 4).
This can then be simplified and solved, and I believe you will be able to do so.
I can confirm that the solution is y = -11/6.
You should also clarify how these expressions follow the order of operations.
For example 2 / y + 3 means divide 2 by y then add 3, whereas 2 / (y + 3) means divide 2 by the sum y + 3. The latter is therefore a fraction with 2 in the numerator and y + 3 in the denominator.
I believe in the text problem the terms are in the latter form.
I suggest putting the two forms of the problem into Wolfram Alpha (just search the Web for 'Wolfram Alpha') to see clearly how they are represented. You can just copy and paste each of the following:
(2/y+3)+(3/y-4)=(5/y+6)
and
2/(y+3)+3/(y-4)=5/(y+6)
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You also have a couple of errors within your simplifications:
2y^2+12y-8y-48 = 2y^2-4y-48
should read
2y^2+12y-8y-48 = 2 y^2 + 4 y - 48
and
5y^2-20y+15y-60 = 5y^2+23y+6
should read
5y^2-20y+15y-60 = 5 y^2 - 5 y - 60.
These expressions won't really occur in solving the problem, but they should help you understand the process of simplification.
Once more Wolfram Alpha can help you confirm how the order of operations applies to your expressions.
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