Asked by Matt O.
I'm trying to perform the indicated operations, and reduce to lowest terms, but I'm not sure if I got the correct answer. The question is y-5/(x^2+5x)+(xy+5y) + 1/x+y - 2/x+5....I got y-6, but am not sure if that is correct.
Answers
Answered by
Reiny
Are you the same James who posted several versions of this yesterday?
http://www.jiskha.com/display.cgi?id=1269297854
If so, it is an improvement, but still not too clear.
I think that (x^2+5x)+(xy+5y) probably should be the entire denominator of the first fraction, you didn't type it that way, so I will read the question this way ....
(y-5)/[(x^2+5x)+(xy+5y)] + 1/(x+y) - 2/(x+5)
= (y-5)/[x(x+5)+y(x+5)] + 1/(x+y) - 2/(x+5)
= (y-5)/[(x+5)(x+y)] + 1/(x+y) - 2/(x+5)
now we have (x+5)x+y) as a common denominator, so
=(y-5)/[(x+5)(x+y)] + (x+5)/[(x+5)(x+y)] - 2(x+y)/[(x+5)(x+y)]
= (y-5 + x + 5 - 2x - 2y)/(x+5)(x+y)]
= (-x - y)/[(x+5)(x+y)]
= -(x+y)/[(x+5)(x+y)]
= -1/(x+5)
http://www.jiskha.com/display.cgi?id=1269297854
If so, it is an improvement, but still not too clear.
I think that (x^2+5x)+(xy+5y) probably should be the entire denominator of the first fraction, you didn't type it that way, so I will read the question this way ....
(y-5)/[(x^2+5x)+(xy+5y)] + 1/(x+y) - 2/(x+5)
= (y-5)/[x(x+5)+y(x+5)] + 1/(x+y) - 2/(x+5)
= (y-5)/[(x+5)(x+y)] + 1/(x+y) - 2/(x+5)
now we have (x+5)x+y) as a common denominator, so
=(y-5)/[(x+5)(x+y)] + (x+5)/[(x+5)(x+y)] - 2(x+y)/[(x+5)(x+y)]
= (y-5 + x + 5 - 2x - 2y)/(x+5)(x+y)]
= (-x - y)/[(x+5)(x+y)]
= -(x+y)/[(x+5)(x+y)]
= -1/(x+5)
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