Expert: Sherman D. - 8/13/2008

Question
QUESTION: 2    2
_  + _

x    y

____________


2       2
__   +  ___

x+2      y

Please simplify the following rational expression.

ANSWER: ((2/x) + (2/y)) / ((2/(x + 2)) + (2/y))

((2y + 2x)/xy) / ((2y + 2(x + 2))/(y(x + 2)))

((2y + 2x)/xy) * ((y(x + 2))/(2y + 2x + 4))

((2y + 2x)(xy + 2y))/((xy)(2y + 2x + 4))

(2xy^2 + 4y^2 + 2x^2y + 4xy)/(2xy^2 + 2x^2y + 4xy)

(2xy^2 + 2x^2y + 4xy + 4y^2)/(2xy^2 + 2x^2y + 4xy)

((2xy^2 + 2x^2y + 4xy) + (4y^2))/(2xy^2 + 2x^2y + 4xy)

1 + ((4y^2)/(2xy^2 + 2x^2y + 4xy))

1 + (2y(2y)/((2y)(xy + x^2 + 2x))

1 + ((2y)/(xy + x^2 + 2x)) or (xy + x^2 + 2x + 2y)/(xy + x^2 + 2x)

---------- FOLLOW-UP ----------

QUESTION: So for this one what would I do?

x-3   -   1     =    x^2-2
____     ____      ________

x + 2    x-2       x^2-4


Answer
((x - 3)/(x + 2)) - (1/(x - 2)) = ((x^2 - 2)/(x^2 - 4))

since (x + 2)(x - 2) gives you (x^2 - 4), you don't really have anything to do with whats after the equal sign, only whats before it.

((x - 3)(x - 2) - (x + 2))/(x^2 - 4) = (x^2 - 2)/(x^2 - 4)

(x^2 - 2x - 3x + 6 - x - 2)/(x^2 - 4) = (x^2 - 2)/(x^2 - 4)

just take out the (x^2 - 4)s and add/subtract like terms

x^2 - 6x + 4 = x^2 - 2

how use inverse of operations

-6x = -6
x = 1

and there you got, the answer is x = 1