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bumblenbumble01
- Newbie | Next Rank: 10 Posts
- Posts: 6
- Joined: Thu Dec 11, 2008 11:18 am
- Thanked: 1 times
Ok the question is as follows
[x+y] / [x-y]
If x does not equal y, and xy does not equal 0, then when x is replaced by 1/x and y is replaced by 1/y everywhere in the expression above, the resulting expression is equivalent to
a. - (x+y)/(x-y)
b. (x-y)/(x+y)
c. (x^2-y^2)/(xy)
d. (y-x)/(x+y)
I chose C by doing the following: [1/x + 1/y]/ [1/x - 1/y] -->
[y/xy + x/xy] / [y/xy-x/xy]-->
[y+x/xy] / [y-x/xy] -->
[y+x/xy] * [xy/y-x-->
[x(y)^2 + y(x)^2] / [x^2-y^2]-->
[y^2+x^2] / [x-y]
Now the correct answer is A and I understand why and how they got that, what I don't understand is why MY method was wrong?
Thanks
[x+y] / [x-y]
If x does not equal y, and xy does not equal 0, then when x is replaced by 1/x and y is replaced by 1/y everywhere in the expression above, the resulting expression is equivalent to
a. - (x+y)/(x-y)
b. (x-y)/(x+y)
c. (x^2-y^2)/(xy)
d. (y-x)/(x+y)
I chose C by doing the following: [1/x + 1/y]/ [1/x - 1/y] -->
[y/xy + x/xy] / [y/xy-x/xy]-->
[y+x/xy] / [y-x/xy] -->
[y+x/xy] * [xy/y-x-->
[x(y)^2 + y(x)^2] / [x^2-y^2]-->
[y^2+x^2] / [x-y]
Now the correct answer is A and I understand why and how they got that, what I don't understand is why MY method was wrong?
Thanks
