VIC Problem Question

[bumblenbumble01]

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

[dumb.doofus]

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

Dude.. mistake in the 4th step.. have highlighted it as Bold.. xy will get cancelled.. and you have squared it..