Take a Walk Simultaneous Avenue

a=1/x+1/y=(x+y)/xy
b=x+y

so ab = (x+y)*(x+y)/xy = (x+y)^2/xy

let ab = c for variable simplicity

c*xy = x^2 + 2xy + y^2
0 = x^2 + xy(2-c) + y^2

only form that will yield a valid solution is c=4 to give (C)

0=x^2 - 2xy + y^2 = (x-y)^2 and x=y

Seems more like a first year engineering question then a GMAT question

m&m wrote:a=1/x+1/y=(x+y)/xy
b=x+y

so ab = (x+y)*(x+y)/xy = (x+y)^2/xy

let ab = c for variable simplicity

c*xy = x^2 + 2xy + y^2
0 = x^2 + xy(2-c) + y^2

only form that will yield a valid solution is c=4 to give (C)

0=x^2 - 2xy + y^2 = (x-y)^2 and x=y

Seems more like a first year engineering question then a GMAT question

But if ab=4 we will have not have an unique solution.

If ab=4

i.e. (x+y)^2 = 4xy

i.e. (x+y)^2-4xy=0

i.e. (x-y)^2=0

Hence ab=4 for all values of x and y if x=y.

For example if x=y=1, we have ab=4
and if x=y=2 we have ab=4.

So if ab=4, we will have infinite number of solutions and not unique solution

Good point - I guess the only unique solution is ab=2 and x=y=0.... so if you get that question on test day press and answer 4 then press the Back button furiously until it lets to go back so you can answer 2

m&m