Picking the same value for x and y when picking numbers

[jackson.b.allan]

deleted by moderator JNelson--banned source

[jnelson0612]

Hi Jackson,
Thanks for your question, but unfortunately the GMAC does not allow us to post any of the official questions from the print books. I'm sorry that we can't help you with this one. Here are our forum guidelines: read-before-you-post-gmatprep-math-guidelines-t2723.html

[jackson.b.allan]

Sorry, ignoring the question then and getting to the nuts and bolts of it.

With x+y=xy
also sometimes shown as x+y/xy=1

Is it possible for x and y to be anything other than 2?

[RonPurewal]

Sorry, ignoring the question then and getting to the nuts and bolts of it.

With x+y=xy
also sometimes shown as x+y/xy=1

Is it possible for x and y to be anything other than 2?

Sure.

* By inspection, it's easy to see that x = y = 0 works, too.

* More generally, you can solve for one of the variables in terms of the other one, to get a formula:

x + y = xy

y = xy - x

y = x(y - 1)

y/(y - 1) = x

So, pick any value at all for y (except y = 1), and let x = y/(y - 1).

[jackson.b.allan]

Thanks Ron!

I didn't think about 0, rookie error. This has come up where the question stipulates xy>0. I think the key take away though is that x and y have to be the same value if x+y=xy.

[RonPurewal]

Thanks Ron!

I didn't think about 0, rookie error. This has come up where the question stipulates xy>0. I think the key take away though is that x and y have to be the same value if x+y=xy.

Well... no.
Go back again and read my previous post: You can pick any value in the world (except 1) for y, and then let x = y/(y - 1). (Since the condition x + y = xy is symmetric, you can also pick a random value for x and let y = x/(x - 1).)

So, the two values most definitely don't have to be the same; in fact, they're only the same if x = y = 2. They're different in every other possible case.

For instance, they could be...
3 and 3/2;
-1 and 1/2;
√2 and 2 + √2;
pi and pi/(pi - 1);
... you get the picture.

[jackson.b.allan]

I should have included in that stipulation 'where x and y are integers greater than zero'.

[RonPurewal]

I should have included in that stipulation 'where x and y are integers greater than zero'.

But, still, no. Unless they are both 2, they are never the same value.