What values of x have a corresponding value of y that satisfies both x

Given: \(xy = x + y\)
Subtract \(y\) from both sides to get: \(xy - y= x\)
Factor: \(y(x-1)=x\)
Divide both sides by \(x-1\) to get: \(y=\frac{x}{x-1}\)

Important: At this point, there are at least two different approaches we can take

APPROACH #1:
Now that we know \(y=\frac{x}{x-1}\), we can readily see that x cannot equal 1, otherwise, the denominator in y is 0, which would make y undefined.
At this point we can eliminate answer choices A, C and E, since they all allow for x to equal 1.
Since we're down to just two answer choices, let's just test some x-value.

For example, if \(x = 2\), then we get: \(y=\frac{x}{x-1}=\frac{2}{2-1}=2\).
So, \(x = 2\) and \(y = 2\) is a possible solution to the system of equations.
Since it's possible for x to equal 2, we can eliminate answer choice B

By the process of elimination, the correct answer is D.

APPROACH #2:
We've already concluded that: \(y=\frac{x}{x-1}\)

Now take the given inequality, \(xy > 0\), and replace \(y\) with \(\frac{x}{x-1}\) to get: \((x)(\frac{x}{x-1})>0\)
Simplify: \(\frac{x^2}{x-1}>0\)

Since \(x^2\) is always greater or equal to 0, it must be the case that the denominator, \(x-1\), is positive
In other words, it must be the case that: \(x-1>0\)
Add \(1\) to both sides of the inequality to get: \(x>1\)

Answer: D

Cheers,
Brent