parkhydel wrote:
What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?
A. \(x \leq 1\)
B. \(-1 < x \leq 0\)
C. \(0 < x \leq 1\)
D. \(x > 1\)
E. All real numbers
PS22680.02
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 takeAPPROACH #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: \(y=\frac{x}{x-1}=
\frac{2}{2-1}=1\). So, \(x = 2\) and \(y = 1\) 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
Thank you for the solution.
Just wanted to point out a small calculation error in the highlighted part. y should be 2.
Could you please amend the solution.