2150.11 – Difference of Reciprocals

For all non-zero real numbers xx and yy such that x−y=xyx - y = xy, 1x−1y=\frac{1}{x}-\frac{1}{y}=

  1. 1xy\frac{1}{xy}
  2. 1x−y\frac{1}{x-y}
  3. 00
  4. −1-1
  5. y−xy - x
Solution

Hang on here. First of all, are there any numbers such that x−y=xyx - y = xy?

Well, x−y=xy→x−xy=y→x(1−y)=y→x=y1−yx - y = xy \rightarrow x - xy = y \rightarrow x(1 - y) = y \rightarrow x = \frac{y}{1-y}.

If y=3y = 3, say, then x=31−3=−32x = \frac{3}{1-3}=-\frac{3}{2}.

Checking, x−y=32−3=−32−62=−9/2x - y =\frac{3}{2}-3=-\frac{3}{2}-\frac{6}{2}=-9/2 .

And, xy=32(3)=−92xy =\frac{3}{2} (3)=-\frac{9}{2} .

Okay, there are such numbers, so let’s figure out the answer:

1x−1y=yxy−xxy=y−xxy=y−xx−y=−1\frac{1}{x}-\frac{1}{y}=\frac{y}{xy}-\frac{x}{xy}=\frac{y-x}{xy}=\frac{y-x}{x-y}=-1

The answer is (d).