2150.11 – Difference of Reciprocals

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

  1. 1xy\dfrac{1}{xy}

  2. 1x−y\dfrac{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 = \dfrac{y}{1-y}

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

Checking, x−y=32−3=−32−62=−9/2x - y =\dfrac{3}{2}-3=-\dfrac{3}{2}-\dfrac{6}{2}=-9/2. And, xy=32(3)=−92xy =\dfrac{3}{2} (3)=-\dfrac{9}{2} .

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

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

The answer is (d).