Bunuel wrote:
If \(\frac{x + y}{x − y} = \frac{1}{2}\), then \(\frac{xy + x^2}{xy − x^2}=\)
(A) –4.2
(B) –1/2
(C) 1.1
(D) 3
(E) 5.3
Method I - Algebra
\(\frac{xy + x^2}{xy − x^2}\) =
\(\frac{x (y + x)}{x (y - x)}\) =
\(\frac{(y + x)}{(y - x)}\)
\((-1)*\frac{(y + x)}{(y - x)}\) =
- \(\frac{(x + y)}{(x - y)}\)
Given \(\frac{x + y}{x − y} = \frac{1}{2}\), then
- \(\frac{(x + y)}{(x - y)}\) = - \(\frac{1}{2}\)
Method II - assign values derived from first equation, plug into second equation
\(\frac{x + y}{x − y}\) = \(\frac{1}{2}\)
Cross multiply:
\(2 (x + y) = (x - y)\)
\(2x + 2y = x - y\)
\(x = -3y\)
Let x = -6 and y = 2, plug into second equation \(\frac{xy + x^2}{xy − x^2}=\)
\(\frac{-12 + 36}{- 12 - 36}\)
= - \(\frac{24}{48}\) or - \(\frac{1}{2}\)
Answer B
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