Carcass wrote:
\(\frac{x}{y}+\frac{y}{x}=8\). What is the value of \(\frac{x+y}{\frac{1}{x}+\frac{1}{y}}\)
(A) 5
(B) 8
(C) 10
(D) 12
(E) 16
Unless I'm missing something, this question is flawed. Here's why:
Given: \(\frac{x}{y}+\frac{y}{x}=8\)
Rewrite with common denominators: \(\frac{x^2}{xy}+\frac{y^2}{xy}=8\)
Combine fractions: \(\frac{x^2+y^2}{xy}=8\)
Multiply both sides of the equation by \(xy\) to get: \(x^2+y^2=8xy\)
Add \(2xy\) both sides of the equation: \(x^2+2xy+y^2=10xy\)
Factor the left side: \((x+y)^2=10xy\)
Finally, divide both sides by \(xy\) to get: \(\frac{(x+y)^2}{xy}=10\)
So,
IF the given expression evaluates to be \(\frac{(x+y)^2}{xy}\), then the correct answer will be \(10\).
The given expression: \(\frac{x+y}{\frac{1}{x}+\frac{1}{y}}\)
Rewrite the denominator with common denominators: \(\frac{x+y}{\frac{y}{xy}+\frac{x}{xy}}\)
Combine terms in the denominator: \(\frac{x+y}{(\frac{x+y}{xy})}\)
Since we're dividing by a fraction, we'll multiply by the reciprocal to get: \((x+y)(\frac{xy}{x+y})\)
Unfortunately this simplifies to be \(xy\), which we don't know the value of.
Please let me know if I'm missing something (or if I made a mistake above)
_________________
Brent Hanneson - founder of Greenlight Test Prep