If \(\displaystyle\frac{x+y}{x-y}+\frac{x-y}{x+y}=\frac{10}{3}\), then \(\displaystyle\frac{x}{y}=\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\pm 2\)
The two fractions are reciprocals of each other. Let
\( \displaystyle t = \frac{x+y}{x-y} \)
Then the equation becomes
\( \displaystyle t + \frac{1}{t} = \frac{10}{3} \)
\(3t^2 - 10t + 3 = 0\)
\((3t - 1)(t - 3) = 0\)
\(t = 3\ \text{ or }\ t = \frac{1}{3}\)
When t = 3:
\(x + y = 3x - 3y\)
\(4y = 2x\)
\(\frac{x}{y} = 2\)
When t = 1/3:
\(3x + 3y = x - y\)
\(2x = -4y\)
\(\frac{x}{y} = -2\)
So \(\frac{x}{y} = \pm 2\).