u^2 + 1 = 3u \quad \Rightarrow \quad u^2 - 3u + 1 = 0.
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- Let \( u = rac{x+y}{x-y} \), so the second term becomes \( rac{1}{u} \). Then the equation becomes:
- u + rac{1}{u} = 3.
- Multiply both sides by \(u\) (assuming \(u
- u = rac{3 \pm \sqrt{9 - 4}}{2} = rac{3 \pm \sqrt{5}}{2}.
- So \( u = rac{3 + \sqrt{5}}{2} \) or \( u = rac{3 - \sqrt{5}}{2} \).
- Now recall \( u = rac{x+y}{x-y} \). Let \( r = rac{x}{y} \), assuming \(y
