It is given that \((\frac {x}{3}+1, \frac {y-2}{3}) = (\frac {5}{3}, \frac {1}{3})\).
Since the ordered pairs are equal, the corresponding elements will also be equal.
\((\frac {x}{3}+1, \frac {y-2}{3}) = (\frac {5}{3}, \frac {1}{3})\)
Therefore,
\(\frac {x}{3}+1=\frac {5}{3}\)
⇒ \(\frac{x}{3} = \frac{5}{3} -1,\frac{y-2}{3} = \frac{1}{3}\)
⇒\(\frac{x}{3} = \frac{2}{3}, y = \frac{1}{3}+\frac{2}{3}\)
⇒ x=2, y=1
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: