From Equation (1)
$y(x^2 + 1) = 17$
$y = \dfrac{17}{x^2 + 1}$
From Equation (2)
$x^4 \left( \dfrac{17}{x^2 + 1} \right)^2 + \left( \dfrac{17}{x^2 + 1} \right)^2 = 257$
$\dfrac{289x^4}{(x^2 + 1)^2} + \dfrac{289}{(x^2 + 1)^2} = 257$
$289x^4 + 289 = 257(x^2 + 1)^2$
$289x^4 + 289 = 257(x^4 + 2x^2 + 1)$
$289x^4 + 289 = 257x^4 + 514x^2 + 257$
$32x^4 - 514x^2 + 32 = 0$
$16x^4 - 257x^2 + 16 = 0$
Using the quadratic formula (a = 16, b = -257, c = 16)
$x^2 = \dfrac{-(-257) \pm \sqrt{(-257)^2 - 4(16)(16)}}{2(16)}$
$x^2 = \dfrac{257 \pm 255}{32}$
$x^2 = 16 ~ \text{ and} ~ \frac{1}{16}$
$x = \pm 4 ~ \text{ and} ~ \pm \frac{1}{4}$
For x = ±4
$y = \dfrac{17}{(\pm 4)^2 + 1} = 1$
For x = ±1/4
$y = \dfrac{17}{\left( \pm \frac{1}{4} \right)^2 + 1} = 16$
The solutions are
(4, 1), (-4, 1), (1/4, 16), and (-1/4, 16) answer
