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*