1. ## Equation system

Solve the following system of equation for $x,y\in\mathbb{R}$:
$x^3+4y=y^3+16x$
$\frac {1+y^2}{1+x^2}=5$

2. Originally Posted by james_bond
Solve the following system of equation for $x,y\in\mathbb{R}$:
$x^3+4y=y^3+16x$
$\frac {1+y^2}{1+x^2}=5$
$x^3-16x = y^3-4y$

$x(x^2-16) = y(y^2-4)
$

$\frac {1+y^2}{1+x^2}=5$

$1+y^2 = 5+5x^2$

$y^2 = 4+5x^2$

$x(x^2-16) = y(5x^2)$

$y = \frac{x^2-16}{5x}$

$\frac{(x^2-16)^2}{25x^2} = 4 + 5x^2$

$x^4 - 32x^2 + 256 = 100x^2 + 125x^4$

$0 = 124x^4 + 132x^2 - 256$

$0 = 31x^4 + 33x^2 - 64$

$0 = (x^2-1)(31x^2 + 64)$

$0 = (x-1)(x+1)(31x^2+64)$

can you finish?