# Equation system

#### james_bond

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

#### skeeter

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

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

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

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

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

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

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

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

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

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

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

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

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

can you finish?

james_bond