I have to solve the following system of equations:
$x^2 + 4y + 2 = 22$
$2y^2 + x + 6 = 40$
I tried to solve for one variable and then substitute it into the other equation, but a problem appears:
$y = \pm \sqrt{18 - \frac{x}{2}}$
And so:
$x^2 \pm 4\sqrt{18 - \frac{x}{2}} +2 = 22$
In order to keep solving this, I need to square both sides of the equation, what means a 4th grade equation, which I am unable to solve.
Is there any other way to solve this?
HINT: solving the second equation for $x$ we get $$x=34-2y^2$$ plugging this equation in the first one we obtain $$(34-2y^2)^2+4y=20$$ this equation has to be solved the last equation is equivalent to $$4\, \left( y-4 \right) \left( {y}^{3}+4\,{y}^{2}-18\,y-71 \right) =0$$