Solving a system of quadratic equations which evaluates to a 4th grade equation

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?

• Well, these are two parabola's, one inverted, and another sideways opening leftward, and in general will intersect at 4 points. So can't really avoid a quartic. Commented Dec 27, 2015 at 14:14
• You probably mean a 4th degree equation (one that involves variables raised to the 4th power), not 4th grade. These equations are also known as "quartic" equations. To an American, at least, a 4th grade equation is one that could be solved by children in the 4th grade (i.e children around 9 years old). Commented Dec 27, 2015 at 14:32

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 $$4\, \left( y-4 \right) \left( {y}^{3}+4\,{y}^{2}-18\,y-71 \right) =0$$