I'm confused why you say in comments that "still $y$ does not get eliminated" -- at a point in the conversation where $y$ should have disappeared long ago.
You can solve your third equation $2x+y=c$ to get $y=c-2x$. When you plug that into the two other equations you get
$$ x^2(c-2x)=a \\
x(x+c-2x)= b $$
No matter what you do subsequently, there won't be any $y$s left to deal with!
What I would do at this point is put off dividing for as long as possible, so rewrite the second equation to
$$ x^2 = cx - b $$
This form lets you reduce any polynomial in $x$ to a first-degree polynomial, by repeatedly using it to eliminate the highest-degree term. We will use it to simplify the third-degree first equation. First substitute the leading $x^2$ to get
$$ (cx-b)(c-2x) = a $$
$$ c^2x + 2bx - 2cx^2 - bc = a $$
and then insert $x^2=cx-b$ once again:
$$ c^2x + 2bx - 2c(cx-b) - bc = a $$
This is finally a linear equation in $x$. You can solve it without running into any square roots, and insert this value for $x$ into $x^2=cx-b$. What is left then is a rational equation in only $a$, $b$, and $c$, as desired.
Unless the coefficient of $x$ in the linear equation was $0$ (which is a case you'll need to handle separately), each step along the way was reversible, so you will know that if you have $a,b,c$ that satisfy the final equation, they will also have a solution for $x$ and $y$.