System of two equations with two unknowns and one parameter

Question

I need to solve this system of two equations for $x\in[0,0.5]$ and $y\in[0.5,1]$, leaving $b\in[-1,1]$ as a parameter in the solutions.

\begin{cases} 6x^2 b-y(2+(-2+y)b)-2x(-2+(2+y)b)=0\\ -2+y(4-6b)+b-x^2 b+6y^2 b-2x(1+(-1+y)b)=0 \end{cases}

WolframAlpha finds only one real solution for $b=0$, $x=1/3$ and $y=2/3$, but there are others.

What I would like to find, if possible, are two functions $y(b)$ and $x(b)$ that yield the solutions.

See WolframAlpha

Geogebra plot: https://www.geogebra.org/calculator/g3592hwy

The only way I was actually able to solve this system is by setting the parameter $b$ to a given number, reducing the system of equation and then solving for $x$ and $y$. See example: for $b=1$ the system of equations becomes

\begin{cases} 6x^2-2xy-y^2=0\\ 6y^2-2xy-x^2-2y-1=0 \end{cases}

whose solutions are $x\approx0.457849, y\approx0.753505$.

Answer 1

You can eliminate $y^2$ from the two equations, then express $y$ in terms of $x$. Plug in one of the initial equations. It seems that you will get a quartic polynomial in $x$. There are analytical formulas for the roots of a quartic, but they are a little scary.

Answer 2

Using resultants we see that the general solution with a parameter $b$ is given by $$ x=\frac{1029 b^2y^3 - 1764b^2y^2 + 938b^2y - 140b^2 + 980by^2 - 1188by + 348b + 212y - 136}{4(7b^2 + 10b + 4)} $$ where $y$ is a root of the quartic equation $$ 1029b^3y^4 - 2352b^3y^3 + 1876b^3y^2 - 608b^3y + 68b^3 + 1568b^2y^3 - 2856b^2y^2 + 1616b^2y - 272b^2 + 732by^2 - 960by + 304b + 96y - 64=0 $$ For $b=0$ we obtain $y=2/3$ and hence $x=1/3$.