How to solve a system of 2 unknows with radical from the Vietnamese University Entrance exam? I am running into a problem of solving the following system of 2 equations and 2 unknows $x,y$ over the real
$x \left(4 x^2+1\right)+(y-3) \sqrt{5-2 y}=0$
and
$4 x^2+2 \sqrt{3-4 x}+y^2-7=0$
This problem belong to the Vietnamese University Entrance exam of Block A (Math, Physic and Chemistry) in 2010
My questions are:
1/ How to solve this challenging system
a/ Exactly in a systematic way
b/ Using numerical method
2/ Is there a general way to solve system with radical like this ?
Note that the solution is $(x,y)=(\frac{1}{2} , 2)$
The link to the full exam paper is
https://toanmath.com/2015/07/de-thi-va-dap-an-mon-toan-khoi-a-nam-2010.html
Thank you for your enthusiasm
 A: First, what can we say about the number of possible solutions? Well, when $y > 0$, the first curve has positive slope and the second negative. So there is at most $1$ solution with $y > 0$. Could there be a solution with $y \le 0?$ From the first equation, that would mean $x(4x^2+1) \ge 3\sqrt{5} > 5$, i.e., $x > 1$. But that would make the radical in the second equation complex, so we can't have $y \le 0$. Thus the problem has at most 1 solution, and it has positive $y$.
Let's get more bounds on the possible solutions. Going back to the first equation, we note that $y \le 5/2$ is required for the radical to be real, and as such  $(y-3)\sqrt{5-2y}$ is never positive. Therefore $x(4x^2+1)$, and hence $x$ itself, is nonnegative. And of course, from the radical in the second equation, we have $x \le 3/4$. The solution is now restricted to the interval $(x, y)\in [0, 3/4]\times(0, 5/2]$.
Now since this is an exam problem, the radicals probably can't be too "bad". They're likely to be in $\mathbb Z$, very likely to be in $\mathbb Q$, and, if it comes to it, almost certainly in $\mathbb Q[\sqrt{n}]$. Starting with $\mathbb Z$, the only options consistent with the bounds are $\sqrt{4-3x} \in \{0, 1\}$ and $\sqrt{5-2y}\in \{0, 1, 2\}$. Trying all 6 possible combinations finds a solution: $(x, y) \in (1/2, 2)$.
