Does the system of two equation in two variables given below have a possible real solution? Consider the following two equations:
$$\sqrt {3x} \left(1 + \frac{1}{x+y} \right) = 2 $$
$$\sqrt{7y}\left(1 -\frac{1}{x+y}\right)= 4\sqrt{2}$$
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It is expected to verify if the equations possess a solution and if yes what is the floor of $ y/x $ for that solution.
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What have I tried:
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Eliminating the $\frac {1}{x+y}$ term from the two equations leads to :
$$ \frac{2}{\sqrt{3x}} + \frac{4\sqrt{2}}{\sqrt{7y}} = 2 $$
Which gives
$$ y = \frac{24x}{7(1- \sqrt{3x})^2}$$
Also from the first equation value of y is :
$$ y = \frac{\sqrt{3x}}{2-\sqrt{3x}} - x $$
Equating the two values of y and simplifying resulted in a polynomial with fractional powers of x. This proved to be a dead end for me.
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Note: some underlying inquisitions:

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*Is there a method for simplifying such equations that I am unaware of?

*Are the given two equations representing hyperbola? (Checked the graph on desmos!). If yes how can one see that without referring to the plot.


Starting from @Moo's comment $$1323 x^4+3024 x^3+2610 x^2-7456 x+147 = 0$$ is the same as
$$\left(9 x^2+30 x+49\right) \left(147 x^2-154 x+3\right)=0$$ The first quadratic does not show real roots and for the second
$$x_\pm=\frac{11\pm4 \sqrt{7}}{21} $$
Similarly
$$3087 y^4-25284 y^3+44716 y^2-38816 y+4032 = 0$$is the same as
$$\left(63 y^2-120 y+112\right)\left(49 y^2-308 y+36\right)=0 $$
The first quadratic does not show real roots and for the second
$$y_\pm=\frac{2\left(11\pm 4 \sqrt{7}\right)}{7} =6 x_\pm$$
