How to solve this system of equations?

Question

How to solve this system of equations? $$\begin{cases} 1+\sqrt{2 x+y+1}=4 (2 x+y)^2+\sqrt{6 x+3 y},\\ (x+1) \sqrt{2 x^2-x+4}+8 x^2+4 x y=4. \end{cases}$$

Answer 1

You have two equations with two variables. $$1+\sqrt{2x+y+1}-4(2x+y)^2-\sqrt{6x+3y}=0$$ $$(x+1)\sqrt{2x^2-x+4}+8x^2+4xy-4=0$$ Solve it with any root finding algortihm.

If you are looking for real solutions $$x=0.5\qquad y=-0.5$$

Answer 2

Hint Define $U=2x+y$ in first equation

You will get $1+\sqrt{U+1}=4 U^2+\sqrt{3U}$ solve $U$

here you need to solve $U$

$$(x+1) \sqrt{2 x^2-x+4}+8 x^2+4xy=4$$

$$(x+1) \sqrt{2 x^2-x+4}+4x(2x+y)=4$$

Then put U in second equation and find $x$

$$(x+1) \sqrt{2 x^2-x+4}+4xU=4$$

after finding $x$ , you can get $y$ from $U=2x+y$