How to solve this system of equations? How to solve this system of equations?
$$\begin{cases}
1+\sqrt{2 x+y+1}=4 (2
   x+y)^2+\sqrt{6 x+3 y},\\
(x+1) \sqrt{2 x^2-x+4}+8 x^2+4
   x y=4.
\end{cases}$$
 A: 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
$$x=0.5\qquad y=-0.5$$
A: 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 $U=2x+y$ 
A: The trick here is to not use both equations at the same time. 
The first equation screams to use the substitution $k=2x+y$. Substituting that, we get $1+\sqrt{k+1}=4k^2+\sqrt{3k}$. So we transform it to $(4k^2-1)+(\sqrt{3k}-\sqrt{k+1})=0$. So $(4k^2-1)(\sqrt{3k}+\sqrt{k+1})+(2k-1)=0$. 
(this is because we multiplied both sides by $\sqrt{3k}+\sqrt{k+1}$. 
Thus we have either $2k-1=0$ or $(2k+1)(\sqrt{3k}+\sqrt{k+1})+1=0$. For the quantity $(2k+1)(\sqrt{3k}+\sqrt{k+1})$ to be real, we need $k$ nonnegative. However, if $k$ is nonnegative, then the quantity is clearly positive, so no roots. 
Thus we need $k=\frac{1}{2}$. $8x^2+4xy=4x(2x+y)=2x$. So $(x+1)\sqrt{2x^2-x+4}+2x=4$. $(x+1)\sqrt{2x^2-x+4}=4-2x$, and squaring we get $(x+1)^2(2x^2-x+4)=(4-2x)^2$. After expanding, we get $2x^4+3x^3+23x-12=0$, so it factors as $(2x-1)(x+3)(x^2-x+4)=0$. Since the discriminant of $x^2-x+4$, which is $1-4(4)$, is negative, it doesn't have a real root. 
Hence the possible values of $x$ are $\frac{1}{2}$ and $-3$, giving values of $y$ as $\frac{-1}{2}$ and $\frac{-11}{2}$. We could just "check" them to see that both work. (just plug back in)
