<s>[INST] Solve system of equations $\begin{align} \begin{cases} |y|=|x-3| \\(2\sqrt{z}+2-y)y=1+4y \\ x^2+z-4x=0\end{cases} \end{align}$ Solve system of equations$$\begin{align}
    \begin{cases}
        |y|=|x-3| \\
        (2\sqrt{z}+2-y)y=1+4y \\
        x^2+z-4x=0
    \end{cases}
\end{align}$$


*

*$$y=3-x;z=4x-x^2$$
Then the second equation is equivalent with $$2\sqrt {4x-x^2}=\frac{x^2-4x+4}{x-3}(0\le x\le 4)$$
This equation has no root but i do not know how to prove it. In 2nd case $y=3-x$  i have done but i am stuck on this case.
Help me, ty.
 [/INST] In this case $4x-x^2\geq0,$ which gives $0\leq x\leq4.$
Now, after substitution $z=4x-x^2$ rewrite the second equation in the form:
$$2(3-x)\sqrt{4x-x^2}=(x-4)^2,$$ which gives $x=4$ or
$$2(3-x)\sqrt{x}=\sqrt{(4-x)^3},$$ which gives also $0<x<3$ and after squaring of the both sides we obtain:
$$4x(3-x)^2=(4-x)^3$$ or $$(x-2)^2(16-5x)=0,$$ which gives also $x=2.$
In the second case we'll prove that the equation $$2(x-3)\sqrt{4x-x^2}=(x-2)^2$$ has no real roots for $3<x<4.$
For which we'll prove that:
$$(x-2)^4>4(x-3)^2(4x-x^2)$$ or
$$5x^4-48x^3+156x^2-176x+16>0$$ for which it's enough to prove that
$$5x^4-48x^3+156x^2-176x+15>0$$ or
$$(x-3)(5x^3-33x^2+57x-5)>0$$ or
$$(5x^3-33x^2+55x)+(2x-5)>0,$$ which is obvious for any $x>3.$
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