Solve the following system of equations ($x,y \in \Bbb R$):
$$\begin{cases} xy-2y-3 &=\sqrt{y-x-1}+\sqrt{y-3x+5} \\ (1-y)\sqrt{2x-y}+2(x-1) &=(2x-y-1)\sqrt{y}. \end{cases}$$
I think this system of equations have no solution, but I can't prove it.
I really appreciate if some one can help me. Thanks!
On
Let me try. One has $y \geq 0$ and $y \leq 2x$.
Then, the second equation can be written as follows:
$$(1-y)\sqrt{2x-y} + (2x-1) -1 - (2x-1)\sqrt{y} + y\sqrt{y} = 0$$
$$(1-\sqrt{y})\left( (1+\sqrt{y})\sqrt{2x-y} + (2x-1) - 1 - \sqrt{y} - y\right) = 0$$
$$(1-\sqrt{y})(\sqrt{y}\sqrt{2x-y}-\sqrt{y} + \sqrt{2x-y} - 1 + (2x-y) -1) = 0$$
$$(1-\sqrt{y})(\sqrt{2x-y}-1)(\sqrt{y}+1+\sqrt{2x-y}+1) = 0$$
Thus, one has $\sqrt{y} = 1$ or $\sqrt{2x-y} = 1$.
The first case: $y = 1$. From this and the first equation, one has
$$x-5 = \sqrt{-x} + \sqrt{6-3x}.$$ But, $x\geq \frac{y}{2} \geq 0$. Then, there is no solution for this case.
The second case: $2x-y = 1$ or $y = 2x-1$. From this and the first equation, one gets:
$$2x^2 - 5x -1 = \sqrt{x-2} + \sqrt{4-x}.$$
On
To finish, let
$f(x)=2x^2-5x-1\quad \text{and}\quad g(x)=\sqrt{x-2}+\sqrt{4-x}$.
Note that $f(3)=g(3)=2$ and $f(x)$ is increasing $\forall x\in [2,4]$, while $g(x)$ is decreasing when $x\in (3,4]$ and increasing when $x\in [2,3)$.
Thus $x=3$ is the unique solution of $f(x)=g(x)$ in $[2,4]$.
Answer: $x=3, \; y=5$.
Hint: Rewrite the second equation in the form
$(2x-y-1)+(y-1)+(1-y)\sqrt{2x-y}=(2x-y-1)\sqrt y$
Which is equivalent to
$(2x-y-1)(1-\sqrt y)+(1-y)(\sqrt{2x-y}-1)=0$
Or
$(1-\sqrt y)(\sqrt{2x-y}-1)(\sqrt{2x-y}+1+1+\sqrt y)=0$
From here we get $2x-y=1$.