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I'm stuck with this system of equations.. How to see through it? $$ \begin{aligned} 2(1+y)(1-x)=(2-x)(1+2y) \\ x(2-x)=y(1+y) \end{aligned}$$

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    $\begingroup$ What is the relation with Cauchy-Riemann equations? $\endgroup$ – lhf Sep 25 '12 at 2:15
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Hint: Expand the first equation, you will get a simple expression which you can then plug into the second, and this will give you the two solutions easily.

(on a side note: in cases like these it is often useful to interpret the equations geometrically, so that you know what you are looking for)

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Trivial solutions of $x(2-x)=y(y+1)$........eqn($1$) are $(0,0),(0,-1),(2,0),(2,-1)$

Now, consider $2(1+y)(1-x)=(2-x)(1+2y)$

If we assume a non-trivial solution $\implies $ we can multiply both sides by $xy$ which gives

$2y(y+1)x(1-x)=x(2-x)y(1+2y)$ where $x(2-x)$ cancels out with $y(y+1)$ (if we exclude above four pairs) which gives

$2x(1-x)=y(1+2y).....$ eqn($2$)

Multiply eqn($1$) by $2$ and subtract eqn($2$) from it gives

$2x=y$ (excluding above four cases)

Putting it back in eqn($2$) gives

$1-x=1+4x\implies x=0\implies y=0$

As only $(0,0)$ and $(2,-1)$ satisfies eqn($2$) out of trivial solution pairs of eqn($1$) and no other solution exists, Thus, there are no solutions other than $(0,0),(2,-1)$

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