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

  • 5
    $\begingroup$ What is the relation with Cauchy-Riemann equations? $\endgroup$ – lhf Sep 25 '12 at 2:15

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)


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 $(0,0),(2,-1)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.