1. ## Find all solutions

Find all solutions to this system of equations
$\displaystyle 3x^2y-3y=0$ (i)
$\displaystyle x^2-3x-2y=0$ (ii)

I take
3(ii) – 2(i) = 0

=> $\displaystyle 6x^2y-6y=3x^3-9x$ (2)
=> $\displaystyle 6y(x^2-1)=3x(x^2-3)$ (3)

Therefor, I have these points.
(0,0)
(±√3, 0)

But x=±1 is also a solution. But that doesn't satisfy the right hand side in (3). I could have seen x=±1 and put it in (i) and (ii). But I didn't see that. Do I have a bad, nonsystematic method for solving this system of equation? How do I do it better?

2. ## Re: Find all solutions

Equation (2) is incorrect. As you have a $\displaystyle x^3$ term which is not present in the original equation.

3. ## Re: Find all solutions

Originally Posted by jacob93
But x=±1 is also a solution. But that doesn't satisfy the right hand side in (3). I could have seen x=±1 and put it in (i) and (ii). But I didn't see that. Do I have a bad, nonsystematic method for solving this system of equation? How do I do it better?
$\displaystyle 3x^2y-3y=0 - (1), x^2-3x-2y=0 - (2)$
From Eq (1) we have
$\displaystyle 3y(x-1)(x+1)=0$ satisfied by $\displaystyle y=0, x= \pm 1$
When $\displaystyle y=0$ have $\displaystyle x=0,3$ from Eq(2)
When $\displaystyle x=1$ have $\displaystyle y=-1$ from Eq(2)
When $\displaystyle x=-1$ have $\displaystyle y=1$ from Eq(2)

So solutions are $\displaystyle (0,0), (0,3), (1,-1), (-1,1)$

4. ## Re: Find all solutions

Originally Posted by kalyanram
When $\displaystyle x=1$ have $\displaystyle y=-1$ from Eq(2)
How? When $\displaystyle x=1$, then $\displaystyle 6y-6y=3-9$ which never happens.

5. ## Re: Find all solutions

Originally Posted by jacob93
How? When $\displaystyle x=1$, then $\displaystyle 6y-6y=3-9$ which never happens.
As I have already indicated in my previous post the equation you have referred to as Eq(2) is incorrect as it should have eliminated the $\displaystyle y$ term and should not have a $\displaystyle x^3$ term.

The Eq(2) I have mentioned is the second of the original two equations.

thanks