Find all solutions

**jacob93** · September 14th 2012, 08:07 AM

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

I take
3(ii) – 2(i) = 0

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

**kalyanram** · September 14th 2012, 08:20 AM

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

**kalyanram** · September 14th 2012, 08:29 AM

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?

$3x^2y-3y=0 - (1), x^2-3x-2y=0 - (2)$
From Eq (1) we have
$3y(x-1)(x+1)=0$ satisfied by $y=0, x= \pm 1$
When $y=0$ have $x=0,3$ from Eq(2)
When $x=1$ have $y=-1$ from Eq(2)
When $x=-1$ have $y=1$ from Eq(2)

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

**jacob93** · September 14th 2012, 09:17 AM

Originally Posted by kalyanram

When $x=1$ have $y=-1$ from Eq(2)

How? When $x=1$ , then $6y-6y=3-9$ which never happens.

**kalyanram** · September 14th 2012, 09:22 AM

Originally Posted by jacob93

How? When $x=1$ , then $6y-6y=3-9$ 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 $y$ term and should not have a $x^3$ term.

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

**jacob93** · September 14th 2012, 09:33 AM

thanks

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