Solve three simultaneous polynomial equations in three variables....

[chwala]

I have found the solution. There are two possible values for z.
From $6z^2 - 12zy =0$
We have $6z(z-2y)=0$
$6z=0$ or $z=2y$
Picking $z=0$ solves the simultaneous equations,
Since $x=3z-2y$
$x=-2y$
$y=1, x=2$
$y=-1, x=-2$
Thanks guys,
Algebra is the best take on this, my thoughts.

 

[ehild]

I have found the solution. There are two possible values for z.
From $6z^2 - 12zy =0$
We have $6z(z-2y)=0$
$6z=0$ or $z=2y$
Picking $z=0$ solves the simultaneous equations,
Since $x=3z-2y$
$x=-2y$
$y=1, x=2$
$y=-1, x=-2$

The result is not quite correct. Recall that $x=-2y$, If y=1, x=-2, and if y=-1, x=2.

Also there is one more solution set, for $z-2y=0$.

 

[Ray Vickson]

I have found the solution. There are two possible values for z.
From $6z^2 - 12zy =0$
We have $6z(z-2y)=0$
$6z=0$ or $z=2y$
Picking $z=0$ solves the simultaneous equations,
Since $x=3z-2y$
$x=-2y$
$y=1, x=2$
$y=-1, x=-2$
Thanks guys,
Algebra is the best take on this, my thoughts.

You have made some sign errors. There are four solutions:
$$\begin{array}{ccc}
x=2, &y=-1,&z=0\\
x=-2, &y=1, &z=0\\
x=4/\sqrt{7}, &y=1/\sqrt{7}, &z=2/\sqrt{7} \\
x=-4/\sqrt{7}, &y=-1/\sqrt{7},& z = -2/\sqrt{7}
\end{array}
$$
You can get the values of $z$ from the last equation in post #12, then substitute those values in the original equations to find $x$ and $y$; or, you can use the other equations in post #12 to find the corresponding values of $x,y$.

 

[chwala]

The result is not quite correct. Recall that $x=-2y$, If y=1, x=-2, and if y=-1, x=2.

Also there is one more solution set, for $z-2y=0$.

What do you mean? I clearly said considering $6z=0$ from my working will give you the solution. Kindly note that there are two values for $z$ and clearly $z-2y=0$ will not give you the solution.

 

[ehild]

What do you mean? I clearly said considering $6z=0$ from my working will give you the solution.

In Post #31, you got $6z(z-2y)=0$. That means two possibilities: either $6z=0$ or $z-2y=0$. Assuming z=0 gives one sets of solutions. Assuming z-2y=0 gives the other set.

 

[chwala]

The result is not quite correct. Recall that $x=-2y$, If y=1, x=-2, and if y=-1, x=2.

Also there is one more solution set, for $z-2y=0$.

Aaaaaaaah I have seen my sign error, guess I was tired, thanks though.

 

[ehild]

What do you mean? I clearly said considering $6z=0$ from my working will give you the solution. Kindly note that there are two values for $z$ and clearly $z-2y=0$ will not give you the solution.

Well, z=2y also gives solutions. See @Ray Vickson's Post #33, he gave the full solution (although it is not allowed here).

 

[chwala]

Well, z=2y also gives solutions. See @Ray Vickson's Post #33, he gave the full solution (although it is not allowed here).

I find no offense with his solution, I found the solution he though gave other possible solutions which is OK, reading my post I was only interested in finding the solutions in the text, which I found... Nothing wrong with anyone giving more solutions, my thoughts

 

[ehild]

I find no offense with his solution, I found the solution he though gave other possible solutions which is OK, reading my post I was only interested in finding the solutions in the text, which I found... Nothing wrong with anyone giving more solutions, my thoughts

It is the policy of this Forum that the Helper must not give full solution, unless the OP produced his one. You did not give the full solution yet.

 

[chwala]

noted...i am able to get the other solution, but i was interested in the solution that i found...thanks a lot guys...

 

[ehild]

noted...i am able to get the other solution, but i was interested in the solution that i found...thanks a lot guys...

There is only one solution, the whole set of possible x, y, z values.

 

[StoneTemplePython]

I found another approach to solving this with elementary symmetric sums, plus a bit of linear algebra mixed in. If @chwala would perhaps post a full solution, I could post my solution approach.

---
edit: maybe not. too many small bugs.

 

[chwala]

I found another approach to solving this with elementary symmetric sums, plus a bit of linear algebra mixed in. If @chwala would perhaps post a full solution, I could post my solution approach.

---
edit: maybe not. too many small bugs.

I thought i had posted the solution, made some sign error... Unless you want me to post step by step which is time consuming, if you insist I will just highlight on the last steps leading to all the four possible solutions..

 

[StoneTemplePython]

I thought i had posted the solution, made some sign error... Unless you want me to post step by step which is time consuming, if you insist I will just highlight on the last steps leading to all the four possible solutions..

No worries. There were just too many small bugs in my solution so I'm not going to post it -- I wouldn't insist anything more from you.

 

[chwala]

There is only one solution, the whole set of possible x, y, z values.

Whichever way you call it, x ,y, z may have different values as indicated by the four possible sets or rather values.

 

[chwala]

No worries. There were just too many small bugs in my solution so I'm not going to post it -- I wouldn't insist anything more from you.

Can I have a look at your symmetric method