How can I solve these equations?

Question

I can't figure it out, please help :S

$$\begin{array}{rcl} 2x + 2xz &=& 0\\ -2y + 2yz &=& 0\\ x^2 + y^2 &=& 4 \end{array}$$ Thanks in advance!

Answer 1

You have $$2x(z+1)=0\iff x=0\ \text{or}\ z=-1,$$ $$2y(-1+z)=0\iff y=0\ \text{or}\ z=1,$$ $$x^2+y^2=4.$$

1) When $x=0\ \text{and}\ y=0$, these don't satisfy the last equation.

2) When $x=0\ \text{and}\ z=1$, $y=\pm 2$.

3) When $z=-1\ \text{and}\ y=0$, $x=\pm 2$.

4) There is no $z$ such that $z=-1\ \text{and}\ z=1$.

As a result, you'll have $$(x,y,z)=(0,\pm2,1),(\pm2, 0,-1).$$

Answer 2

ok so you have

$x(1+z)=0$

$y(z-1)=0$

$x^2+y^2=4$

now clearly it can't be both $x$ and $y$ be zero,could you say something about roots? for example from first two,what if $z=-1$? in this case what is $y$?

what if $x=0$? and

$(z=1)$?

Answer 3

Factor the first two. That gives four easy solutions, two with $x=0$ and two with $y=0$ Now assume $x,y \neq 0$ Then find a contradiction because each of the first two require a different value of $z$

Answer 4

Szuppose that $x=0$. Then $y=\pm 2$, hence $z=1$. Otherwise we may assume that $x\neq 0$. Then $z=-1$ by the first equation, and $y=0$.