Solving nonlinear system with three equations and three variables

Question

Hope someone can help me with this:

Solve the next nonlinear equations system over $\mathbb{R}$ $$ \left\{ \begin{array}{} x^2+y^2+z^2=6 \\ x^2-y^2+2z^2=2 \\ 2x^2+y^2-z^2=3 \end{array} \right. $$ Hint: Use auxiliary variables.

I tried to solve this using the hint like that: $a = x^2, b = y^2, c = z^2$ Then the system become: $$ \left\{ \begin{array}{} a+b+c=6 \\ a-b+2c=2 \\ 2a+b-c=3 \end{array} \right. $$ And there is one solution: $(1, 3, 2)$, but how to proceed from here?

Thanks in advance

Answer 1

Multiply the second equation by $2$, the third one by $3$ and subtract them from the first one. Then we obtain: $$(x^2+y^2+z^2)-2(x^2-y^2+2z^2)-3(2x^2+y^2-z^2)=6-2\cdot2 -3\cdot 3$$ that is $-7x^2=-7$ or $x^2=1$ and we get $x=1$ or $x=-1$.

Can you take it from here? Finally you should obtain $8$ solutions.

Answer 2

multiplying the first equation by $-1$ and adding to the second we get $$-2y^2+z^2=-4$$ (I) multiplying the first by $-2$ and adding to the third we have $$-y^2-3z^2=-9$$ (II) and $-2\times$ (II)+(I) we have $$7z^2=14$$ can you finish?

Answer 3

Set $a:=x^2$, $b:=y^2$, and $c:=z^2$. Then the system becomes $$ \begin{cases} a+b+c &= 6 \\ a-b+2c &= 2 \\ 2a+b-c &= 3. \end{cases} $$

Solving this system yields $(a,b,c)=(1,3,2)$. Hence we have the equations $$ x^2=1, \quad y^2=3, \quad z^2=2.$$ From these we deduce $x=\pm 1$, $y=\pm\sqrt{3}$, and $z=\pm\sqrt{2}$. Consequently the solution set is $$ \{ (1,\sqrt{3},\sqrt{2}), (1,\sqrt{3},-\sqrt{2}), (1,-\sqrt{3},\sqrt{2}), (1,-\sqrt{3},-\sqrt{2}), (-1,\sqrt{3},\sqrt{2}), (-1,\sqrt{3},-\sqrt{2}), (-1,-\sqrt{3},\sqrt{2}), (-1,-\sqrt{3},-\sqrt{2}) \}. $$ We could also write this set as $$ \{(x,y,z) \mid x=\pm 1,\,y=\pm\sqrt{3},\,z=\pm\sqrt{2}\} $$ or $$ \{1,-1\}\times\{\sqrt{3},-\sqrt{3}\}\times\{\sqrt{2},-\sqrt{2}\}. $$

Answer 4

obs : that's not a non-linear equation. A non-linear would be if it involved both x^2 and x at the same time.