Finding range of variables with defined relations Q: Given that $\{x,y,z\}\in\mathbb{R}$ , and
$$\begin{cases}
x+y+z=6\\
xy+yz+zx=7
\end{cases}$$  and  ; determine range of $x$ ,$y$ , and $z$.
I thought that defining $x$ in terms of $y$ and $z$ : $x=6-y-z$ , using it in the second provided equation , then assume it to be a quadratic in $y$ or $z$ and then using $D\ge 0$ could solve it ; am i right?
Can someone provide a simpler solution; i will be grateful for it.
 A: As you are looking for something different, here is a mainly geometrical approach.
Due to relationship:
$$\underbrace{(x+y+z)^2}_{36}=x^2+y^2+z^2+\underbrace{2(xy+yz+zw)}_{14}$$
the initial issue is equivalent to the system:
$$\begin{cases}x^2+y^2+z^2&=&22\\x+y+z&=&6\end{cases}$$
Therefore, the locus is the intersection of the sphere centered in $(0,0,0)$ with radius $\sqrt{22}$ and a plane .
As a consequence , the set of solutions is a circle whose center is $C(x_c=2,y_c=2,z_c=2)$ (due to symmetry in coordinates $x,y,z$) and radius $\sqrt{10}=\sqrt{22-12}$ by Pythagoras ; indeed,  length OC = $\sqrt{12}$).
A parametrization of the circle is as follows:
$$\begin{pmatrix}x\\y\\z\\\end{pmatrix}=\underbrace{\begin{pmatrix}2\\2\\2\\\end{pmatrix}}_C+\sqrt{10}\cos\theta \underbrace{\begin{pmatrix} \ \ \ \tfrac{1}{\sqrt{2}}\\ -\tfrac{1}{\sqrt{2}}\\0\\\end{pmatrix}}_U+\sqrt{10}\sin\theta \underbrace{\begin{pmatrix}\ \ \ \tfrac{1}{\sqrt{6}}\\ \ \ \ \tfrac{1}{\sqrt{6}} \\ -\tfrac{2}{\sqrt{6}}\end{pmatrix}}_V$$
(please note that vectors $U$ and $V$ constitute an orthonormal basis of the vector plane with equation $x+y+z=0$ parallel to affine plane $x+y+z=6$).
Let us now take a look at coordinate
$$z=2+2\tfrac{\sqrt{10}}{\sqrt{6}}\sin \theta$$
whose values are taken in the interval:
$$z \in [2-2\sqrt{\tfrac53},2+2\sqrt{\tfrac53} ]\tag{1}$$
Coordinate ranges for $x$ and $y$ are identical, on account of symmetry.
