<s>[INST] A simple 2 grade equations system If we have:
$$x^2 + xy + y^2 = 25 $$
$$x^2 + xz + z^2 = 49 $$
$$y^2 + yz + z^2 = 64 $$
How do we calculate $$x + y + z$$
 [/INST] $$x^2 + xy + y^2 = 25 \dots (1)$$
$$x^2 + xz + z^2 = 49 \dots(2)$$
$$y^2 + yz + z^2 = 64 \dots(3)$$
$(2)-(1)$ 
$x(z-y)+(z+y)(z-y)=24$
$\Rightarrow (x+y+z)(z-y)=24$
Similarly we get ,
$(x+y+z)(y-x)=15$ by $(3)-(2)$
$(x+y+z)(z-x)=39$ by $(3)-(1)$
Clearly Let $1/\lambda=(x+y+z)\ne 0$
Then we have, 
$(z-y)=24\lambda$
$(y-x)=15\lambda$
$\Rightarrow x+z-2y=9\lambda$
$\Rightarrow 3y=1/\lambda+9\lambda\Rightarrow y=1/3\lambda+3\lambda$
Now solve for x and put in the first equation to find the value of $\lambda$.
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