Q. Solve for $x$, $y$ and $z$ if
$$(x+y)(x+z)=15, $$ $$(y+z)(y+x)=18, $$ $$(z+x)(z+y)=30. $$
Solution: I expanded each equation above as :
$$x^2+xz+yx+yz=15, \tag{1}$$ $$y^2+xz+yx+yz=18, \tag{2}$$ $$z^2+xz+yx+yz=30. \tag{3}$$
Then I subtracted $(1)-(3)$, $(2)-(1)$ and $(3)-(2)$; so I got the equations as below: $$x^2-z^2=-15 \tag{4}$$ $$y^2-x^2=3 \tag{5}$$ $$z^2-y^2=12 \tag{6}$$
I then tried solving $(4)$,$(5)$,$(6)$ by using matrices, but I couldn't reach any solution.
Please advise. Thank You.