Algebra and matrices. 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.

Take $x+y=a,y+z=b,x+z=c$. Then you get 
$ac=15\quad[1]$
$ab=18\quad[2]$
$bc=30\quad[3]$ 
Multiply to get $(abc)^2=8100$, thus getting $abc=90$.
Divide by $[1],[2],[3]$ separately to get
$a=3$
$b=6$
$c=5$
Now, substitute to get 
$x+y=3\quad[4]$
$y+z=6\quad[5]$
$x+z=5\quad[6]$
Add to get $x+y+z=7$
Subtract separately to get 
$x=1,y=2,z=4$.
EDIT
As @hkBst pointed out, we could also get $abc=-90$, thus giving 
$x=-1,y=-2,z=-4$.
