Algebra and matrices.

Question

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.

Answer 1

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

$x=-1,y=-2,z=-4$.