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.

• This is a nightmare to read. Try typesetting using MathJax to make it easier for us to read. – J. Redman Jun 24 '18 at 8:50

Take $x+y=a,y+z=b,x+z=c$. Then you get

$ac=15\quad$

$ab=18\quad$

$bc=30\quad$

Multiply to get $(abc)^2=8100$, thus getting $abc=90$.

Divide by $,,$ separately to get

$a=3$

$b=6$

$c=5$

Now, substitute to get

$x+y=3\quad$

$y+z=6\quad$

$x+z=5\quad$

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

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

• Thank You very much @MalayTheDynamo – Math Tise Jun 24 '18 at 9:00
• nice answer. @MalayTheDynamo – sameera lakshitha Jun 24 '18 at 11:04
• or abc = -90 with all three variables negative... no idea if it leads to an additional solution – hkBst Jun 24 '18 at 11:08
• This doesn't really answer OP's request "Please advise." It's just another answer to the original question. – JiK Jun 24 '18 at 18:45
• I tried solving again by matrices; and I got infinite number of solutions. – Math Tise Jun 25 '18 at 7:27