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.

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

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




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

Divide by $[1],[2],[3]$ separately to get




Now, substitute to get




Add to get $x+y+z=7$

Subtract separately to get



As @hkBst pointed out, we could also get $abc=-90$, thus giving


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

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