Question

Solve the following equations :
$(x+y{)}^2-z^2=-9,(y+z{)}^2-x^2=15,(z+x{)}^2-y^2=3.$

Answer 1

$(x+y{)}^2-z^2=-9$----------------1

$(y+z{)}^2-x^2=15$-----------------2

$(z+x{)}^2-y^2=3$-------------------3

By adding Equation 1,2 & 3, we get

$x^2+y^2+2xy-z^2+y^2+z^2+2yz-x^2+z^2+x^2+2zx-y^2=9$

$x^2+y^2+z^2+2(xy+yz+zx)=9$

$(x+y+z{)}^2=9$

$x+y+z=3$

$x+y=3-z,y+z=3-x$

$(x+y{)}^2-z^2=-9$

$\Rightarrow (3-z{)}^2-z^2=-9$

$\Rightarrow 9+z^2-6z-z^2=-9$

$\Rightarrow 6z=18$

$\Rightarrow z=3$

$z+x=3-y$

$\Rightarrow (3-x{)}^2-x^2=15$

$\Rightarrow 9+x^2-6x-x^2=15$

$\Rightarrow 6x=-6$

$\Rightarrow x=-1$

$x+y+z=3$

$\Rightarrow y=3-x-z=3+1-3=1$

$\therefore x=-1,y=1,z=3$