Question

Solve for $x,y$ and $z$, if $xy+x+x+y=23,yz+y+z=31,zx+z+x=4$

Answer 1

Given,

$xy+x+y=23\Rightarrow (1+x)(1+y)=24\to \left(1\right)$

$yz+y+z=31\Rightarrow (1+y)(1+z)=32\to \left(2\right)$

$zx+x+z=47\Rightarrow (1+z)(1+x)=48\to \left(3\right)$

$\left(1\right)\times \left(2\right)\times \left(3\right)\Rightarrow \left[\right(1+x\left)\right(1+y\left)\right(1+z){]}^2=24\times 32\times 48$

$(1+x)(1+y)(1+z)=\pm 192\to \left(4\right)$

$\frac{\left(4\right)}{\left(1\right)}\Rightarrow 1+z=\frac{\pm 192}{24}$

$1+z=\pm 8$

$\therefore z=-9\left(or\right)7$

$\frac{\left(4\right)}{2)}\Rightarrow 1+x=\frac{\pm 192}{32}$

$1+x=\pm 6$

$x=-7\left(or\right)5$

$\frac{\left(4\right)}{\left(3\right)}\Rightarrow 1+y=\frac{\pm 192}{48}$

$1+y=\pm 4$

$y=-5\left(or\right)3$

$\therefore (x=-7,y=-5,z=-9)\left(or\right)(x=5,y=3,z=7)$