Is this diophantine?

$xy +x+y=23...(1)$ $\rightarrow$ $xy+x+y+1=24$

$xz +x+z=41...(2)$ $\rightarrow$ $xz+x+z+1=42$

$yz +y+z=27...(3)$ $\rightarrow$ $yz+y+z+1=28$

Now using Simon's favourite Factoring trick we can factorise the equations into:

$(x+1)(y+1)=24...(1)$

$(x+1)(z+1)=42...(2)$

$(y+1)(z+1)=28...(3)$

Now dividing the second equation by the first equation we get:

$\large\frac{y+1}{z+1}=\frac{24}{42}=\frac{4}{7}$

By cross-multiplying we get $7(y+1)=4(z+1)$ and solving we get $z+1=\large\frac{7(y+1)}{4}$

Substituting this into Equation 3:

$\frac{7(y+1)}{4}\cdot(y+1)=28$

Hence $(y+1)^{2}=16$

$\boxed {y=3}$ (since the question only asks for positive solutions)

Substituting into Equation 1:

4(x+1)=24

$\boxed{x=5}$

Substituting into Equation 3:

$4(z+1)=28$

$\boxed{z=6}$

Therefore $x+y+z=5+3+6=\boxed{14}$

$given\quad equations\quad are\\ xy+x+y=23\\ xz+x+z=41\\ yz+y+z=27\\ Now\quad adding\quad 1\quad on\quad both\quad sides\\ 1+(x+y)+xy=24\\ 1+(x+z)+xz=42\\ 1+(y+z)+yz=28\\ Now\quad facorizing\quad we\quad get\\ (1+x)(1+y)=24--(1)\\ (1+x)(1+z)=42--(2)\\ (1+y)(1+z)=28--(3)\\ Taking\quad H.C.F\quad of\quad (1)\& (2)\\ (1+x)=6\quad (note\quad that\quad since\quad x\quad is\quad positive\quad 1+x\neq -6)\\ so\quad x=5\quad so\quad y=3\\ sustituting\quad this\quad in\quad (2)\quad we\quad get\\ z=6\\ (you\quad can\quad take\quad H.C.F\quad of\quad other\quad equations\quad also\\ but\quad the\quad values\quad of\quad x,y,z\quad don't\quad satisfy\quad all\quad the\quad 3\quad equations)\\ So\quad the\quad answer\quad is\quad 5+3+6=14$