How to solve these $3$ equations for three unknowns $x$,$y$,$z$? Question:

Solve:

$xy+x+y=23\tag{1}$

$yz+y+z=31\tag{2}$

$zx+z+x=47\tag{3}$

My attempt:

By adding all we get 

$$\sum xy +2\sum x =101$$

Multiplying $(1)$ by $z$, $(2)$ by $x$, and $(3)$  by $y$ and adding altogether gives

$$3xyz+ 2\sum xy =31x+47y+23z$$

Then, from above two equations after eliminating $\sum xy$ term we get 

$$35x+51y+27z=202+3xyz$$

After that subtracting $(1)\times 3z$ from equation just above (to eliminate $3xyz$ term) gives 

$$35x +51y-3z(14+x+y)=202\implies (x+y)[35-3z]+16y-42z=202$$

I tried pairwise subtraction of  $(1),(2)$ and $(3)$ but it also seems to be not working.

Please give me some hint so that I can proceed or provide with the answer.
Hint: Put
$$X=x+1$$
$$Y=y+1$$
$$Z=z+1$$

Then we have

$$XY=24$$
$$YZ=32$$
$$ZX=48$$

Can you take it from there?
Consider first equation :

XY + X + Y = 23
=> X ( Y + 1 ) = 23 -Y
=> X = (23-Y)/(Y+1) -(a)

Consider second equation :

YZ + Y +Z = 31
=> Y (Z+1)=31-Z
=>Y=(31-Z)/(Z+1)-(b)

Consider third equation :

XZ +Z+X = 47
=> Z(X+1)=47-X
=>Z=(47-X)/(X+1)-(c)

Now put the value of z from eq c into eq b , then simply that equation , then put the simplified value of Y from eq b into eq A , then from there you will get the value of X , then put that value into eq c you will get the value of Z and finally put the value of Z into eq b to get the value of Y .