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.
 A: 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?
A: We can use Simon's favorite factoring trick.
$$xy+x+y+1=(x+1)(y+1)$$
This tells us
$$(x+1)(y+1)=24$$$$(y+1)(z+1)=32$$$$(x+1)(z+1)=48$$So, we know that $x+1 = \pm\frac{\sqrt{24\cdot32\cdot48}}{32}\to x=5,-7$. Likewise, you can find the other variables.
A: You can convert the first equation into an equation that expresses y in terms of x.
You can convert the third equation into an equation that expresses z in terms of x.
You can substitute these formulas for y and z into the second equation.  This gives a single equation, in a single variable (x).
The single equation can be simplified, by letting v = x + 1, and substituting v-1 in for x.  Then you can multiply through by v².  Notice that you are assuming that v ≠ 0. Then you can solve for v.  Since this is a second order equation, you will get two solutions (which might be equal to each other).
Now you can solve for x.  (It is v-1).  Notice that you are assuming that x ≠ -1.
Now you can solve the first equation for y, and the third equation for z.
Now you need to make sure you don't have a divide-by-zero error.  In other words, check what values you get for y and z if x were equal to -1.  Since these are the asymptotes of the hyperbolas that are hinted at by marshal craft's answer, y and z turn out to be ±∞.  This demonstrates that it was okay to assume that x ≠ -1.
Now you can perform your check-by-substitution, to verify that both solutions are correct.
A: hint
$(2)-(1)$ gives
$$(z-x)(y+1)=8=(z+1-(x+1))(y+1)$$
$(3)-(2)$ becomes
$$(x-y)(z+1)=16=(x+1-(y+1))(z+1)$$
$(3)-(1)$ yields to
$$(z-y)(x+1)=24=(z+1-(y+1))(x+1)$$
From here, we put
$$X=x+1,\;Y=y+1,\; Z=z+1$$
thus
$$YZ-YX=8$$
$$ZX-ZY=16$$
$$XZ-XY=24$$
A: 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 . 
