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xy  +  x +  y  =  23  ⇔     y =  (23 - x) / (x + 1)       (1)

yz  +  y + z   =   31              (2)

zx  + z +  x     =  47  ⇒   z   =  ( 47 - x)  / ( x + 1)    ( 3)

Sub

(1)  and (3)  into (2)

(23 - x) / ( x + 1) *  ( 47 - x) / (x + 1)  +  (23 - x)/(x + 1)  +  (47 - x) / ( x + 1)  =   31

Multiply  through by ( x + 1)^2

(23 - x) ( 47 - x)  +  (23 - x) ( x + 1)   + (47 - x) (x + 1)  = 31  (x + 1)^2

Simplifying this we  get  that

-32x^2  -64x + 1120  =  0

32x^2  + 64x  -  1120  =  0        divide through by  32

x^2  + 2x  - 35  =  0         factor

( x + 7) ( x - 5)  =  0

Seting  each  factor to 0 and  solving for  x  we  get   that

x =  -7   y = -5  , z  = -9

      or

x  =  5    y =  3    z  = 7

Solve
$xy + x + y = 23 \\ yz + y + z = 31 \\ zx + z + x = 47$
in real numbers.

$\begin{array}{|rcll|} \hline xy + x + y &=& 23 \\ (x+1)(y+1) -1 &=& 23 \\ \mathbf{(x+1)(y+1)} &=& \mathbf{24} \qquad (1) \\ \hline yz + y + z &=& 31 \\ (y+1)(z+1) -1 &=& 31 \\ \mathbf{(y+1)(z+1)} &=& \mathbf{32} \qquad (2) \\ \hline zx + z + x &=& 47 \\ (z+1)(x+1) -1 &=& 47 \\ \mathbf{(z+1)(x+1)} &=& \mathbf{48} \qquad (3) \\ \hline \end{array}$

$\text{Let $a=x+1,~b=y+1,~c=z+1$}$

$\begin{array}{|rcll|} \hline ab &=& 24 \qquad (4) \\ bc &=& 32 \qquad (5) \\ ca &=& 48 \qquad (6) \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline \dfrac{(4)*(6)}{(5)}: & \dfrac{ab*ca}{bc} &=& \dfrac{24*48}{32} \\ & a^2 &=& 36 \\ & \mathbf{a} &=& \pm \mathbf{6} \\ & x+1 &=& \pm 6 \\ & x_1 &=& 5 \\ & x_2 &=& -7 \\ \hline & \mathbf{(x+1)(y+1)} &=& \mathbf{24} \\ & (\pm6)(y+1) &=& 24 \\ & 6(y_1+1) &=& 24 \\ & y_1 +1 &=& 4 \\ & y_1 &=& 3 \\\\ & (-6)(y_2+1) &=& 24 \\ & y_2 +1 &=& -4 \\ & y_2 &=& -5 \\ \hline &\mathbf{(z+1)(x+1)} &=& \mathbf{48} \\ & (\pm6)(z+1) &=& 48 \\ & 6(z_1+1) &=& 48 \\ & z_1 +1 &=& 8 \\ & z_1 &=& 7 \\\\ & (-6)(z_2+1) &=& 48 \\ & z_2 +1 &=& -8 \\ & z_2 &=& -9 \\ \hline \end{array}$