+128468 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
+26367 Solve
\(xy + x + y = 23 \\ yz + y + z = 31 \\ zx + z + x = 47\)
in real numbers.
\(\begin{array}{|rcll|} \hline xy + x + y &=& 23 \\ (x+1)(y+1) -1 &=& 23 \\ \mathbf{(x+1)(y+1)} &=& \mathbf{24} \qquad (1) \\ \hline yz + y + z &=& 31 \\ (y+1)(z+1) -1 &=& 31 \\ \mathbf{(y+1)(z+1)} &=& \mathbf{32} \qquad (2) \\ \hline zx + z + x &=& 47 \\ (z+1)(x+1) -1 &=& 47 \\ \mathbf{(z+1)(x+1)} &=& \mathbf{48} \qquad (3) \\ \hline \end{array}\)
\(\text{Let $a=x+1,~b=y+1,~c=z+1$}\)
\(\begin{array}{|rcll|} \hline ab &=& 24 \qquad (4) \\ bc &=& 32 \qquad (5) \\ ca &=& 48 \qquad (6) \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline \dfrac{(4)*(6)}{(5)}: & \dfrac{ab*ca}{bc} &=& \dfrac{24*48}{32} \\ & a^2 &=& 36 \\ & \mathbf{a} &=& \pm \mathbf{6} \\ & x+1 &=& \pm 6 \\ & x_1 &=& 5 \\ & x_2 &=& -7 \\ \hline & \mathbf{(x+1)(y+1)} &=& \mathbf{24} \\ & (\pm6)(y+1) &=& 24 \\ & 6(y_1+1) &=& 24 \\ & y_1 +1 &=& 4 \\ & y_1 &=& 3 \\\\ & (-6)(y_2+1) &=& 24 \\ & y_2 +1 &=& -4 \\ & y_2 &=& -5 \\ \hline &\mathbf{(z+1)(x+1)} &=& \mathbf{48} \\ & (\pm6)(z+1) &=& 48 \\ & 6(z_1+1) &=& 48 \\ & z_1 +1 &=& 8 \\ & z_1 &=& 7 \\\\ & (-6)(z_2+1) &=& 48 \\ & z_2 +1 &=& -8 \\ & z_2 &=& -9 \\ \hline \end{array}\)
