+9519 Add 1 to each equation.
\(\begin{cases}xy + x + y + 1 = 24\\yz + y + z + 1= 32\\zx + z + x + 1 = 48\end{cases}\)
We do this because now the left-hand side is factorable.
\(\begin{cases}(x + 1)(y + 1) = 24\qquad --- (1)\\(y + 1)(z + 1) = 32\qquad --- (2)\\(z + 1)(x + 1) = 48\qquad --- (3)\end{cases}\)
Now consider (1) * (2) / 3, we have
\(\dfrac{(x + 1)(y + 1)^2(z + 1)}{(z + 1)(x + 1)} = \dfrac{24 \cdot 32}{48}\\ (y + 1)^2 = 16\\ y = 3\text{ or }y = -5\)
When y = 3,
(x + 1)(4) = 24
x = 5
(z + 1)(5 + 1) = 48
z = 7
Therefore (x, y, z) = (5, 3, 7) is a solution.
When y = -5,
(x + 1)(-4) = 24
x = -7
(z + 1)(-7 + 1) = 48
z = -9
Therefore (x, y, z) = (-7, -5, -9) is another solution.
Solutions are (x, y, z) = (5, 3, 7) and (x, y, z) = (-7, -5, -9).
