Just elegancy required 2

Algebra Level 4

$\begin{cases} xy+x+y=23 \\ xz+x+z=41 \\ yz+y+z=27 \end{cases}$

Let the two solution set of $x,y,z$ be $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ , find $x_1 y_1 z_1 + x_2 y_2 z_2$ .

1 solution

Brian Charlesworth
Jan 8, 2016

The given equations can be rewritten as

$(x + 1)(y + 1) = 24, (x + 1)(z + 1) = 42$ and $(y + 1)(z + 1) = 28$ .

Then $(x + 1)(y + 1) * \dfrac{1}{(x + 1)(z + 1)} * (y + 1)(z + 1) = 24*\dfrac{1}{42}*28$

$\Longrightarrow (y + 1)^{2} = 16 \Longrightarrow y + 1 = \pm 4.$

Solving the initial two equations in turn gives us that $x + 1 = \pm 6$ and $z + 1 = \pm 7.$

Thus $(x_{1}, y_{1}, z_{1}) = (5,3,6)$ and $(x_{2},y_{2},z_{2}) = (-7, -5, -8)$ , and so

$x_{1} \times y_{1} \times z_{1} + x_{2} \times y_{2} \times z_{2} = 90 + (-280) = \boxed{-190}.$