# If xy+yz+zx=12-x^2=15-y^2=20-z^2 then x+y+z=?

Jul 29, 2018

The answer is $= \pm 6$

#### Explanation:

Firstly,

$x y + y z + z x = 12 - {x}^{2}$

$\implies$, $x y + {x}^{2} + y z + z x = 12$

$\implies$, $x \left(x + y\right) + z \left(x + y\right) = 12$

$\implies$, $\left(x + y\right) \left(x + z\right) = 12$

Therefore,

$\left\{\begin{matrix}x + y = \pm 3 \\ x + z = \pm 4\end{matrix}\right.$

Similarly,

$x y + y z + z x = 15 - {y}^{2}$

$x y + z x + {y}^{2} + y z = 15$

$\left(x + y\right) \left(y + z\right) = 15$

Therefore,

$\left\{\begin{matrix}x + y = \pm 3 \\ y + z = \pm 5\end{matrix}\right.$

And finally,

$x y + y z + z x = 20 - {z}^{2}$

$x y + z x + y z + {z}^{2} = 20$

$\left(y + z\right) \left(x + z\right) = 20$

Therefore,

$\left\{\begin{matrix}y + z = \pm 5 \\ x + z = \pm 4\end{matrix}\right.$

So,

$x + y + x + z + y + z = \left(\pm 3\right) + \left(\pm 4\right) + \left(\pm 5\right) = \pm 12$

$2 \left(x + y + z\right) = \pm 12$

$x + y + z = \pm \frac{12}{2} = \pm 6$

Before, I considered that $\left(x , y , z\right) \in {\mathbb{N}}^{3}$ but $\left(x , y , y\right) \in {\mathbb{Z}}^{3}$ is also valid.

Jul 29, 2018
• $x y + y z + z x = 12 - {x}^{2}$

$\implies x y + y z + z x + {x}^{2} = 12$

$\implies y \left(x + z\right) + x \left(z + x\right) = 12$

$\implies \left(x + y\right) \left(z + x\right) = 3 \times 4. \ldots . \left[1\right]$

• $x y + y z + z x = 15 - {y}^{2}$

$\implies x y + z x + y z + {y}^{2} = 15$

$\implies x \left(y + z\right) + y \left(z + y\right) = 15$

$\implies \left(x + y\right) \left(y + z\right) = 3 \times 5. \ldots \left[2\right]$

• $x y + y z + z x = 20 - {z}^{2}$

$\implies y \left(x + z\right) + z \left(x + z\right) = 20$

$\implies \left(y + z\right) \left(x + z\right) = 5 \times 4. \ldots \left[3\right]$

It is obvious from [1], [2] and [3]

${\left(\left(x + y\right) \left(y + z\right) \left(z + x\right)\right)}^{2} = {3}^{2} \times {4}^{2} \times {5}^{2}$

$\implies \left(x + y\right) \left(y + z\right) \left(z + x\right) = \pm \left(3 \times 4 \times 5\right) \ldots . . \left[4\right]$

By [1] and [4]
$y + z = \pm 5$

By [2] and [4]
$z + x = \pm 4$

By [3] and [4]
$x + y = \pm 3$

Summing up we get

$2 \left(x + y + z\right) = \pm 12$

$\implies x + y + z = \pm 6$