Question

if$x^{2} - y^{2}- z^{2}-xy-yz-zx=0$
then find
$\frac{x+y}{z}$

Answer 1

Since we are given one equation with three variables, we can find the answer in two variables.

     x² - y² - z² - x y - y z - x z  = 0

     z² + z (x+y) + (y² - x² + xy) = 0

taking it as a quadratic equation in z, we get

           z = [ - (x+y) + -  √(5 x² - 3 y² - 2 x y) ] / 2

$\frac{x+y}{z} = \frac{2 (x+y)}{-(x+y) + - \sqrt{5 x^2 - 3 y^2 - 2 x y}}\\\\=\frac{2}{-1 +- \sqrt{5-\frac{8y}{x+y}-\frac{x}{(x+y)^2}}}$

Answer 2

Answer:

Since we are given one equation with three variables, we can find the answer in two variables.

x² - y² - z² - x y - y z - x z = 0

z² + z (x+y) + (y² - x² + xy) = 0

taking it as a quadratic equation in z, we get

z = [ - (x+y) + - √(5 x² - 3 y² - 2 x y) ] / 2

$\begin{gathered}\frac{x+y}{z} = \frac{2 (x+y)}{-(x+y) + - \sqrt{5 x^2 - 3 y^2 - 2 x y}}\\\\=\frac{2}{-1 +- \sqrt{5-\frac{8y}{x+y}-\frac{x}{(x+y)^2}}}\\\end{gathered} </p><p>$