I'm showing no solutions.
The 3rd equation gets you
$$(x+y)^2=100 \Rightarrow \pm (x+y)=10$$
$$(x=10-y) \vee (x=-(10+y))$$
when you try substituting those into equations 1 and 2 the result is no solution.
Is this system solvable? I can solve for z in terms of x and y, but I can't separate x and y to get a solution. Its three unkowns and three equations, but I don't know if all three unkowns have to be in all three equations. Anyway, if I solve for z I get but that's as far as I can get. Any suggestions?
Thanks
$Case\ I:\ x \ne - y\ and\ x \ne - 2y$
$x = zx + zy = z(x + y) \implies z = \dfrac{x}{x + y}.$
$y = zx + 2zy = z(x + 2y) \implies z = \dfrac{y}{x + 2y}.$
So $\dfrac{x}{x + y} = \dfrac{y}{x + 2y} \implies x^2 + 2xy = xy + y^2 \implies x^2 + xy - y^2 = 0 \implies$
$x = \dfrac{- y \pm \sqrt{y^2 - 4 * 1 * y^2}}{2} = \dfrac{- y \pm y\sqrt{-3}}{2}.$ No real solution.
$Case\ II:\ x = - y.$
$x = - y \implies x^2 + 2xy + y^2 = x^2 + 2x(-x) + (-x)^2 = x^2 - 2x^2 + x^2 = 0 \ne 100 \implies x \ne - y.$ No solution.
$Case\ III:\ x = - 2y.$
$x = - 2y \implies y = zx + 2zy = z(x + 2y) = z * 0 = 0 \implies x = - 2 * 0 = 0 = y = - y,\ which\ is\ case\ II.$ No solution.
It is an inconsistent system (at least in the reals).