elimination part of a mathematical question During solving a math question I got stuck in the elimination part and I couldn't proceed! Can someone help me, please? 
$$2(x+y)-3^\frac 1{2}(x-y)=0$$
$$4(x+y)+2(3)^{\frac 1{2}}(x-y)+9z=1$$
$$x+y+z=0$$
I want to get the values for $x$, $y$, and $z$. 
 A: $$2(x+y)-\sqrt3(x-y)=0$$
$$4(x+y)+2\sqrt3(x-y)+9z=1$$
$$x+y+z=0$$
Multiply the first equation by $2$, so$$2(x+y)-\sqrt3(x-y)=0\implies 4(x+y)-2\sqrt3(x-y)=0$$
Add this to $4(x+y)+2\sqrt3(x-y)+9z=1$.
$$8(x+y)+9z=1$$
$$8x+8y+9z=1$$
Multiply the $x+y+z=0$ by $8$.
$$x+y+z=0\implies 8x+8y+8z=0$$
Subtract this from $8x+8y+9z=1$, so $z=1$.
Can you figure out the rest?
A: From the last equation we get: $$z=-x-y$$ Plugging this in the second equation we obtain:
$$4(x+y)+2\cdot 3^{1/3}(x-y)+9(-x-y)=1$$
And from the first equation we get:
$$y=\frac{x(3^{1/2}-2)}{3^{1/2}+2}$$
Can you proceed?Finally we get $$x=-\frac{1}{2}-\frac{1}{3}\sqrt{3},y=-\frac{1}{2}+\frac{1}{3}\sqrt{3},z=1$$
A: I would like to change the variables here.
let:
$u = x+y\\
v = \frac {\sqrt 3}{2} (x-y)$
then
$2u-2v = 0\\
4u + 4v + 9z = 1\\
u+z = 0$
$u=-1, v = -1, z = 1$
Now back to $x,y$
$x+y = u\\
x-y = \frac {2}{\sqrt 3}v$
$x+y = -1\\
x-y = -\frac {2}{\sqrt 3}$
$2x = -1-\frac {2}{\sqrt 3}\\
2y = -1 + \frac {2}{\sqrt 3}\\
x = -\frac {1}{2} - \frac {\sqrt 3}{3}\\
y = -\frac {1}{2} + \frac {\sqrt 3}{3}\\
z = 1$
