elimination part of a mathematical question

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 $z$.

Answer 1

$$2(x+y)-\sqrt3(x-y)=0$$

$$4(x+y)+2\sqrt3(x-y)+9z=1$$

$$x+y+z=0$$

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?

Answer 2

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 $$x=-\frac{1}{2}-\frac{1}{3}\sqrt{3},y=-\frac{1}{2}+\frac{1}{3}\sqrt{3},z=1$$

Answer 3

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 $x,y$

$x+y = u\\ x-y = \frac {2}{\sqrt 3}v$

$x+y = -1\\ x-y = -\frac {2}{\sqrt 3}$

$2x = -1-\frac {2}{\sqrt 3}\\ 2y = -1 + \frac {2}{\sqrt 3}\\ x = -\frac {1}{2} - \frac {\sqrt 3}{3}\\ y = -\frac {1}{2} + \frac {\sqrt 3}{3}\\ z = 1$