How to solve this system of equations

I had a system of differential equations and I ended up in this.

$$4x+2y+4z=0$$ $$2x+y+2z=0$$ $$4x+2y+4z=0$$

I don't know what to do from here.

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$(x,y,z)=(x,-2x-2z,z)$ –  pedja Jan 26 '12 at 12:35
There's always taking the null space... –  Guess who it is. Jan 26 '12 at 12:35
Would you know what to do if you ended up with just one equation, $2x+y+2z=0$? –  Gerry Myerson Jan 26 '12 at 12:37

$$4x+2y+4z=0 \iff 2(2x+y+2z=0) \iff 4x+2y+4z=0$$ Since there are three unknowns, and only one equations, two parameters are required to fix a solution. We can have a parametric solution as follows:
Set $x=\lambda$ and $y=\mu$. Then, $$2\lambda+\mu+2z=0 \implies z=\dfrac{-2\lambda-\mu}{2}$$
Hence, the triplet $(\lambda,\mu,\dfrac{-2\lambda-\mu}{2})$ with $\lambda,\mu \in \mathbb R$ is a solution to your system.