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Note that $4x \in \mathbb Z$ and hence $x= k +\frac{r}{4}$ for some $k \in \mathbb Z$ and $r \in \{ 0,1,2,3\}$.

Then, the equation becomes $$4k+r= k+[2k+\frac{r}{2}]+[3k+\frac{3r}{4}]=6k+[\frac{r}{2}]+[\frac{3r}{4}]$$ which is equivalent to $$r=2k+[\frac{r}{2}]+[\frac{3r}{4}].$$

Now, just solve this for each $r \in \{ 0,1,2,3\}$: