Here I read in a book that:
All homomorphic images of a solvable Lie algebra are solvable as well.
How to prove such a statement? Let's say we have a homomorphism $$\phi: \mathcal{G}\rightarrow \mathcal{H},$$ where $\mathcal{G}$ is a solvable Lie algebra and $\phi(0)=\mathcal{I}\in \mathcal{G}$ is the non-trivial kernel of the homomorphism. Then there is $g^{\{i\}}=[g^{\{i-1\}},g^{\{i-1\}}]=0$ for a number $i$. $[\phi(0)\equiv\mathcal{I},\phi(0)\equiv\mathcal{I}]=0$.
Let $L$ be a solvable Lie algebra. Then all homomorphic images $\phi(L)$ are again solvable, because for the derives series we see inductively that $$ \phi(L^ {(n)})=(\phi(L)^ {(n)}. $$ So if we have $L^{(n)}=0$, because $L$ is solvable, then this implies that also $\phi(L)$ is solvable.