Homomorphic image of modular lattice is modular?

A lattice is modular if and only if it satisfies the modular law: $$a\le b\implies a\lor(x\land b)=(a\lor x)\land b.\tag1$$ Equivalently, a lattice is modular if and only if it satisfies the modular identity: $$(c\land b)\lor(x\land b)=[(c\land b)\lor x)\land b.\tag2$$ Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $a\le b$ if and only if $a=c\land b$ for some $c$.

Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.

P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":

Given $a,b,x\in L$ with $f(a)\le f(b)$, we want to show that $$f(a)\lor(f(x)\land f(b))=(f(a)\lor f(x))\land f(b).\tag3$$ We don't know if $a\le b$, so let $a_0=a\land b$. Then $a_0\le b$, so by the modular law we have $$a_0\lor(x\land b)=(a_0\lor x)\land b.\tag4$$ Moreover, $f(a_0)=f(a\land b)=f(a)\land f(b)=f(b)$ since $f(a)\le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.