Equality of submodules satisfying given conditions - Verification of solution Question:
Let $M$ be a left $R$-module($R$ is a unitary ring). let $M_1,M_2,N$ be $R-$submodules of $M$ such that $M_1\subseteq M_2$, $M_1+N=M_2+N$ and $M_1 \cap N=M_2 \cap N$. Are $M_1,M_2$ necessarily equal?
Attempt: I think I have a proof. I figured that $M_1=M_2$ if and only if $\frac{M_1}{M_1 \cap N}=\frac{M_{2}}{M_{2}\cap N}$. I also see that $\frac{M_1+N}{N}=\frac{M_2+N}{N}$ and hence the canonical isomorphisms $\natural_1:\frac{M_1+N}{N}\rightarrow \frac{M_1}{M_1\cap N}$ and $\natural_2:\frac{M_2+N}{N}\rightarrow \frac{M_2}{M_2\cap N}$ are identical. This proves the result.
It would be great if this can be verified. If it is correct, I don't have a good intuitive understanding as to why this holds. Hence, any interpretation/intuition or alternative simpler argument is highly appreciated. If wrong, please point out the error(s). Thank you.

Yes, that's correct.
You can also prove it directly. Take $x\in M_2$; then you know that $x\in M_2+N=M_1+N$, so there are $y\in M_1$ and $z\in N$ such that $x=y+z$. Since $M_1\subseteq M_2$ you have $z=x-y\in M_2\cap N$, so $z\in M_1\cap N$ and therefore $z\in M_1$. Hence $y=x+z\in M_1$.

This is also a consequence of the modular identity, which holds for the lattice of submodules of a module (and guess why it's called this way…)

for all submodules $X,Y,Z$ of $M$, if $X\subseteq Z$, then $X+(Y\cap Z)=(X+Y)\cap Z$

In your case $X=M_1$, $Z=M_2$ and $Y=N$. Then
$$
M_1=M_1+(M_1\cap N)=M_1+(N\cap M_2)=(M_1+N)\cap M_2=(M_2+N)\cap M_2=M_2
$$
