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.

This is unfortunately not true even for vector spaces. For example, in $\mathbb{R}^2$ you can let $N=\mathrm{span}(e_1)$, $M_1=\mathrm{span}(e_2)$, and $M_2=\mathrm{span}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Thus $N$ is the $x$-axis, $M_1$ is the $y$-axis, and $M_2$ is the line $y=x$. Then $$M_1+N=\mathbb{R}^2=M_2+N$$ and $$M_1\cap N = \{0\} = M_2\cap N.$$ But $M_1\neq M_2$.
Edit: It IS true if you require $M_1\subseteq M_2$. The proof is purely set-theoretic, no homomorphisms required; it goes as follows. Let $m_2\in M_2$. Then $m_2$ is in $M_2+N$, so it must also be in $M_1+N$ by assumption, so $m_2=m_1+n$ for some $m_1\in M_1$ and $n\in N$. But then $m_1$ is also in $M_2$ by assumption, so $n=m_2-m_1$ belongs to $M_2$. This shows $n\in M_2\cap N$, so $n\in M_1\cap N$. Thus $n$ belongs to $m_1$, so $m_2=m_1+n$ also belongs to $M_1$.
