If $I$ is ideal of $R$, $N$ is submodule of $M$, then $I\frac MN=\frac{IM+N}N$ $I$ is ideal of ring $R$. $M,N$ are $R$-modules and $N$ is submodule of $M$, how to prove:
$$I\cdot\frac MN=\frac{IM+N}N  \quad? $$
Conclusion:
It's not a "problem" to solve, just a notion. If you meet the same confusion, see answers below :)
 A: Let $p_1:M\rightarrow M/N$, $I(M/N)$ is the submodule of $M/N$ generated by $p_1(im), i\in I, m\in M$. $IM+N$ is a submodule of $M$ which contains $N$, we can define the quotient $p_2:IM+N\rightarrow (IM+N)/N$.
Consider $i:IM+N\rightarrow M$ the canonical injection and $p_3=p_1\circ i$. $p_3(im+n)=p_1(im+n)=p(im)$. The kernel of $p_3$ contains $N$. This implies that $p_3$ induces a morphism $f: (IM+N)/N\rightarrow I(M/N)$. $f$ is surjective, and $f(p_2(im+n))=0$ is equivalent to $p_1(im)=0$ this is equivalent to saying that $im\in N$ and $im+N\in N$, this  implies that $p_2(im+n))=0$.
A: Isn't  the equality obvious if you observe that
\begin{alignat}{2}
I\,(M/N)&=\Bigl\{\sum_\text{finite}i_k(m_k+N)\Bigr\}, \quad \rlap{i_k\in I, m_k\in M}\\
&=\Bigl\{\sum_\text{finite}i_km_k+N\Bigr\} &&\text{by definition of the product by $i_k$ in $M/N$ }\\
&=\Bigl\{\Bigl(\sum_\text{finite}i_km_k\Bigr)+N\Bigr\}=(IM+N)/N.
\end{alignat}
A: $I(M/N)= (IM+N)/N$ is just a definition(don’t overthink it!), because every element in $M/N$ is of the form $m+N$ so when you multiply by an $i \in I$ you just get $im + N \in IM+N$ 
A: Pedro Tamaroff's comment inspires my a lot.
$N$ may not be contained in $IM$, but it is contained in $IM+N$. 
Eg. $R=\mathbb Z$, $I = p\mathbb Z$, $M = \mathbb Z$, $N = q\mathbb Z$, then $IM = p \mathbb Z$. However $N = q\mathbb Z$ may not contained in $ IM  = p\mathbb Z$. 
Notion ${IM}/N$ doesn't make much sence while $(IM+N)/N$ does, so we treat $im+N(i\in I, m\in M)$ as element in $(IM+N)/N$. And it's just a matter of notion. .
Homomorphisms are:
$\phi: I\cdot M/N\to (IM+N)/N,\quad i\cdot(m+N)\mapsto im+N$.
$\psi:(IM+N)/N \to I\cdot M/N,\quad im+N\mapsto i\cdot(m+N)$.
