Let $M,N$ be linear sub-spaces of a vector space $V$with $x,y \in V$: if $z \in (x+M) \cap (y+N)$ then $(x+M) \cap (y+N) = z+ M \cap N$ Let $M,N$  be linear sub-spaces of a vector space $V$with $x,y \in V$:
Trying to show if $z \in (x+M) \cap (y+N)$ then $(x+M) \cap (y+N) = z+  M \cap N$:
If $z' \in (x+M) \cap (y+N)$ then $z' = x+m = y+n$ for some $m \in M, n \in N$. 
Not sure how to proceed. Hints appreciated. 
 A: Put $A = (x + M) \cap (y + N)$ and $B = z + M \cap N$. Let $v \in B$. It follows that $v = z + w$ for some $w \in M \cap N$. Since $w \in M \cap N$, $w \in M$ and $w \in N$. Since $z \in A$, $z = x + m$ for some $m \in M$, and it follows that $m = z - x$. Therefore, put $z = x + (z - x) = x + m$. Hence $v = x + (m + w)$, so $v \in (x + M)$. The argument to show that $v \in (y + N)$ is similar, and we get the desired inclusion.
Partial answer for the other inclusion to get you started.
Let $u \in A$. Then, we want to show that $u = z + r$ for $r \in M \cap N$. Since $u \in A$, $u = x + m = z + (x - z) + m$ and $u = z + (y - z) + n$ for $m \in M$ and $n \in N$.
A: Suppose $z \in (x+M) \cap (y+N)$, then there exists $m \in M, n \in N$ such that $z= x+m = y+n$.
First to show that $(x+M) \cap (y+N) \subset z + M \cap N$, take any $t \in (x+M) \cap (y+N)$, then there exists $m' \in M, n' \in N$ with $t = y+n' = x+m'= z + (x-z) + m'$ where $(x-z) + m'$ is in $M$ since $-m = x-z\in M$ and $m' \in M$, and so $z + (x-z) + m' \in z + M$.
Similarly $t = y+n' = z+ (y-z) + n'$ where $-n = (y-z) \in N$ and $n' \in N \implies (y-z)+n' \in N \implies z+ (y-z)+n' \in (z+N)$, then $t \in z+ M \cap N$
Now to show that $z + M \cap N \subset (x+M) \cap (y+N)$, take any $l \in z+M\cap N$. Then there exists some $a' \in M, b' \in N$ with $l = z+a'=z+b'$, but $z = x+m$ and so $l = x+m+a'$ where $a' \in M \implies m+a' \in M$ and so $l=x+(m+a') \in x+M$.
Similarly $l = z+b' = y+(n+b')$ where $(n+b') \in N$, therefore $l \in y+N \implies
l \in (x+M) \cap (y+N)$
