Domination problem with sets For nearly two years, I have been struggling with the next task I have already published on MSE, but unfortunately with no respond. 

Let $M$ be a non-empty and finite set, $S_1,...,S_k$ subsets
  of $M$, satisfying:
(1) $|S_i|\leq 3,i=1,2,...,k$
(2) Any element of $M$ is an element of at least $4$ sets among
  $S_1,....,S_k$.
Show that one can select $[\frac{3k}{7}] $ sets from $S_1,...,S_k$
  such that their union is $M$.


Partial solution with probabilistic method: I can find a family of ${13\over 25}k$ such sets that no element in $X$ is in more then 3 set from that family. Thus we have a family of the size ${13\over 25}k$ instead of ${4\over 7}k$. 
Let' s take any set independently with a probability $p$. Let's mark with $X$ a number of a chosen sets and with $Y$ a number of elements that are ''bad''
i.e. elements which are in at least 4 sets among a chosen sets. Note that $4n\leq 3k$. Then we have $$E(X-Y)=E(X)-E(Y) \geq  kp-np^4 \geq kp (1-3p^3/4)$$Since a function $x \mapsto  x(1-3x^3/4)$ achives a maximum at $x=\sqrt[3]{1/3}$ we have $E(X-Y)\geq {\sqrt[3]{9}\over 4}k> {13\over 25}k$.
So with the method of alteration we find constant ${\sqrt[3]{9}\over 4}$ which is about $0,051$ worse then ${4\over 7}$.
For a solution using probabilistic method will be awarded with 200pt of bounty.

Alternative formulation.
 A: Clearly we can assume that each element in $M$ appears in exactly $4$ subsets. Let $|M| =n$. 
Stage 1. Take maximal subfamily $\mathcal{A} \subseteq
\{S_1,....,S_k\} =:\mathcal{S} $ such that:
$\bullet$ every member of that family $\mathcal{A}$ has 3 elements;
$\bullet$ all sets in $\mathcal{A}$ are pairwise disjunct.
Let $|\mathcal{A}| =a$ and let $A= \cup _{X\in \mathcal{A}} X$.
Then $|A|=3a$. Now, since each $a\in A$ appears exactly $4$ times
it must appear exactly $3$ times in sets not in $\mathcal{A}$. So by double counting between $M$ and $\mathcal{S}\setminus \mathcal{A}$ we have (elements in $A$ have a degree $3$ and other $4$) $$ 3\cdot 3a +4(n-3a)\leq 3\cdot (k-a)\;\;\Longrightarrow \;\;4n \leq 3k\;\;\;...(1)$$
Let us now erase all the elements in $M$ which appears in $A$ and let this new set be $M_1$, so $M_1 = M\setminus A$ (so $|M_1| = n-3a$) and do the same thing in remaining sets in $\mathcal{S}\setminus \mathcal{A}$ and we get new family of sets $\mathcal{S}_1$. 
Notice that each element in $M_1$ appears still $4$ times in sets from $\mathcal{S}_1$ and that each set in $\mathcal{S}_1$ has at most $2$ elements. (Why? If some of it, say $X$, has $3$ elements, that means that no element in $X$ was erased, so no element in $X$ is in $A$. But then we could put $X$ in $\mathcal{A}$ and we would get bigger family than $\mathcal{A}$ which is already maximal.) Also, let $k_1=|\mathcal{S}_1|$
Stage 2. Now take a maximal subfamily $\mathcal{B} \subseteq
\mathcal{S}_1 $ such that:
$\bullet$   every member of that family $\mathcal{B}$ has 2 elements;
$\bullet$    all sets in $\mathcal{B}$ are pairwise disjunct.
Let $|\mathcal{B}| =b$ and let $B= \cup _{X \in \mathcal{B}} X$.
Then $|B|=2b$. Now, since each $b\in B$ appears exactly $4$ times
it must appear exactly $3$ times in sets not in $\mathcal{B}$. So by double counting between $M_1$ and $\mathcal{S}_1\setminus \mathcal{B}$ we have (elements in $B$ have a degree $3$ and other $4$) $$ 3\cdot 2b +4(n-3a-2b)\leq 2\cdot (k_1-b)\;\;\Longrightarrow \;\;2n \leq k+5a\;\;\;...(2)$$
Let us now erase all the elements in $M_1$ which appears in $B$ and let this new set be $M_2$, so $M_2 = M_1\setminus B$ (so $|M_2| =n-3a-2b$) and do the same thing in remaining sets in $\mathcal{S}_1\setminus \mathcal{B}$ and we get new family of sets $\mathcal{S}_2$. 
Notice that each element in $M_2$ appears still 4 times in sets from $\mathcal{S}_2$ and that each set in $\mathcal{S}_2$ has at most 1 elements. (Why? If some of it, say $X$, has 2 elements, that means that no element in $X$ was erased, so no element in $X$ is in $B$. But then we could put $X$ in $\mathcal{B}$ and we would get bigger family than $\mathcal{B}$ which is already maximal.) Also, let $k_2=|\mathcal{S}_2|$.
Final stage. Now take a maximal subfamily $\mathcal{C} \subseteq
\mathcal{S}_2 $ such that:
$\bullet$ every member of that family $\mathcal{C}$ has 1 element;
$\bullet$ all sets in $\mathcal{C}$ are pairwise disjunct.
Let $|\mathcal{C}| =c$ and let $C= \cup _{X \in \mathcal{C}} X$.
Then $|C|=c$. Now, since each $c\in C$ appears exactly 4 times
it must appear exactly 3 times in sets not in $\mathcal{C}$. So by double counting between $M_2$ and $\mathcal{S}_2\setminus \mathcal{C}$ we have (elements in $C$ have a degree $3$ and other $4$) $$ 3\cdot c +4(n-3a-2b-c)\leq 1\cdot (k_2-c)\;\;\Longrightarrow \;\;4n \leq k+11a+7b\;\;\;...(3)$$
Clearly $C=M_2$ so we are finish with the process, that is $c+2b+3a = |M|$. All we have to check if resurrected sets (that is, we refile all the sets with erased elements) satisfies $$a+b+c\leq {3\over 7}k\;\;\;\; {\bf ?}$$ 
Using (1) and $3a+2b+c=n$ we get:
$$ 12a+8b+4c\leq 3k$$
Using (2) and $3a+2b+c=n$ we get:
$$ a+4b+2c\leq k$$
Using (3) and $3a+2b+c=n$ we get:
$$ 2a+2b+8c\leq 2k$$
If we add these three inequalites we get
$$ 14(a+b+c)<15a+14b+14c\leq 6k$$ and thus a conclusion.
A: 59k/140 sets suffice
Using your three equations, it is possible to get an improved upper bound of $59k/140$ (instead of $3k/7 = 60k/140$).
$$12a+8b+4c \leq 3k$$
$$ a+4b+2c \leq k$$
$$ a+b+4c \leq k$$
Multiply the first equation by 9, the second one by 12, and the last one by 20. 
$$108a+72b+36c \leq 27k$$
$$12a+48b+24c \leq 12k$$
$$20a+20b+80c \leq 20k$$
Then add them all and get:
$$140a+140b+140c \leq 59k$$
$$a+b+c \leq \frac{59}{140}k$$
This is the smallest solution that is attainable by combining the three equations. I found it by solving the linear optimization problem
$$3x_1 + x_2 + x_3 \rightarrow max$$
$$12x_1+x_2+x_3 \geq 1$$
$$8x_1+4x_2+x_3 \geq 1$$
$$4x_1+2x_2+4x_3 \geq 1$$
The optimal solution is $x_1=9/140, x_2=12/140, x_3=20/140$, which yields the factors for the optimal linear combination of the tree equations.
