TITLE: Domination problem with sets
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: 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.