My attempt:-
Two pairs of cells :- (c1,c3) and (c1,c2); therefore, c1+c3=c1+c2 or c2=c3=x
Similary, another two pairs of cells I took were :- (c4,c5) and (c4,c6) with common cell having 5 in it, I took c4 as y , now c4+c5=c4+c6 , c4=c6=a
How to proceed further ?
Under the interpretation I presented in the comments, and piggybacking your diagram, the three cells surrounding $5$ should all be $a.$ Of course, this means that each cell sharing a common edge with those containing $a$ should themselves contain $5$; those sharing a common edge with $5$-cells should contain $a$, and so on until all cells are filled. It turns out - if my interpretation is correct (where did you get this question?) - you only have two numbers in the diagram: $5$ and $7$. The sum of all the $7$'s is equal to $7\sum_{n=1}^{5}n$, or $\frac{5\cdot6\cdot7}{2}$, which is $105$. Likewise, the sum of all the $5$'s is $5\sum_{n=1}^{4}n$, which is $\frac{4\cdot5\cdot5}{2} = 50.$ The sum of everything is therefore $155.$
Note that you can start the process from the cell containing $7$ instead, and reach the same answer. I've drawn a diagram of my own; I hope it helps (and I apologize for the quality). Note that the $a$'s in the first diagram correspond to the same cells containing $7$ in the second, while the $x$'s in the second diagram correspond to the cells containing $5$ in the first. It is this correspondence that allows us to conclude $a=7$ and $x=5$.