Solve for X in the graph and matrix below

Using the provided matrix we can see the distance from D to C is 5 and the only path is D=>E=>C, and the distance from E to C is 3. If the total distance from D to C is 5 and the distance from E to C is 3, then X has to be 2.

The answer is available from the first row. Take respectively the nodes B, C and D as intermediate nodes to reach E from A and the respective distances stand to be 10+X for B, 24+X for C and 5+X for D, out of which 5+X is the minimum distance, hence 5+x must be equal to 7(the given minimum distance from A to E) and here you have it! x=7-5=2 :-)

The only way to go from $A$ to $D$ without repeating edges is to take the edge from $A$ to $D$ , so that the edge $A \rightarrow D$ has weight $5$ . Similarly, the edge $C \rightarrow B$ has weight $4$ . Finally, the two real choices from $A$ to $C$ is to either take the direct edge, or to go via $A \rightarrow D \rightarrow E \rightarrow C$ . The direct edge has cost $12$ , which does not match the given matrix. Thus the shortest path from $A$ to $C$ is $A \rightarrow D \rightarrow E \rightarrow C$ , which has cost $5 + x + 3 = 10$ . Thus ${\bf x = 2}$ .