I'd want to use two variables if I use algebra, which may be beyond you.
For trial and error, just make a table (keeping in mind that it appears the number of tables is even, since they can arrange them in pairs!):
tables ... chairs (single) ... chairs (double)
2 ............. 8 ............................. 6
and so on. Do you see that when the two numbers of chairs differ by 10 (needing 6 more then there are vs having 4 too many), you'll have your answer?
You might find a pattern that lets you get to the answer more quickly.