Rohan have some books and some tables . if he puts 9 books into each table he will have 2 tables?

First you have to remember that this is only an exercise. Math is only valid when it describes reality, and what you have here is not a description of any real situation.

Call the number of books B. Call the number of tables T.

(B/9)(T - 2) = B "if he puts 9 books into each table he will have 2 tables left vacant."
6T + 3 = B "f he puts 6 books on each table he have 3 books left over."

Now I rewrite this because my math program likes to use x and y instead of B and T.

(x/9)(y - 2) = x
6y + 3 = x
Evaluating: (x/9)(y - 2) = x
xy - 2x = 9x
xy = 11x
y = 11 tables
x = 69 books

You have two lines. They look like this: www.wolframalpha.com/input/?i=solve+(x%2F9)(y+-+2)+%3D+x,+6y+%2B+3+%3D+x

Where they cross, the values of x and y satisfy both equations simultaneously, so it is called a simultaneous system. There are two ways to find that point. Substitution is when you solve one equation for one variable and substitute that into the other equation. However, the problem went to a solution without ever applying this method.

Elimination is when you add the equations in a way to eliminate one variable. Otherwise the process is the same.

I said math is only valid when it describes reality. The problem has two solutions, but a negative number of tables is not reality, so we disregard it.