Given that ‘*’ is an operation that is valid on the set S which consists of all real numbers except – 1 i.e., R – { – 1} defined as a*b = a + b + ab 

Let us assume a + b + ab = – 1

⇒ a + ab + b + 1 = 0

⇒ a(1 + b) + (1 + b) = 0 

⇒ (a + 1)(b + 1) = 0 

⇒ a = – 1 or b = – 1 

But according to the problem, it is given that a≠ – 1 and b≠ – 1 so, 

a + b + ab≠ – 1, so we can say that the operation ‘*’ defines a binary operation on set ‘S’. 

We know that commutative property is p*q = q*p, where * is a binary operation. 

Let’s check the commutativity of given binary operation: 

⇒ a*b = a + b + ab 

⇒ b*a = b + a + ba = a + b + ab 

⇒ b*a = a*b

∴ Commutative property holds for given binary operation ‘*’ on ‘S’. 

We know that associative property is (p*q)*r = p*(q*r) 

Let’s check the associativity of given binary operation: 

⇒ (a*b)*c = (a + b + ab)*c 

⇒ (a*b)*c = a + b + ab + c + ((a + b + ab)×c) 

⇒ (a*b)*c = a + b + c + ab + ac + bc + abc ...... (1) 

⇒ a*(b*c) = a*(b + c + bc)

⇒ a*(b*c) = a + b + c + bc + (a×(b + c + bc))

⇒ a*(b*c) = a + b + c + ab + bc + ac + abc ...... (2) 

From (1) and (2) we can clearly say that associativity holds for the binary operation ‘*’ on ‘N’. We need to also solve for x in the given expression: 

⇒ (2*x)*3 = 7 

⇒ (2 + x + 2x)*3 = 7 

⇒ (2 + 3x)*3 = 7 

⇒ 2 + 3x + 3 + ((2 + 3x)×3) = 7 

⇒ 5 + 3x + 6 + 9x = 7 

⇒ 11 + 12x = 7 

⇒ 12x = – 4

⇒ x = \(\frac{4}{12}.\)

⇒ x = \(\frac{-1}{3}.\)

∴ The value of x is \(\frac{-1}{3}.\)