Constructing an addition operation to a multiplication so that we get a field structure I am given a "multiplication" acting on real numbers such that \begin{equation} x * y = 2xy - 6x - 6y + 21.\end{equation}
So far I have proven that this is commutative, associative, closed on the set of real numbers, the identity element with respect to it is $\frac{7}{2}$.
I have to construct an "addition" operation so that the multiplication will be distributive with respect to the addition, ie. \begin{equation}x * (y \oplus z) = x * y \oplus x*z.\end{equation}
I've tried $x \oplus y = \frac{x+y}{2}$, but it is not associative. Please help me find an "addition" that is associative.
 A: Instead of $*$, let us use $\otimes$ to denote the "multiplication" at hand. 
Let $+, - $ and $\times$ be the ordinary addition, subtraction and multiplication of $\mathbb{R}$.
Let $g(x) = 2 \times x - 6$, we have
$$x \otimes y = g^{-1}( g(x) \times g(y) ) = 2 \times x \times y - 6 \times x - 6 \times y + 21$$
If we define
$$x \oplus y = g^{-1}( g(x) + g(y) ) = x + y -3 $$
We will find $\otimes$ and $\oplus$ satisfy following distributive law:  
$$\begin{align}
x \otimes ( y \oplus z )
&= g^{-1}( g(x) \times g(g^{-1}( g(y) + g(z) )) )\\
&= g^{-1}( g(x) \times ( g(y) + g(z) ) )\\
&= g^{-1}( (g(x) \times g(y) ) + (g(x) \times g(z) ) )\\
&= g^{-1}( g( x \otimes y) + g( x \otimes z) )\\
&= ( x \otimes y) \oplus ( x \otimes z )
\end{align}
$$
The main point is the function $g$ provides an isomorphism between the algebra
$( R, +, \times )$ and $( R, \oplus, \otimes )$. Other properties of the algebraic structure $( R, \oplus, \otimes )$ can be derived from those of $(R, +, \times )$ by repeat application of $g$ and unwinding with $g^{-1}$. The distributive law above is an example of this general procedure.
Another example is the associativity of our "addition" $\oplus$:
$$\begin{align}
x \oplus ( y \oplus z ) 
&= g^{-1}(g(x) + g(g^{-1}(g(y) + g(z)))\\
&= g^{-1}(g(x) + (g(y) + g(z)))\\
&= g^{-1}((g(x) + g(y)) + g(z))\\
&= g^{-1}(g( x \oplus y ) + g(z) )\\
&= (x \oplus y ) \oplus z
\end{align}$$
Update 
About the question how to discover $g(x)$.
Given a complicated expression, one strategy to simplify it is repeatly get rid of pieces that look most ugly. In an expression like 
$$z \stackrel{def}{=} x \otimes y = 2xy - 6x - 6y + 21$$
the most ugly pieces are the $-6x$ and $-6y$, so I start by getting rid of them first.
$$\begin{array}{rrl}
& z &=  2xy - 6x - 6y + 21\\
\iff & z &= 2(x-3)(y-3)+3\\
\end{array}\tag{*1}$$
Since I suspect the underlying algebra will be ordinary addition/multiplication, I try to transform the expression to the form $g(z) = g(x)\times g(y)$.
With this as guideline, what need to do is sort of obvious.
$$\begin{array}{rrl}
(*1) \iff & z - 3 &= 2(x-3)(y-3)\\
\iff & 2(z-3) &= (2(x-3))(2(y-3))\\
\iff & 2z - 6 &= (2x - 6)(2y - 6)
\end{array}$$
and the last equality leads to the choice $g(x) = 2x - 6$.
