show no equilibrium pairs exist in a non cooperative game using pay-off set? I am trying to understand the following exersice from the solutions of my professor and I really don't understand what she is doing.  The exersice is the following:
Suppose the matrix below is a pay off matrix of  a two person non cooperative game where the left pay offs are A's pay offs and right pay offs are B's pay offs and  A has the horizontal strategies and B the vertical.  The question is two find all the equilibrium pairs of randomized strategies.
$\begin{bmatrix}(1,2) & (4,3)\\(3,1) & (4,2)\end{bmatrix}$  
The solution is the following. We first plot the pay-off set which is the whole of the convex set determined by the four points representing pairs of pure strategies. Then we note that the pairs (ai, b2)(i = 1, 2) are
both equilibrium pairs and so therefore is any pair of the form (α, b2) where α is a randomized strategy .Then the solution says that by considering x = −2pq − q + 4 and y = p − q + 2  we must show that there are no other strategies in equilibrium. This is the part I don't understand... How can I use this x and y to show that there is no strategy in equilibrium? Any help would be highly appreciated
Thanks in advance for any help
 A: First, let's see directly from the bimatrix why there are no other equilibria. Notice that the right column strictly dominates the left column, since for the top row we have $3>2$ and for the bottom row we have $2>1$. Thus, player 2 only plays the right column in equilibrium.
Since our reasoning can deal only with player 2's payoff, let's now derive in detail the payoff to player 2 when both players play general mixed strategies.  Let $p$ be the probability for player 1 playing the top row, and $q$ be the probability for player 2 playing the left column. Then, when the players play $((p,1-p),(q,1-q))$, the resulting payoff $y$ for player 2 is:
$$
\begin{align}
y = & \quad p \cdot (q\cdot2 +(1-q)\cdot3) + (1-p) \cdot (q\cdot 1 + (1-q) \cdot 2) \\
= &  \quad 2pq + 3p - 3pq + q +2 - 2q - pq - 2p + 2pq \\
= & \quad p - q + 2 
\end{align}
$$
So, how does this equation for player 2's payoff tell us that 2 never plays anything except the right column in equilibrium? Well it reads as follows:
For any mixed strategy of player 1 (that is for all $p$), the payoff of player 2 is decreasing in $q$, the amount of probability that player 1 places on the left column, since the only term that depends on $q$ is $-q$. Thus, for all $p$, player 2's payoff is maximized when $q=0$, i.e., when player 2 plays the right column.
