I am unsure how to solve problems involving several markets and optimizing the price across all my markets. Note: I am looking to be pointed in a specific direction of study, not a solution to the problem below.
A simple example: Imagine there are $n = 2$ markets. $M1$ and $M2$. Each market has a supply and demand function, where:
$y : price $
$x : quantity $
$M1$
$Supply_{M1} : y = -2 + x$
$Demand_{M1} : y = 8 - x$
$M2$
$Supply_{M2} : y = -2 + 2x$
$Demand_{M2} : y = 10 - x$
Now solving for each market is where the lines intersect, very straight forward. But how do you do this, if you allow for markets to trade with each other? Meaning $M1$ can supply units to $M2$? And vice versa.
The real problem has at around 20 markets, that are connected to $1 - 4$ other markets. They can only trade a certain limit.
The real-life example is electricity prices that are calculated across several areas, but each have a specific price determined by interconnections.
In your example, you are correct that it is easy to solve each market. The first operates at $x=5,y=3$ and the second at $x=4,y=6$. The fact that the price is higher in the second says we should buy in the first and sell in the second. We no longer have the fact that the demand quantity equals the supply quantity in each market, only that the totals are the same. Let $x_1$ be the amount produced in market $1$, $x_2$ the amount produced in market $2$ and $z$ be the amount shipped from market $1$ to market $2$. We still have a common price (we assume no transport cost). Now the equations become $$y=-2+x_1\\y=8-(x_1-z)\\y=-2+2x_2\\y=10-(x_2+z)\\ 2y=18-(x_1+x_2)\\\frac 32y=-3+(x_1+x_2)\\2y=-4+\frac43(x_1+x_2)\\x_1+x_2=\frac {66}7$$ so the total quantity has increased. That makes sense, as we are producing more in the low cost place and selling more in the high value place. We can solve for the other variables. If you have a limited transport capacity (so a maximum $z$) you can solve while ignoring the limit. If the limit is exceeded, set $z$ to the maximum and solve again.