Consider the following linear programme: minimise $x - 2y$
subject to $x+3y\geq 3$ , $3x + y\geq 3$ and $x + y \leq 3$
I solved this question graphically and the answer is (0,3). But is there a way to solve it with equations and be absolutely certain that my answer is correct?
You can solve it using Lagrangian multipliers. It's the basis for many optimization methods and rather tedious to do it by hand, but the analytical way you are looking for. In this case since there are 3 inequality constraints you have to check 8 different cases.
You can read about it http://mat.gsia.cmu.edu/classes/QUANT/NOTES/chap4/node6.html
The ST of equations to solve is, following the nomenclature of the page on the link:
$$1 + \mu_1 + 3\mu_2 -\mu_3 = 0 $$ $$-2 + 3\mu_1 + \mu_2 -\mu_3 = 0 $$ $$ \mu_1(-x -3y + 3) = 0 $$ $$ \mu_2(-3x -y + 3) = 0 $$ $$ \mu_3(x + y - 3) = 0 $$ $$\mu_1,\mu_2,\mu_3 \geq 0$$
And the 8 cases you need to check are for all possible combination of zeros and non-zero multipliers. That is:
$$\mu_1,\mu_2,\mu_3 = 0$$ $$\mu_1 \ne 0 , \mu_2,\mu_3 = 0$$ ... $$\mu_1 , \mu_2 = 0, \mu_3 \ne 0$$ ... $$\mu_1,\mu_2,\mu_3 \ne 0$$
If you find a solution to the system where $$\mu_1,\mu_2,\mu_3 \geq 0$$ and the point is feasible (i.e., complies with the inequality constraints) then it is the solution to your minimization problem. Hint: try the $\mu_1 = 0, \mu_2, \mu_3 \ne 0$ case