How to solve this system of equations systematically?

Question

This might seem a trivial problem, but I have some trouble in arranging the data. So suppose you are given $f(x,y)=x^2y^2(1+x+2y)$ and you want to find it's critical points. Thus we find $$\frac{\partial f}{\partial x}(x,y)=xy^2(2+3x+4y)\textrm{ and }\frac{\partial f}{\partial y}(x,y)=2x^2y(1+x+3y).$$

Now we set $f_x= 0$ and $f_y=0.$ Thus we get a system $$ \begin{split} xy^2(2+3x+4y) &=0\\ 2x^2y(1+x+3y) &=0 \end{split} $$

Now we have a bunch of cases. The way I think about this is as follows: $$((x=0)\lor(y=0)\lor(3x+4y=-2))\land((x=0)\lor(y=0)\lor(x+3y=-1)).$$ Then I consider each possibility separately, but this seems to be slow and sometimes I forget some solutions. Thus I was wondering if there are any other methods which one can use to solve such type of problems.

Answer 1

Use the converse of the distributive property:

$((x=0)\lor(y=0)\lor(3x+4y=-2))\land((x=0)\lor(y=0)\lor(x+3y=-1))\\\equiv(x=0)\lor(y=0)\lor[(3x+4y=-2)\land(x+3y=-1)]$

$3x+4y+2=0=x+3y+1$ is just a pair of straight lines (linear equations) intersecting at $(-2/5,-1/5)$. Therefore, you have $(x=0)\lor(y=0)\lor(x=-2/5\land y=-1/5)$

Answer 2

For $xy^2(2+3x+4y) =0$ we have the set of solutions

$$S_1 = \{x = 0, y = 0, 2+3x+4y = 0\}$$

For $2x^2y(1+x+3y) =0$ we have the set of solutions

$$S_2 = \{x = 0, y = 0, 1+x+3y = 0\}$$

So for the system of equations we have

$$S_1 \cap S_2 = \{x = 0, y = 0, (2+3x+4y = 0)\cap(1+x+3y = 0)\} = \{x = 0, y = 0\}\cup \{x=-\frac 25, y = -\frac 15\}$$