Consider a system of equations: $$ f(x,y) = g(x,y) \tag{1} $$ $$ h(x,y) = I(x,y) \tag{2} $$
Equation $(1)$ has some solution set. Equation $(2)$ has some solution set. The system as a whole has a solution set which is the intersection of the two.
If I multiply equation $(1)$ by equation $(2)$: $$ f(x,y)h(x,y) = g(x,y)I(x,y) \tag{3} $$ This is valid only under the context that $h = I$, otherwise I'd be multiplying one side of equation $(1)$ by something and the other side by something else - an invalid operation. Anyways, equation $(3)$ has the solution set of the original system I presume. I know when you multiply, you can sometimes create solutions: namely multiplying by $0$, which would give any equation an infinite amount of solutions, or multiplying an equation by itself (squaring,etc). Here I'm not interested in the solution set of equation $(1)$ by itself, but of the entire system. So I'll drop this worry for now.
Question
Let's assume that equations $(1)$ and $(2)$ are "independent" in the sense that they aren't "parallel to each other" or "multiples of each other." In other words, lets assume a finite number of solutions to the system. Instead of solving equations $(1)$ and $(2)$, could I solve equations $(1)$ and $(3)$ or $(2)$ and $(3)$? Or even if the original system has no solution or and infinite number of solutions, could I still use equation $(3)$ in place of one of the others?
Question of Interest
If I divide the equations to yield $$ \frac{f}{h} = \frac{g}{I} \tag{4} $$ Could I use equations $(1)$ and $(4)$ or $(2)$ and $(4)$ to solve the original system? Division could potentially delete solutions of the original problem: name you can't divide by $0$. Therefore it seems like division is more problematic. For instance, you can't just divide $x^2 + x = 0$ by $x$ as this assumes $x = 0$ is not a solution, when indeed it is. Anyways, I assume we just have to check the $(x,y)$ pair that makes equation $(2)$ zero. If this $(x,y)$ pair also satisfies equation $(1)$ in must be included along with any other solutions.