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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.

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  • $\begingroup$ The answer by the way for anyone interested (from my point of view) is a yes. Draw an $x-y$ "solution space". Equation 1 takes up some region in this plane (some bubble or multiple bubbles or maybe just discrete dots). Equation 2 takes up some space as well. The intersecting regions is the goal. I'm looking for the intersection. What regions does $fg = hI$ or $f/g = h/I$ take up? They inhabit the intersecting regions. But they can also inhabit other regions which aren't solutions of either Equation 1 or Eq 2. For instance consider the pair (2,3) which is not a solution of either $\endgroup$ – DWade64 Apr 10 '18 at 15:23
  • $\begingroup$ $f(2,3) g(2,3) = h(2,3)I(2,3)$ might yield $2(3) = 6(1)$ which is a true statement. So if you think carefully about it, yes. Solving Equations 1 and 3 or 2 and 3 is equivalent to solving the original 1 and 2. Using Division to create $f/g = h/I$ is okay as well. Just be careful with points that cause division by 0. You might have to find those by whatever means necessary and just add them in after solving 1 and 3 or 2 and 3. $\endgroup$ – DWade64 Apr 10 '18 at 15:27

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