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I am wondering whether this viewpoint is a good way to think of the process of solving equations on a logical level:

That when we have an equation such as $2x= 8$, "solving" the equation is just seeing what is implied of $x$ as far as its value if $2x=8$ is true of it- so $x$ is a generic number (with restrictions by context of the equation), and if $x$ by the context is always a natural number, we essentially prove that $\forall x\in \mathbb{N} (2x=8)\implies (x=4))$ using a direct proof if we were to divide by $2$, and the antecedent of the implication ($x=4$) is the result which constitutes the solution. So solving an equation just amounts to deducing all of the necessary conditions of a number which constitutes a solution, and any number for which the necessary conditions of a solution aren't false is part of the solution set?

Is this viewpoint good/valid, or should I just think of it like this: that solving an equation such as $2x = 8$ is trying to find the set such that all elements therein render the statement $2x=8$ true when $x$ is assigned to this element.

Princess Mia
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    An equation may have multiple solutions... Thus, we may say that, in principle, to find the solutions of an equation $\varphi(x)$ [en expression with $x$ free, like the above $2x=8$] amounts to check if the formula $\exists x \varphi(x)$ holds in a suitable domain $D$ (naturals, reals, etc.). If yes, we are usually asked to find the solution set, i.e. the set $S = { x \in D \mid \varphi(x) }$. – Mauro ALLEGRANZA Aug 31 '23 at 09:13
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    I think both views you describe are the same. – 5xum Aug 31 '23 at 09:14
  • Your last sentence is all that you need. We are looking for the set of elements that make the statement true. – John Douma Aug 31 '23 at 09:35
  • "So, solving an equation just amounts to deducing all of the necessary* conditions of a number which constitutes a solution"* $\quad$ Not quite comprehending. In any case, noting that $$x+2=0\implies x=-2\implies x^2=4\implies x=\pm2\implies x\in {-2,2},$$ we can say that the equation $x+2=0$ necessitates that $x\in {-2,2};$ yet, ${-2,2}$ is certainly not the required solution set, which contains precisely every solution of the given equation. – ryang Sep 01 '23 at 01:49
  • You may find the discussion here relevant. – Noah Schweber Sep 03 '23 at 00:57

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