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I am trying to understand why, when solving a system of equations in two variables and have already solved for, say, $x$, it doesn't matter whether I plug into the first equation or the second to find $y$. I don't understand why it's not possible that the first equation and second equation can give me a different value of $y$ or, worse, why it isn't possible that when I plug the $(x,y)$ pair into the other equation, I won't get a contradiction.

Could someone please explain why this works?

JohnT
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  • You need to specify what type of system. Is it linear? Then there are theorems that specify conditions when a unique solution exist. If you know that a unique solution exist, it does not matter what equation to use to get the second value. Also, by expressing one variable thru the other and substituting, you are restricting possible values for the second variable. You do understand that the system will have the same solution if we swap equations (put the second into the first place), right? – Vasili Aug 30 '22 at 03:17
  • @Vasili I should have been clearer. Yes, a linear system with, let's say, real coefficients, and let's say it's $ax + by = C_1$ and $cx + dy = C_2$ and $ad - bc \neq 0$. The theorem is then that a unique solution exists, I understand. I'm mainly thinking of the proof of this and why it doesn't matter when solving for the second variable why I can plug into either equation. I do understand that swapping the equations will have no effect, yes. – JohnT Aug 30 '22 at 03:25

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The reason that you can substitute into either equation is that the solution is the intersection of two straight lines. Both of the lines meet at the intersection, so the values of $x$ and $y$ are the same there for either line. You have already got the $x$ value for the intersection, so the $y$ value is unique, regardless of which line you choose. Intersection of two lines

Suzu Hirose
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  • This perspective is extremely helpful. Would you say that if there is no intersection, I wouldn't be able to solve for a value of $x$ at all? This is the last thing I'm not able to fully convince myself of. – JohnT Aug 30 '22 at 04:16
  • @JohnT: that's true. – ultralegend5385 Aug 30 '22 at 04:42
  • If there is no intersection then the lines are parallel. That is the same thing as the two equations being linearly dependent. – Suzu Hirose Aug 30 '22 at 04:45
  • @SuzuHirose I suppose what confuses me, though, is why it wouldn't be possible to solve for an erroneous solution, $x$. But I suppose if I get an $x$, it uniquely determines a $y$, so I get some solution, which I can't if the lines are parallel. In that case, I think I understand. Thank you very much. – JohnT Aug 30 '22 at 05:40
  • @JohnT I think you understand too. – Suzu Hirose Aug 30 '22 at 06:11