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Consider the equations $x^2 +y^2=1$ and $y=x^2-1$. If $y=x^2-1$ is substituted into $x^2 +y^2=1$ to obtain $x^2 +(x^2-1)^2=1$, then the $x$ values that result from solving this equation will be the $x$ values that produce equal $y$ values in the original equations $y=x^2-1$ and $x^2 +y^2=1$. I don't understand why this is true on conceptual a level, but rather just accept it as a process. I understand that if I am trying to solve the system $y=x+1$ and $y=2x-1$, I can set these equations equal to each other and $2x-1=x+1 \implies$ that the $y$ values are the same when $x=2$. This works because $x+1=2x-1 \iff x+2=2x \iff x=2$. I don't understand, however, why this sort of substitution works in general. Is there a proof that this technique works for higher order polynomials and more complicated equations? Please not only answer the questions posed above but also help me identify additional things that I may not understand.

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    My thought process: suppose $x$ and $y$ are numbers and $x^2 + y^2=1$ and $y = x^2 -1$. Replacing $y$ with $x^2-1$ in the first equation (which is allowed because they are the same number, after all), we discover that $x^2 + (x^2 -1)^2 = 1$. Finding the solutions to this equation gives a list of possible values for $x$. For each possible value of $x$, we can compute $y = x^2 -1$ and then check if $x^2 + y^2=1$ is satisfied. If it is, then we have found a solution to the original system of equations. – littleO Mar 15 '22 at 11:08
  • I think one way to understand this is the following: the meaning of the assertion that $A=B$ is that anywhere you see the expression $A$, you can replace it with $B$ without changing anything (and vice versa). Or, to say it another way: suppose there were some fact you could assert about $A$, but not about $B$ -- maybe, $A$ is an odd prime bigger than 30 but $B$ is not. In that world, how would you feel about the claim $A=B$? And if, on the other hand, all the facts you could state about $A$ were also true of $B$, then on what grounds could you object to the claim that $A=B$? – Daniel Wagner Mar 15 '22 at 19:42

2 Answers2

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Let $(a,b)$ be a solution of the system which consists of the equations$$x^2+y^2=1\tag1$$and$$y=x^2-1.\tag2$$Asserting that $(a,b)$ is a solution of both $(1)$ and $(2)$ means that $a^2+b^2=1$ and that $b=a^2-1$. But then $a^2+(a^2-1)^2=1$, and therefore $a$ is indeed a solution of the equation $x^2+(x^2-1)^2=1$. And $b=a^2-1$ means that if, in the equality $y=x^2-1$, if you replace $x$ by $a$, then $y$ becomes $b$.

José Carlos Santos
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In terms of sets

Let $S$ be the set of $(x,y)$ pairs solving $x^2 +y^2 = 1$, and, $S'$ be the set of $(x,y)$ pairs solving $y=x^2 -1$. When we substitute, we are searching for solutions in $ S \cap S'$.

Why it works is that if $(a,b) \in S \cap S'$ then it solves both equations simultaneously, allowing us to add, subtract equatiosn without worry.

tryst with freedom
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