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.
Why substitution works in general
algebra-precalculus
Best Answer
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 $b$.