I'm having trouble understanding the logic underlying the proof for this problem (from Hefferon's Linear Algebra):
Prove that, where $a, b, c, d, e$ are real numbers with $a \ne 0$, if this linear equation $ax + by = c$ has the same solution set as this one $ax + dy = e$, then they are the same equation.
The proof provided is fairly straightforward:
The solution set of the first equation is this: $$ \left\{(x, y) \in \mathbb{R}^2 \left| x = \frac{c − by}{a} = \frac{c}{a} − \frac{b}{a} \cdot y\right.\right\} \tag{∗} $$ Thus, given $y$, we can compute the associated $x$. Taking $y = 0$ gives the solution $(c/a, 0)$, and since the second equation $ax + dy = e$ is supposed to have the same solution set, substituting into it gives that $$a(c/a) + d \cdot 0 = e,$$ so $c = e$. Taking $y = 1$ in (∗) gives $$a((c − b)/a) + d · 1 = e,$$ and so $b = d$. Hence they are the same equation.
Why are we justified in knowing that $c = e$ in all solution sets, when we have only demonstrated it in one (when $y = 0$)? What assures us that this is the rule, and not a coincidence? I feel like this is fairly basic, so would love any help understanding what is going on here.
I'm going to do my best to clear up the confusion. If something that I'm saying is not clear, please leave a comment, and I'll do my best to make it clearer.
Saying that $ax+by=c$ and $ax+dy=e$ have the same solution set is actually quite a strong statement. If $ax+by=c$ and $ax+dy=e$ have the same solution, then that means that whenever $(x,y)$ is a solution to $ax+by=c$, we automatically get that $(x,y)$ is also a solution to $ax+dy=e$, and vice versa.
Hence the argument goes like this: assume that $a\ne0$ and that $ax+by=c$ and $ax+dy=e$ have the same solution set. Since $\left(\frac{c}{a},0\right)$ is a solution to $ax+by=c$, we automatically get that $\left(\frac{c}{a},0\right)$ is also a solution to $ax+dy=e$. Hence it follows that $a\cdot\frac{c}{a}+d\cdot0=e$, which gives us that $c=e$.
Similarly, we can note that $\left(\frac{c-b}{a},1\right)$ is a solution to $ax+by=c$. Since $\left(\frac{c-b}{a},1\right)$ is a solution to $ax+by=c$, and $ax+by=c$ and $ax+dy=e$ have the same solution set, we automatically get that $\left(\frac{c-b}{a},1\right)$ is also a solution to $ax+dy=e$. Hence it follows that $a\cdot\frac{c-b}{a}+d\cdot1=e$, which gives us that $c-b+d=e$. Since we already have that $c=e$, this gives us that $b=d$.