I started working through Hefferon's Linear Algebra book, and am stuck on one of the exercises (exercise 1.30). The question is as follows:
Prove that, where $a, b, c, d, e$ are real numbers with $a≠0$, if this linear
equation $ax+by=c$ has the same solution set as this one $ax+dy=e$, then
they are the same equation.
After struggling a bit I think I understood the logic of the proof, which goes like this:
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 $a(c/a)+d⋅0=e$, 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.
However, after experimenting a bit with some simple examples, I found the following two linear equations: $x+y=2$, and $x+2y=3$. Here, $a$ is the same in both equations and the two equations have the same solution set, namely: $(1,1)$. However, in this example $b\ne d$ and $c \ne e$, which seems to counter the proof. I feel like I'm overlooking something basic here, but can't find what it is.
PS I'm fairly new to math.
Best Answer
As alex said, $(1,1)$ is just one solution among many others. $x+y=2$ has $(x,y)=(2,0)$ as a solution, but $x+2y=3$ does not have that same solution. Hence, the sets of solutions for the two equations are not the same.