If $a, b, ... e$ $\in \mathbb{R}$ and $a \neq 0$, if $ax + by = c$ has the same solution set of $ax + dy = e$, then these equations are the same I have an exercise in my last assignment for liner algebra class, where I have to prove that 2 equations are the same. 

The problem is the following:

Prove that, if $a, b, ... e$ are real numbers and $a \neq 0$, if $ax + by = c$ has
  the same solution set of $ax + dy = e$, then these equations are the
  same.
What if $a = 0$?

First of all, I have some problems understanding exactly the problem.


*

*For solution set of an equation, I suppose it indicates all the possible values for $x$ and $y$, such that the equation is true. But I am don't exactly know how would I turn this in mathematical terms that can help me solving this problem. We only know that if we insert the same pair $x$ and $y$ in both equations, we obtain an equality, like $1 = 1$ or $0 = 0$.

*I have also problems understanding what does it mean to 2 equations be the same. Does this mean that they have the same left and right side expression of the equal sign, or it simply means that they have the same solution set (then I would not have anything to prove, since this is the assumption).
What I have tried to do is to make a system of equations (even though I am not seeing why should this help me):
$$
\begin{cases}
ax + by = c \\
ax + dy = e
\end{cases}
$$
I can for example use the first equation to find $x$ respect to $y$:
$$x = \frac{c - by}{a}$$
Replace this $x$ in the second equation to find $y$:
$$a(\frac{c - by}{a}) + dy = e$$
$$c - by + dy = e$$
$$y(b + d) = e - c$$
$$y = \frac{e - c}{(b + d)}$$
If I replace this $y$ in the first equation, I obtain (if I did not do any mistake):
$$x = \frac{cd - be}{a(d - b)}$$
Sincerely, I am not seeing why should this help me, and even if I am on the right track or not.

For $a = 0$, both equations result in:
$$
\begin{cases}
by = c \\
dy = e
\end{cases}
$$
$$
\begin{cases}
y = \frac{c}{b} \\
y = \frac{e}{d}
\end{cases}
$$
So, when $a = 0$, I suppose that for both equations being equal, $\frac{c}{b}$ must be equal to $\frac{e}{d}$.

If $a\ne 0$ we have for the solutions:
$$
x_1=\dfrac{c-by}{a} \; \land \;x_2=\dfrac{e-dy}{a} \quad \forall y 
$$
since we want $x_1=x_2$ we have:
$$
(c-by = e-dy) \Rightarrow y(b-d)=c-e \quad \forall y
$$
so, for $y=0$ gives: $c=e$ and for $y\ne 0$ gives $b=d$.
The case $a=0$ is solved in OP.
