If the solution sets of two linear equations are equal, then the equations are identical. Show that if the linear equations
$$x_1 +kx_2 = c \tag{1}$$
and
$$x_1+lx_2 = d \tag{2}$$
have the same solution set, then the equations are identical.
I attempted this beforehand, but a user pointed out a logical flaw with the proof. Here is my new attempt.
We wish to show that $c=d$ and $k=l$.
We note that the solution sets of each of the equations are nonempty, because we can find two points that satisfy their respective equations, as expressed below.
Clearly the point $(c-kc,c)$ satisfies $(1)$, and because the solution sets of the two equations are equal, it follows that this point also satisfies $(2)$. Substituting this point into the second equation, we have
$$d=c-kc+cl \tag{3}$$
Similarly, the point $(d-dl, d)$ satisfies $(1)$, so we substitute this point into the equation and solve for $d$ to get
$$d=c+dl-dk \tag{4}$$
Equating $(3)$ and $(4)$, we have
$$c-kc+cl = c+dl -dk $$
Simplification renders
$$(c-d)(l-k)=0$$
If $c=d$, then if $x_2\neq 0$, we have 
$$x_1+kx_2 = x_1+lx_2$$
from which we deduce $k=l$, so the equations are identical.
Now if $k=l$, then for all $(x_1,x_2)\in \mathbb{R}^2$,
$$x_1 + kx_2 = x_1+lx_2$$
which implies that $c=d$. Hence the equations are identical.
Therefore for all $(x_1,x_2)\in \mathbb{R}^2$, the equations $x_1+kx_2 = c$ and $x_1+lx_2=d$ are identical whenever their respective solution sets are equal.
 A: A rough sketch of the proof:
Rewriting the equations as:
$$x_1 = -kx_2 + c$$
$$x_1 = -lx_2 + d$$
For $x_2 = 0$, (1) gives $x_1 = c$. The pair $(0,c)$ also satisfies (2):
$$c = 0 + d$$
$$\mathrm{or}\ c=d$$
Therefore
$$x_1+kx_2=x_1+lx_2$$
$$\mathrm{or}\ k=l$$
Since $k=l$ and $c=d$, the equations are identical.
A: I will criticize your attempt as follows. The first two objections are minor while the third is more serious. Then I will show a similar but more stream-lined proof.

*

*I think you meant to say that $(d-dl,d)$ satisfies $(2)$, not $(1)$.


*immediately after that, you substitute into "the equation" without specifying which one. You want to substitute into the first one.


*You correctly arrive at two cases, the first of which is $c=d$. But then you assume (without justification) that there is a solution for which $x_2\not=0.$
Here is a proof in the same style:
Let equation (1) be $x_1+kx_2=c$ with $S_1\subset \mathbb{R}^2$ its solution set. Let equation (2) be $x_1+\ell x_2=d$ with $S_2\subset \mathbb{R}^2$ its solution set.
We are given $S_1=S_2$, and we immediately note that this set is not empty, for $(c,0)$ belongs to it. Since $(c,0)\in S_2,$ $$c=c+\ell \cdot 0=d.$$ Hence all that remains is to show $k=\ell$. This follows easily once we note that $(c-k,1)\in S_1=S_2,$ so that $$c-k+\ell\cdot 1 = c.$$
