If $ad - bc = 0$, then $\begin{cases} ax + by = j \\ cx + dy = k \end{cases}$ has no unique solutions Problem:
I have an exercise in an assignment, which is the following:

Show, if $ad - bc = 0$, then $$ax + by = j \\ cx + dy = k$$ has no unique
  solutions.

I first have problems understanding what unique solutions means, but now I think  have to demonstrate that the system above has not just one solution but more.
I had a hint from one of my professors to use forward/backward substitution to prove the following problem, but I must admit I would not know how to do it.

What I have tried?
I have tried to isolate an $x$ in one equation and replace it in the other, for example, isolating $x$ of the first equation and replace it in the second:
$$x = \frac{j - by}{a}$$
$$c\left(\frac{j - by}{a} \right) + dy = k$$
If I try to solve this, I think I will not arrive at any helpful conclusion.
I have tried to apply the hint of one of the professors to use forward/backward substitution. I must admit I am not so familiar with these concepts.
I am not sure if this is backward/forward substitution, but maybe one example where I use these concepts was to solve a system of equations like this:
$$\begin{matrix}7x + 7y - 7z = 0 \\ 0 + y + z = 7 \\ 0 + 0 -z - 7 \end{matrix}$$
(Zeros are there just to align the system.)
What I think I have tried to do is a backward substitution. I have started by getting $z$ from the last equation: $$z = -7$$
Then, I replaced $-7$ in the equation in the middle, we obtain $$y - 7 = 7 \\ y = 14$$
Finally, we can replace $y$ and $z$ in the first equation and we obtain 
$$7x + (7 * 14) - (7 * -7) = 0$$
$$x = -21$$

Questions


*

*Is my process of backward substitution in my example correct?

*How would I solve my original problem using backward/forward substitution? 

I remember that what I have to demonstrate is that it has no unique solutions, so probably I have to show it has more than one solution. For example, that for different $x$ values we get different $y$ values...
 A: Continue with your own approach, as you said, you get 
$$y(ad - bc) = ak - cj \\ y * 0 = ak - cj$$
This means either it has no solution, if $ak-cj\neq 0$, which gives you a contradiction; or it has infinitely many solution, if $ak-cj=0$, since this equation becomes $0=0$, in which case you have only one equation for two variables. 
A: Hint $\ $ If $\,ad = bc\,$ then $\,(x,y)\,$ a solution $\,\Rightarrow\, (x+d,y-c)\,$ a solution.
Key Idea $\ $ The map  $\,{\bf  x}\mapsto \bf A  x\,$ is linear, therefore  if $\, \color{#c00}{{\bf A}\begin{bmatrix}x_0\\ y_0\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}}\,$ then   
$\qquad\ \ {\bf A} \begin{bmatrix}x\!+\!x_0\\ y\!+\!y_0\end{bmatrix} = {\bf A}\left( \begin{bmatrix}x\\ y\end{bmatrix}\! +\! \begin{bmatrix}x_0\\ y_0\end{bmatrix}\right) = {\bf A}\begin{bmatrix}x\\ y\end{bmatrix} + \color{#c00}{{\bf A}\begin{bmatrix}x_0\\y_0\end{bmatrix}} =\, {\bf A} \begin{bmatrix}x\\y\end{bmatrix}$
i.e. $\quad \color{#c00}{{\bf A x_0 = 0}}\,\Rightarrow\, {\bf A}(\bf x+x_0) = A {\bf x} + \color{#c00}{A{\bf x_0}} = A{\bf x}$ 
hence  $\ {\bf Ax} = {\bf b}\,\Rightarrow\, {\bf A}({\bf x+x_0)} = {\bf Ax} = \bf b\, $
thus  $\,{\bf x}\,$ a solution $\,\Rightarrow\, {\bf x+x_0}\,$ a solution, for any $\,{\bf x_0}\,$ such that $\, {\bf A x_0 = 0}.$  
This is a fundamental property of linear equations, e.g. differential equations, recurrences, etc.
