Let $S$ be the free algebra $k\langle x,y\rangle$ and $R$ be the quotient $S/\langle xy-1\rangle$. In talking with someone, they mentioned they thought this algebra is left primitive. So, I set about looking how to prove that by finding a faithful simple left $R$ module.
Here's my candidate: let $M$ be the abelian group that is a countable direct sum of copies of $k$. Let $y$ act as a right shift on the entries: $y(f_0)=0$ and $y(f_i)=f_{i-1}$ for $i>0$, and $x$ as the left shift $x(f_i)=f_{i+1}$, and extend the action to all of $S$.
Obviously $xy-1$ annihilates $M$, so we can consider $M$ as a left $R$ module, and I convinced myself it's simple. The stumbling block for me is decisively concluding that the annihilator is no bigger than $(xy-1)$, proving that $_RM$ is faithful.
A little digression: I'm speaking more generally of a problem where one knows an ideal contains a desired set of relations, but needs also to conclude that the relations generate the entire ideal. I have heard that Bergman's Diamond lemma is a good thing to learn with regards to this, but I have not found a good introduction to it.
This seems tricky sometimes. Here's an example of intuition going wrong: Take the algebra $\mathbb Q\langle x,y\rangle/(xy-yx-1)$ and mod out by the ideal generated by $yx$. Alternative description of the algebra above, right? Except there's this inconvenient fact that $\mathbb Q\langle x,y\rangle/(xy-yx-1)$ is a simple ring, and therefore the quotient is the zero ring
Modulo $\langle xy-1\rangle$ we'd just be left with sums of monomials of the form $p(y)x^m$ where $p(y)$ is a polynomial in $y$. I want to reason that one can always produce an element of $M$ not annihilated by such an element, but I haven't gotten any traction with this. The $x^m$ doesn't cause any trouble, of course, but how do you prevent $p(y)$ from scrambling your element to produce $0$?
I'll only adress the first part of the question, i.e. find a simple faithful module for your ring $R$.
Let $V = \bigoplus_{n=1}^\infty e_n k$ and let $E= \operatorname{End}(V_k)$.
Define $f(e_i) = 0$, $f(e_i)= e_{i-1}$ for $i \ge 2$.
Let $g \in E$ such that $g^m e_1 = e_{r(m)}$ and $\lim_m r(m) = \infty$. Now, let $S$ be the $k$-algebra generated by $g$ and $f$. Then we can prove that $V$ is a simple faithful left $S$-module.
How is this useful?
For example, choosing $g$ such that $g(e_i) = e_{i+1}$, the map
$$k\langle x,y\rangle \to S: x \mapsto f, y \mapsto g$$
is a ring epimorphism with kernel $(xy-1)$ so we get $$k\langle x,y \rangle/(xy-1) \cong S$$
and we can thus view the module above as simple faithful $R$-module.
For more information and details, see "A first course in non-commutative rings" by Lam, starting p195, where this example is discussed in more detail.