Thanks for adding all that work. I think you followed the ideas well and got an answer within reasonable error. Good work!
I know you may not yet have been introduced to nodal analysis. But this is another way to imagine and I thought I may as well show you how your answer might be approached in a different way -- one I expect you will soon see, if you haven't already.
The way I use nodal analysis is to put all the "outgoing currents" on the left side and all of the incoming currents on the right side of each equation. You have two unknown node voltages, assuming your bottom node can be considered \$0\:\text{V}\$, so you'd need to set up the following two:
$$\begin{align*}
\frac{V_a}{2200\:\Omega}+\frac{V_a}{2700\:\Omega}+\frac{V_a}{2200\:\Omega}&=\frac{5.5\:\text{V}}{2200\:\Omega}+\frac{0\:\text{V}}{2700\:\Omega}+\frac{V_b}{2200\:\Omega}\\\\
\frac{V_b}{2200\:\Omega}+\frac{V_b}{1800\:\Omega}&=\frac{V_a}{2200\:\Omega}+\frac{0\:\text{V}}{1800\:\Omega}
\end{align*}$$
This solves out as \$V_a\approx 2.3258\:\text{V}\$ and \$V_b\approx 1.0466\:\text{V}\$, with the difference of \$V_{ab}\approx 1.2792\:\text{V}\$. This is also the Thevenin voltage.
This is a way of not having to go through individual steps, one at a time, transforming the circuit. You can get the answer, somewhat more directly. But it does involve the simultaneous solution of two linear equations. (Not hard, at all, though.)
The equivalent resistance can be achieved by simply shorting out the voltage supply and then looking at the resistor network from the perspective of nodes \$a\$ and \$b\$. Here, the answer is easily seen as:
$$\left[\left( 2200\:\Omega\mid\mid 2700\:\Omega\right)+1800\:\Omega\right]\mid\mid 2200\:\Omega\approx 1271.42\:\Omega$$
So there are other ways to get there.
Just a heads-up.