Electronic – Thevenin voltage for a simple circuit

This statement and what follows is false:

Then the current through the \$30\Omega\$ resistor is \$.567 \times 100/130 = .436\$ A

It looks like you are attempting to use a current divider (which is a correct approach) but you are using the wrong resistance. You need to use the equivalent resistance looking into the \$30\Omega\$ resistor, not just the \$30\Omega\$ resistor itself. That equivalent resistance is \$30\Omega + 40\Omega = 70\Omega\$ so the correct current divider is $$0.567\text{A} \times \frac{100\Omega}{100\Omega + 70\Omega} = 0.333\text{A}$$

You should be able to repeat this process to find the current through the leftmost \$20\Omega\$ resistor and then arrive at \$V_{\text{TH}}\$. Doing so I arrived at \$V_{\text{TH}} = 10\$V.

I have no idea how you got to the second to last equation of yours. However, let's assume you have the circuit with a Thevenin equivalent circuit substituted in only for R2 and I (your \$R_{th}\$ and \$V_{th}\$ are correct).

Several ways to solve this new circuit:

Nodal Analysis

^{simulate this circuit – Schematic created using CircuitLab}

Assuming current flows "clockwise" (in the direction of the drawn arrow), performing nodal analysis at node e gives:

\begin{gather} I_{R1} = I_{R2}\\ \frac{V - e}{R_1} = \frac{e - V_{th}}{R_2}\\ R_2 V + R_1 V_{th} = R_1 e + R_2 e\\ e = \frac{R_2 V + R_1 V_{th}}{R_1 + R_2} = \frac{R_2 V + R_1 R_2 I}{R_1 + R_2} \end{gather}

Which is the same answer as you got without using the Thevenin equivalent.

Superposition

See Weezveez's answer. It's the same result.

Mesh Analysis

\begin{gather} V - V_{R1} - V_{R2} - V_{th} = 0\\ V - I_1 R_1 - I_1 R_2 - V_{th} = 0\\ I_1 = \frac{V - V_{th}}{R_1 + R_2}\\ e = V_{th} + V_{R2} = V_{th} + I_1 R_2\\ e = I R_2 + \frac{V - V_{th}}{R_1 + R_2} R_2\\ e = I R_2 + \frac{V - I R_2}{R_1 + R_2} R_2\\ e = \frac{R_2 V + R_1 R_2 I}{R_1 + R_2} \end{gather}

As you can see, all of these methods give the same answer.