I'm trying to find the Thevenin voltage using the super position method. I'm having trouble with the 8A source and finding the Vab due to it. Here is the problem and what I got so far. Also suggestions if I'm going about this the wrong way about this. Just noticed I titled this Thev Resistance, meant to put voltage. Sorry about that mods.

The first part where you calculate the contribution from the 50 volt source is wrong; you did the divider calculation wrong. As far as the 8A source, you can indeed replace the 8A in parallel with 12Ω with a voltage source of 96 volts in series with 12Ω. Just finish the calculation.

So for the 50V source would it be V = 50 ( (176/19) / (176/19) +24) and with the 8A to 96V would it be V = 96 ( (176/19) / (176/19) +24)

Still not right. Have a look at: http://en.wikipedia.org/wiki/Voltage_divider In your circuit the part of Z1 is played by the top 12Ω resistor. The part of Z2 is played by 176/19Ω in series with the bottom 12Ω resistor. No, that's wrong. Sometimes it helps to redraw your circuit with various series and parallel combinations replaced with a single resistor; this can make it easier to visualize what's going on. Note that in your reduced circuit using the 8A as a source, terminal b is at the junction of the two 12Ω resistors; you will use this fact at the end of your calculations. The two 12Ω resistors on the left are in series, and then that combination is in parallel with the 16Ω resistor. That combination (24||16) plays the part of Z2 in the Wikipedia circuit. The 96 volt source is in series with 12Ω, 2Ω, and 8Ω, so the Z1 resistor in the Wikipedia circuit would be 22Ω. Now you have a voltage divider whose solution is the voltage across the two 12Ω resistors on the left. But the voltage seen at the a-b terminals is half that because the b terminal is connected to the junction between the two 12Ω resistors.