Can source transformation be used to internal circuit (source) while finding thevenin equivalent of the source

At first, I'm very sorry because I don't have money for circuit lab so there is not much visual circuit that clarify my steps

I have practiced source-transformation as well as Thevenin equivalent separately but this is the first time I combine them in one circuit. Does doing it like this still cause no difference as doing it separately? For example, this circuit I am trying to find the Thevenin equivalent

I don't know how to calculate the short voltage between a-b terminals with this design so I intend to simplify it by using source transformation and circuit transformation.

To be more detailed, I will transform the \$Y\$ circuit \$8\Omega-12\Omega-5.2\Omega\$ into \$\Delta\$ circuit then the \$30\Omega\$ Resistor will finally parallel with the \$\frac{12*8+5.2*12+8*5.2}{12}=\frac{50}{3}\Omega\$ Resistor which make the

$$R_{eq}=\frac{30+\frac{50}{3}}{30+\frac{50}{3}}=\frac{75}{7}\Omega$$

Then I will transfer the \$10A\$ current source which is parallel with \$\frac{75}{7}\Omega\$ resistor into equivalent voltage source whose value is \$10*\frac{75}{7}=\frac{750}{7}V\$

Since \$500V\$ voltage source is parallel with \$\frac{12*8+5.2*12+8*5.2}{5.2}=\frac{500}{13}\Omega\$ resistor. I will ignore this resistor because it will not change the load's behavior. Now two voltage source is in series as well as two resistors. Just using voltage divider to calculate the Thevenin Voltage and deactive all source to calculate the Thevenin Resistance

So the circuit is finally simplified that Thevenin Voltage and Thevenin Resistance can be calculate. But what I am not sure is if this way of simplification ok. Will there be any change in the load comparing to the origin circuit.