Thévenin's theorem exercise, resistances
I think I got this : (the voltage source has a value \$E\$)
\$R_2\$ will vanish because apparently that's what happens when you have a resistor in parallel with a voltage source hence we can transform the voltage source which is is now in series with a resistor into a current source \$I_o=\frac{E}{R_1}\$, right after that we find the equivalent resistance \$R=R_1//R_3\$ and now we can transform the whole thing into another voltage source \$E'=\frac{E.(R_1//R_3)}{R_1}\$ in series with \$R\$.
\$E'=E_{th}\$ and \$R=R_{th}\$