Thevenin equivalent (verification of answer)
Redrawing originial circuit for clarity:







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Norton to Thevenin of I and R3 lead to:







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Thevenin voltage is equal to open circuit voltage. Since there is no current through R1, its voltage drop is zero, leading to the following:







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Circuit equations:

$$
\frac{V_{th} - (V_x + V_1 - R_3I)}{R_3} + \frac{V_{th} - V_x}{R_4} = 0
$$
$$
\frac{(V_x + V_1 - R_3I) - V_{th}}{R_3} + \frac{V_x - V_2}{R_5} + \frac{V_x - V_{th}}{R_4} + \frac{V_x}{R_2} = 0
$$

Solution is:

$$
V_{th} = \frac{1}{(R_2+R_5)(R_3+R_4)}\times\\\{V_1(R_2R_4 + R_4R_5) + V_2(R_2R_3 + R_2R_4) - I(R_2R_3R_4 + R_3R_4R_5)\}
$$