Applying Kirchhoff's 1st law to the node at x;
-10 + Vx/4 + (Vx - Vy)/3 = 0
This reduces to ---> 7.Vx - 4.Vc = 120 --- Eq 1
Now apply K's 1st law to the node at y;
(Vy - Vx)/3 + Vy/5 + (Vy - 40)/2 = 0
This reduces to ---> 31.Vy - 10.Vx = 600 --- Eq 2
Eq 1 and Eq 2 form a simultaneous pair thus Vx = 34.576 Volts and Vy = 30.508 Volts.
Assuming conventional current flow through 3Ω resistor its value is (34.576 - 30.508)/3 = 1.356 Amps going eastbound.
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