Node 1:

I1=I2+I3 And,

I1=5, I2= V1/10, (V1-V2)/3 On substituting we get,

5 = V1/10 + (V2+V3)/3

150 = 13V1- 10V2 $\hspace{2cm}$ ……. (1)

Node 2:

I3 + I5 =I4

And,

I3 = (V1-V2)/3, I5 = (10¬-V2)/1, I4=V2/5

On substituting we get,

(V1-V2)/3 + (10¬-V2)/1 = V2/5

5V1-23V2=-150 $\hspace{2cm}$ …… (2)

On solving eq 1 and 2 simultaneously we get,

V2= 10.84Volts.

V1= 19.88Volts.

Since,

I3 = (V1-V2)/3

I3 = (19.88-10.84)/3

I3 = 3.013A ($\rightarrow$)