Node Voltage Analysis

I mainly just wanted to know if my equations were right, I think that I've done it correctly, but at the same time I'm not sure, haha. Thank you in advance!

1. The problem statement, all variables and given/known data
Solve for V1, V2, V3 and V4 (in decimals) using node voltage analysis method for the following:
http://i52.tinypic.com/108csn4.png


2. Relevant equations:
Node Voltage Analysis

3. The attempt at a solution:
For Node #1:

V1 = 6 V


For Node #4:

V4 = 16 V


For Node #2:

(V2) − (V3)3300 + (V2)1000 + (V2) − 61500 = 0

0.00197(V2) - 0.000303(V3)= 0.004 --> equation 1


For Node #3:

(V3) − (V2)3300 + (V3)4700 + V3 −162200 = 0

-0.000303(V2)+0.00097(V3) = 0.007273

V3= 0.3124(V2) + 7.4979 --> equation 2


Solve for V2 by substituting V3 into equation 1:

0.00197(V2) - 0.000303(0.3124(V2) + 7.4979) = 0.004

0.001875(V2) - 0.006272 = 0

V2 = 3.3451 V


Solve for V3 by substituting V2 into equation 2:

V3= 0.3124(3.3451) + 7.4979

V3 = 8.5428 V

V1 = 6 V it's ok

V4 = 16 V it's ok

BUT " V2−V33300 + V21000 + V2−61500 = 0 " where did u get that !!

ok let's do it again step by step :

at Node #2 it should be :

(V1 - V2)/1500 = V2/1000 + (V2 - V3)/3300 .... get why ?

=> (6 - V2)/1500 = V2/1000 + (V2 - V3)/3300

=> 0.004 - V2/1500 = 0.001 V2 + V2/3300 - V3/3300

simplifying :

=> 0.002 V2 - 0.0003 V3 = 0.004 ... (1) [Note : some numbers are approximated]


For Node #3 :

=> (V2 - V3) /3300 = (V3 - 16) /2200 + V3 /4700

simplify :

=> 0.0003 V2 - 0.0003 V3 = 0.00045 V3 - 0.00727 + 0.000213 V3

=> 0.00036 V3 - 0.0003V2 = 0.00727 .... (2)

.....0.002 V2 - 0.0003 V3 = 0.004 ... (1)

Solving : from (1) u get : V2 = (0.004 + 0.0003 V3)/0.002 = 2 + 0.15 V3

sub this in (2) => 0.00036 V3 - 0.0003 (2 + 0.15 V3) = 0.00727

=> 0.000315 V3 = 0.00787

=> V3 = 25 V (approx) <<<

and so V2 = 2 + 0.15 V3 = 5.75 V (approx) <<