Find the currents in the circuit branches

[endeavor]

http://img335.imageshack.us/img335/8383/physics4a8ua.th.png http://img335.imageshack.us/img335/8000/physics4b6vo.th.png
"Find the currents in the circuit branches in [the first figure on the left above]."

V₁ = 20V, V₂ = 10V, R₁ = 5.0 ohms, R₂ = 4.0 ohms, R₃ = 6.0 ohms, R₄=R₅=R₆= 2.0 ohms

R_p^-1 = 3/2 (from R₄,R₅,R₆)
R_p = 2/3 ohms

I assigned loops in the second figure above. There are 6 different currents.
Using Kirchhoff's first rule:
(C1) I₁ = I₂ + I₃
(C2) I₃ = I₄ + I₅
(C3) I₁ = 3I₆
Then using the Kirchhoff's second rule, I get a bunch of equations (L1 = LOOP 1):
(L1) -V₁ +I₂R₁ + I₁R_p = 0
(L2) +V₂ +I₅R₂ - I₂R₁ = 0
(L3) +I₄R₃ - I₅R₂ = 0
(L7) -V₁ + V₂ +I₅R₂ +I₁R_p = 0
(L8) +V₂ - I₂R₁ +I₄R₃ = 0
(L9) -V₁ +I₁R_p +I₄R₃ + V₂ = 0
Then substituting values for the V's and R's:
(L1) 5*I₂ + (2/3)*I₁ = 20
(L2) 4*I₅ - 5*I₂ = -10
(L3) 6*I₄ - 4*I₅ = 0
(L7) 4*I₅ + (2/3)*I₁ = 10
(L8) -5*I₂ + 6*I₄ = -10
(L9) (2/3)*I₁ + 6*I₄ = 10
I then found that I₃ = (5/2)*I₄. Then using this equation, and equations C1 and L9, I finally got:
I₁ = - 90/462 A = -0.19 A

I think the fact that my answer is negative only shows that I chose the wrong loop direction. But the absolute value of my answer is still wrong... help

 

[lightgrav]

You should simplify a bit more before assigning loops ... R2 || R3.
Then you have a loop on the left and a loop on the right.

- I_L R_P - (I_L - I_R) R_1 - V_1 = 0 ... because I_1(upward) = I_L - I_R
- I_R R_{2||3} + V_2 - (I_R - I_L) R_1 = 0 .

 

[endeavor]

thanks! Simplifying really worked. After I simplified the parallel parts, I only had 3 loops, which was a lot easier to work with.