Solve I(x) in the current using supermesh concept?

[asdf12312]

Homework Statement

Homework Equations

4A = I3 - I2 (supplementary eq.)

Mesh 1:
6(I1)-6(I2)=2A

Mesh 2,3 (supermesh):
6(I2-I1)+6(I2-I4)+12(I3)=0
-6(I1)+12(I2)+12(I3)-6(I4)=0
substitute supp. eq.:
-6(I1)+12(I2)+12(I2+4A)-6(I4)=0
-6(I1)+24(I2)-6(I4)=-48

Mesh 4:
6(I4-I2)=-3A
-6(I2)+6(I4)=-3A

The Attempt at a Solution

|6 -6 0| |I1| = | 2 |
|-6 24 -6| |I2| = |-48|
|0 -6 6| |I4| = | -3|

used cramer's rule (not sure if there was an easier way) and used online matrix calculator http://ncalculators.com/matrix/matrix-determinant-calculator.htm cause i was lazy :(

I2 was all i needed to calculate:
I2 =-1764/432= -4.08A

supp. eq.:
4A+I2=I3
I3= -0.08A

since I(x) is opposite to I3, i got I(x)=0.08A as my answer.

 

[The Electrician]

Homework Statement

Homework Equations

4A = I3 - I2 (supplementary eq.)

Mesh 1:
6Ω(I1_Amps)-6Ω(I2_Amps)=2Amps

Mesh 2,3 (supermesh):
6(I2-I1)+6(I2-I4)+12(I3)=0
-6(I1)+12(I2)+12(I3)-6(I4)=0
substitute supp. eq.:
-6(I1)+12(I2)+12(I2+4A)-6(I4)=0
-6(I1)+24(I2)-6(I4)=-48

Mesh 4:
6Ω(I4_Amps-I2_Amps)=-3Amps
-6(I2)+6(I4)=-3A

The Attempt at a Solution

|6 -6 0| |I1| = | 2 |
|-6 24 -6| |I2| = |-48|
|0 -6 6| |I4| = | -3|

used cramer's rule (not sure if there was an easier way) and used online matrix calculator http://ncalculators.com/matrix/matrix-determinant-calculator.htm cause i was lazy :(

I2 was all i needed to calculate:
I2 =-1764/432= -4.08A

supp. eq.:
4A+I2=I3
I3= -0.08A

since I(x) is opposite to I3, i got I(x)=0.08A as my answer.

Right away I see a problem in the two equations I've marked in red. I inserted some units and I don't think it's working out very well, is it?

The equation for mesh 1 should be I1_Amps = 2_Amps and similarly for mesh 4.

 

[asdf12312]

i see my mistake now. i was trying to treat the current source as a voltage source but that's defnitely not right. so the equations are:

I1=2A
I4=-3A

so now this makes everything much simpler. all i need now is the supp. eq. and the eq. for the supermesh:

4A = I3 - I2

-6(I1)+12(I2)+12(I3)-6(I4)=0

-6(2A)+12(I2)+12(I3)-6(-3A)=0
12(I2)+12(I3)=-6V
12(I3-4A)+12(I3)=-6V
24(I3)=42
I3=1.75A
I(x)=-1.75A