Kirchhoff's law problem solving

Question from Physics for the IB Diploma page 228 number 53.

Determine the magnitudes and directions of the currents in the circuit  shown below

Answer:

Kirchhoff's first law: "Total current in to a node/junction = total current leave the node/junction."
I₃ = I₁ + I₂     ................. (1)

Kirchhoff's second law: "Total emf = total potential differences."
Loop kiri
4 = 3I₃ + 2I₁  ................ (2)

Loop kanan
6 = 4I₂ + 3I₃   ............... (3)

Substitute (1) to (2)
4 = 3I₃ + 2I₁ = 3(I₁ + I₂) + 2I₁ = 3I₁ + 3I₂ + 2I₁
4 = 5I₁ + 3I₂   ............... (4)

Substitute (1) to (3)
6 = 4I₂ + 3I₃ = 4I₂ + 3(I₁ + I₂) = 4I₂ + 3I₁ +3I₂
6 = 3I₁ + 7I₂   ............... (5)

Eliminate (4) with (5)
4 = 5I₁ + 3I₂  (x3)
6 = 3I₁ + 7I₂  (x5)
----------------------

12 = 15I₁ + 9I₂
30 = 15I₁ + 35I₂    _

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-18 = -26I₂

I₂  = 18 : 26 = 0.69 A (I₂ direction to the left)

Substitute I₂ value to (3)

6 = 4(0.69) + 3I₃

6 = 2.76 + 3I₃

6 - 2.76 = 3I₃

3.24 = 3I₃

I₃ = 3.24 : 3 = 1.08 A (I₃ direction upwards)

Substitute I₂ and I₃ values to (1)

I₃ = I₁ + I₂

1.08 = I₁ + 0.69

I₁ = 1.08 - 0.69 = 0.39 A (I₁ direction to the right)