Kirchhoff's law problem solving - alternative

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 = 2I₁ + 3I₃  ................ (2)

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

Substitute (1) to (2)
-4 = 2I₁ + 3I₃ = 2I₁ + 3(I₁ + I₂) = 2I₁ + 3I₁ + 3I₂
-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   _

---------------------

18 = -26I₂

I₂  = 18 : -26 = -0.69 A (I₂ = 0.69 A 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 = 1.08 A 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 = 0.39 A direction to the right)