a) Use the node voltage method to determine the six branch currents shown.

Node voltage variables selected are shown in brown.
Voltages added to assist in the solution are shown in blue.

I will sum the currents out of the nodes.

Since we have voltage sources in series with resistors connected to node A, we need to determine the voltage across each of the resistors in order to find the currents.

For V_x

V_B - V_A = - V_x + 110
V_x = V_A - V_B + 110

For V_y

V_A = 110 - V_y
V_y = 110 - V_A

KCL now gives

Multiply through by 6

2 V_A - 2 V_C + 3 V_A - 3 V_B + 330 - 330 + 3 V_A = 0
8 V_A - 3 V_B - 2 V_C = 0 (Equation 1)

Using the expression for V_y derived above, KCL at node B gives

Multiply through by 16

2 V_B - 2 V_C - 8 V_A + 8 V_B - 880 + V_B = 0
- 8 V_A + 11 V_B - 2 V_C = 880 (Equation 2)

KCL at node C gives

Multiply through by 24

8 V_C - 8 V_A + 3 V_C - 3 V_B + V_C = 0
- 8 V_A - 3 V_B + 12 V_C = 0 (Equation 3)

V_A = 82.5
V_B = 157.143
V_C = 94.286

i₁ = V_x / 2 = (V_A - V_B + 110) / 2 = 35.357 / 2
i₁ = 17.679 A

i₂ = (V_C - V_A) / 3= 11.786 / 3
i₂ = 3.929 A

i₃ = V_y / 2 = (110 - V_A) / 2 = 27.5 / 2
i₃ = 13.75 A

i₄ = (V_B - V_C) / 8 = 62.857 / 8
i₄ = 7.857 A

i₅ = V_C / 24= 94.286 / 24
i₅ = 3.929 A

i₆ = V_B / 16 = 157.143 / 16
i₆ = 9.827 A