Colin's solution is correct in principle but I am not so sure about his math..

Making the same assumption that Ig =17mA and R1 = 10kΩ and I agree that there is no need to consider the 2k and 4k resistors, since no current flows through them, and thus they drop no voltage.

Lets see

Vg = Ig * Rtot = Ig * (20||15) = 17*10^-3 * [(20*15*10^6)/(15+20)*10^3] = 5100/35 V

VR1 = IR1 *R1 = Vg/(R1+10) * 10 = Vg/2 V

VR12 =IR12 *R12 = Vg/(12+3) * 12 = 12Vg/15 V

Vo = VR1 - VR12 = Vg (1/2 - 12/15) = -0.3Vg = 43.71 V

You need to specify units...Is R1 10 ohms?...or 10 Kilohms? Is Ig 17 milliamps?...or perhaps 17 Coulombs per hour?

A number without units is pure gibberish. If you do that in an exam paper the examiner will mark you down.

Resistance across Ig = Reff = (10k + 10k) in parallel with (3k + 12k)

(I'm assuming you mean "R1 = 10k")

1/Reff = 1/20 + 1/15

= (1/300)*(15 + 20) = 45/300 = 3/20

Reff = 20/3 k

V across Ig = Reff*I = 17*20/3 v

(I'll assume Ig is in mA)

Volts across R1 = (17*20/3)*(10/(10 + 10)) = 17*10/3

Volts across 12k = (17*20/3)*(12/(12 + 3))

= 17*20*(4/5)*(1/3) = 17*4*4/3

Thus Vo = (17/3)*(10 - 16)

= (17/3)*(-6)

= -34 <<<

Do check my work!

(There is no need to consider the 2k and 4k resistors, since no current flows through them, and thus they drop no voltage.)

Branch currents are inversely proportional to branch resistance and V = IR, then;

Vo = [(V of R1) - (V of 12k Ohm)] = [(15/35)*(.017A)*(10, 000 Ohm)] - [(20/35)*(.017A)*(12,000 Ohm) = (-43.71) Volts

R1 = 10 and ig =17.

I'm only allowed to use circuit and current division with equivalent resistances. No thevenin or norton circuits or nodal or mesh analysis.