figred out

 

I would take a different approach - provided you have discussed op amp theory in class.

Top LH amplifier is a x2 gain inverting stage
Bottom LH amplifier is a x3 gain non-inverting amplifer
The RH amplifier is a summing / inverting stage with gain of x2.

Anyway whatever your method, I have Vo=-2V

 

Apologies - Vo=+2V

 

I don't get 2..

 

Current heading up from 20,000 resistor in (bottom) Op-amp = 1/20,000
Vout2 for second (bottom) Op-amp = -(1/20,000)*40,000

This is where you went wrong.

Vout2 = 1V+(1/20k)*40k

 

nope got it

 

got it cool

 

isn't it 1 minus the drop across the resistor from the logic shown here:

http://www.youtube.com/watch?v=2SwT...es/detail/embed20.htm&feature=player_embedded

@ 22:30

Vout2 = 1 - (1/20k)*40k

For the amp (lower left hand) in question....

The voltage at the negative terminal is 1V which we are agreed on.

The current in the 20k from neg. terminal to ground is 50uA towards ground. Given this is an ideal op-amp, the same current must flow in the 40k feedback path - from the output terminal to the negative terminal. So there is a voltage drop of 2V across this resistor with the polarity being additive to the voltage at the neg. terminal. So for this amp, Vo = 1V + 2V = 3V.