First, the input voltages to the op-amps positive terminals are not -1 and -2 volts, they're +1 and +2 volts.
In general, the process of working these is pretty straightforward. Start with what you know, solve for what you don't. You know that an op-amp with negative feedback (a path from the output to the negative input) tries to make its inputs equal. (In the real world there would be concerns about the voltage rails of the op amp, but this is obviously a textbook problem so we won't worry about such.) The positive input voltages are given to you, fixed by the voltage sources. From this fact, we also know the negative input voltages.
So that means we know the voltage on both sides of the 10k resistor: 1V and 0V. There's 1V across this resistor, and it's 10k, so the current through it is 100 uA, flowing left. We have current leaving a circuit node. It has to also enter that node somewhere, and there are only two possible paths: through the 20k resistor, or through the input of the op-amp.
Another assumption we make about op-amps is that current can't flow into or out of their input terminals. In the real world, that's not true, but for simple analysis of DC circuits with resistances in this range it's probably close enough.
So we know there's 100 uA going through the 20k as well, flowing left. We know the left side of the 20k is at 1V, and it's got 100 uA x 20k = 2V drop across it, so we get a voltage of 3V on the right side of the 20k.
Continue the process. The left side of the 30k resistor has 3V on it, and the right side has 2V. That means there's one volt across it, giving 33 uA through it, flowing right.
That 33 uA can't flow into the input of the op amp, so it has to go through the 40k. 33 uA through a 40k resistor gives a voltage drop of 1.33 volts. The left side of the 40k resistor is at 2V, and the right side is 1.33V lower, giving .66 volts.