Voltage drop, direction of current, extra voltage source
My solution to the problem is:
48mA (409/8500) going through the 6.75V and its 45ohm resistor.
-8.4mA (-71/8500) going through the 4V and the 70ohm resistor.
-40mA (169/4250) going through the 2V, the 20ohm and the 45ohm resistors.
The 5V and the 37ohm can be ignored as their ends are connected, hence no overall P.D across them.
These all add up to give a voltage (taking the top right corner to be 0V) of 4.58V in the "H" shaped bit in the bottom left corner.

As Filkor and I have both got -40mA for the current, it's probably safe to assume the textbook got this part right.
Where it got -3.18V from is a mystery as the correct is -0.75V (V=IR=-0.040*45), which will always be the voltage across any 45ohm resistor when -40mA is flowing through it, regardless of how ridiculous the rest of the circuit is.

My conclusion therefore: I say you should stuff the textbook