Electric circuit. Slight problem with the sign on voltages

Question

Translation first!
The figure shows an electric circuit.
a) Use Kirchhoffs 2.law on the part-circuit abed and show that $I_3$ = 0,50A.
b) Explain that $I_2 + I_5 = 2,0A$. Use Kirchhoffs 2.law on the part-curcuit befc to find another equation with $I_2$ and $I_5$ as unknows. Solve the set of equations.
c) Find $I_1$ and $I_4$.
d) Find the resitance of R.

a) Assume clockwise rotation of circuit. That makes 8v positive and 5,5v negative. 8-5,5=2,5v. This would be the V over the resitance at $I_3$. $\frac{2,5v}{5 ohm} = 0,5A$ I think this is correct.

b) My problem is that I can't seem to find the equation properly. If I assume a clockwise current, that makes V at $I_3$ negative. And V at $I_2$ positive and V at $I_5$ negative.
$-2,5 = 5I_2 - 2,5I_5$
Solution in the book is $-2.5 = - 5I_2 + 2,5I_5 $ with $I_2 + I_5 =2$ It gives $I_2=1$ and $I_5 = 1$
My logic is wrong, but wrong how?

Answer 1

The book defined the 2.law as follows.
The sum of all emf around a closed circuit is equal to the voltage over all the resistors.

I just think it fails to press that if you do just sum up all the voltages it should indeed be 0. Which is a nice thing if you don't really have an emf in a part-circuit you are analyzing. And it should be mentioned that an emf value is positive if you pass trough it from negative to positive. And negative if you go the other way around.

It also helps writing the part equations in the right order so you don't plot the wrong variables into the calculator. Something I didn't do for starts. Bad me.