Resistance in a circuit

[kelsow]

1. The problem statement, all variables and given/known data
http://i33.tinypic.com/4tu9f5.jpg
I have to find the equivalent resistance of that series and parallel circuit.

2. Relevant equations
for resistors in parallel: 1 / R = 1 / R1 + 1 / R2 + 1 / R3 +...
for resistors in series: R = R1 + R2 + R3 + ...

3. The attempt at a solution
The good answer has been given, its 8,5 Ω
I tried but i cant have that answer.
What I made:
15Ω + 9 Ω=24
and 19Ω+5Ω=24
1/ (1/24+1/8) + 1/ (1/24) + 2 +0,2
I know, it is wrong but i cant do the right thing.

[Doc Al]

What I made:
15Ω + 9 Ω=24
Those resistors are not simply in series, so this step is incorrect.
and 19Ω+5Ω=24
This is good.

Start by finding instances where resistors are simply in series or parallel (the 19Ω and 5Ω resistors are a good place to begin) and replace them by their equivalent resistances. Then reanalyze the circuit, looking for more opportunities to combine resistances. Keep doing this until you've found the equivalent resistance of the entire pile.

[kelsow]

okay. thanks. (I made a mistake in the drawing of the circuit, its 2 instead of 20).
So i should start with 19Ω and 5Ω. I add up those 2 resistors 19Ω+5Ω=24
then i place 24Ω in parallel with 8Ω.
1/ (1/24+1/8)=6
i add 6 with 15 cause its in series 6+15=21 and i do the same thing
1/(1/21+1/9)=6,3
6,3+2+0,2=8,5Ω
YES, thank you. i didnt know where to start and i mixed up everything.