Circuit analysis - Potential differences of resistors

[Baou]

Homework Statement

Spoiler

R_1 = 70 \Omega, R_2 = 144 \Omega, R_3 = 237 \Omega, R_4 = 117 \Omega, R_5 = 246 \Omega; V_{ab}=43 V
I am looking for the voltage of each resistor as well as the current passing through each one.

Homework Equations

Kirchoff's Voltage/Current Laws: The sum of any potential differences across a closed path is zero; current (charge) is conserved, so any current exiting a node is equal to the sum of the currents entering it.
V = IR
I = \frac VR

The Attempt at a Solution

I know R_{eq}=339\Omega. Thus, I = \frac VR = \frac{43}{339} = .127 A.

V_5 = I*R_5=.127*246=31.2 V

V_1+V_2+V_3-V_4=0. This is Kirchoff's voltage equation.

I_1=I_2=I_3 b/c current is conserved, so \frac{V_1}{70}=\frac{V_2}{144}=\frac{V_3}{237}. I solve for V_1 and V_2 in terms of V_3: V_1=\frac{70V_3}{237} and V_2=\frac{144V_3}{237}.

I know the current through V_3 and V_4 should sum to .127, so: \frac{V_3}{237}+\frac{V_4}{117}=.127.
V_4=117\left(.127-\frac{V_3}{237}\right).

V_4=V_1+V_2+V_3 from above, so plugging in everything I get:
117\left(.127-\frac{V_3}{237}\right)=\frac{70V_3}{237}+\frac{144V_3}{237}+V_3.

I solve for V_3 and then plug that into everything else.
My final answers are: V_1 = 1.83, V_2=3.77, V_3=6.2, V_4=12.8, V_5=31.2. However, these answers are incorrect, so I was hoping to find out what I'm doing wrong.

 

[rl.bhat]

Hi Baou, welcome to PF.
Voltage across R4 = V4 = 43 - V5 = 43 - 31.2 = 11.8 V

 

[Baou]

Could you explain how you arrived at that? I don't quite follow...

Edit: Oh, you mean it should be 11.8 instead of 12.8? Was everything else that I did correct?

Edit2: Okay, I was just reading the answer on the calculator wrong. Thanks for the help!