Series Circuit Resistance: Finding R1 and R2 | Step-by-Step Guide

[Sino87]

Homework Statement

Two resistances, R1 and R2 are connected in a series circuit across a 12V battery. The current increases by 0.20A when R2 is removed leaving R1 connected across the battery. However, the current increases by just 0.10A when R1 is removed, leaving R2 connected across the battery. Find (a) R1 and (b) R2.

Homework Equations

I = 12/(R1+R2)
I + 0.20A = 12 / R1
I + 0.10A = 12 / R2

The Attempt at a Solution

I = 12 / (12/I + 0.20A) + 12/ (12/I + 0.10A)
I = I + 0.20A + I + 0.10A
I = 2I + 0.30A
-I = 0.30A
I = -0.30A.
Then, I plugged them back into equation 2, 3 but got the wrong answers.

The correct answers are 35 OHM and 50 OHM. PLease help!?

 

[silvashadow]

V=IR
Both: 12=(R1+R2)I;
R2 removed: 12=R1(I+.2)
R1 removed: 12=R2(I+.1)

Solve for R's
R1=12/(I+.2)
R2=12/(I+.1)

Plug into original equation
12=(12/(I+.2)+12/(I+.1))I

Solve for I
I=.1414

Plug I back into R1 and R2
R1=12/(.1414+.2)=35.15ohm
R2=12/(.1414+.1)=49.7ohm