Physics HW pleaaaase help mmeeeeeeeeeeeeee?

Question

Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.26 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.099 A when R1 is removed, leaving R2 connected across the battery. Find (a) R1 and (b) R2.

Answer 1

I = E/R

12/(R1 + R2) = (12/R1) - 0.26

12/(R1 + R2) = (12/R2) - 0.099

two equations in 2 unknowns, should be solvable.

Multiply by 1000 or 100 to drop fractions

12000/(R1 + R2) = (12000/R1) - 260

12000/(R1 + R2) = (12000/R2) - 99

(12000/R1) - 260 = (12000/R2) - 99

(12000/R1) - 161 = (12000/R2)

mult by R1R2

12000R2 - 161R1R2 = 12000R1

R2(12000 - 161R1) = 12000R1

R2 = 12000R1/(12000 - 161R1)

getting too complicated, need a different technique, perhaps.

back later.