V= IR
in this case V = I(R1+R2)
so we have 15 = IR1 + IR2
(15 - IR1)/I = R2 we will use this later in our subsitution.
We also know that...
15 =(I+.05)R1 and 15 = (I+.025)R2
since both equal 15, they must equal each other:
(I+.05)R1 = (I+.025)R2
Use the substitution for R2 from above and you will be able to solve for R1. by eliminating the I.
Two resistances, R1 and R2, are connected in series across a 15-V battery. The current increases by 0.050 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.025 A when R1 is removed, leaving R2 connected across the battery.
Find R1 and R2
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