For Series: R₁ + R₂ = 9 or, R₁ = 9 – R₂ — (i)

For Parallel: 1\over R_\text{parallel} = 1\over R_1 + 1\over R_2

or, 1\over 2 = R_1×R_2\over R_1 + R_2

or, 1\over 2 = (9 - R_2)×R_2\over 9

or, 9R₂ – R₂² = 18

or, R₂² – 9R₂ – 18 = 0

or, R₂² – (6 + 3)R₂ – 18 = 0

or, R₂(R₂ – 6) – 3(R₂ – 6) = 0

or, (R₂ – 6)(R₂ – 3) = 0

or, R₂ = 6Ω or 3 Ω

Put R₂ in eq (i) we get: R₁ = 6Ω or 3 Ω

The resistances of the two conductors are 3Ω and 6Ω.

OR

The 2 Ω and 3 Ω resistors are connected in parallel.

1\over \text{R}_{\text{Parallel}}= 1\over \text{R}_1 + 1\over \text{R}_2

1\over \text{R}_{\text{Parallel}}= 1\over 2 + 1\over 3

1\over \text{R}_{\text{Parallel}}= 3 + 2\over 6 = 5\over 6

or, R_parallel = 6\over 5 = 1.2 Ω

∴ R_total = R_parallel + R₃ = 1 Ω + 1.2 Ω = 2.2 Ω