Question

The solution to
15
a2 − 1
=
5
2a − 2
is

a =
.

The extraneous solution is a =
.

Answer 1

Solve 15/(a^2 − 1) = 5/(2a − 2).

Factor denominators: a^2 − 1 = (a − 1)(a + 1), 2a − 2 = 2(a − 1).

Cross-multiply: 15·2(a − 1) = 5(a^2 − 1) → 30(a − 1) = 5(a^2 − 1) → divide by 5 → 6(a − 1) = a^2 − 1.

Rearrange: a^2 − 6a + 5 = 0 → (a − 1)(a − 5) = 0 → a = 1 or a = 5.

Check domain: a ≠ 1 (denominators zero), so a = 1 is extraneous.

Thus: a = 5. The extraneous solution is a = 1.