The solutions to 15(2a – 2) = 5(a2 – 1) are:

Expand both sides: 15(2a − 2) = 30a − 30 and 5(a² − 1) = 5a² − 5. Rearranging and dividing by 5 gives a² − 6a + 5 = 0, which factors as (a − 1)(a − 5) = 0. Thus a = 1 or a = 5, matching choice C: a = 5, a = 1.

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