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

Divide both sides by 5 to get 3(2a − 2) = a² − 1. Expanding and rearranging gives a² − 6a + 5 = 0, which factors as (a − 1)(a − 5) = 0. Thus a = 1 or a = 5, matching the choice “a = 5, a = 1.”

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