15
a2 − 1
=
5
2a − 2
Which equation results from cross-multiplying?

15(a2 – 1) = 5(2a – 2)

15(2a – 2) = 5(a2 – 1)

15(a2 – 1)(2a – 2) = 5(2a – 2)(a2 – 1)

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

a = 1, a = –1

a = 5, a = –1

a = 5, a = 1

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