Answer:
The solution is a = 5
The extraneous solution is a = 1
Step-by-step explanation:
The solution to the equation is obtained as follows:
15/(a² - 1) = 5/(2*a - 2)
3/(a² - 1) = 1/(2*a - 2)
3*(2*a - 2) = (a² - 1)
6*a - 6 = a² - 1
0 = a² - 6*a + 5
which can be solved with the quadratic formula:
[tex] a = \frac{-b \pm \sqrt{b^2 - 4(a)(c)} }{2(a)} [/tex]
[tex] a = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(5)} }{2(1)} [/tex]
[tex] a = \frac{6 \pm 4}{2} [/tex]
[tex] a_1 = \frac{6 + 4}{2} [/tex]
[tex] a_1 = 5 [/tex]
[tex] a_2 = \frac{6 - 4}{2} [/tex]
[tex] a_2 = 1 [/tex]
The first one is the solution to the problem. The second one is extraneous because make the denominators of the original equation equal to zero, that is, replacing a = 1 in a² - 1 = 0 and in 2*a - 2 = 0.
