which solution to the equation?

answer:

a = - 2 is extraneous

step-by-step explanation:

consider the denominators of the fractions

if these are zero then they will be undefined, thus

from a + 2 we have that a ≠ - 2

similarly

a² - 4 = (a - 2)(a + 2), thus a ≠ ± 2

solving the equation

+ =

multiply through by a(a - 2)(a + 2) to clear the fractions

3a(a - 2) + 2(a - 2)(a + 2) = a(4a - 4) ← distribute/ expand parenthesis

3a² - 6a + 2(a² - 4) = 4a² - 4a

3a² - 6a + 2a² - 8 = 4a² - 4a

5a² - 6a - 8 = 4a² - 4a ← subtract 4a² - 4a from both sides

a² - 2a - 8 = 0 ← in standard form

(a - 4)(a + 2) = 0 ← in factored form

equate each factor to zero and solve for a

a - 4 = 0 ⇒ a = 4

a + 2 = 0 ⇒ a = - 2

thus a = 4 is the solution and a = - 2 is extraneous