Question

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

Answer 1

Answer: The solution is a=5

Step-by-step explanation: For this case the correct expression is:

15 / (a ^ 2 - 1) = 5 / (2a - 2)

Rewriting we have:

3 / (a ^ 2 - 1) = 1/2 (a - 1)

6 / (a ^ 2 - 1) = 1 / (a - 1)

6 (a - 1) = (a ^ 2 - 1)

6 (a - 1) = (a-1) (a + 1)

(a + 1) = 6

a = 6-1

a=5

Answer 2

The solution is a = 5

The extraneous solution is a = 1

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:

`a = (-b \pm √(b^2 - 4(a)(c)) )/(2(a))`

`a = (6 \pm √((-6)^2 - 4(1)(5)) )/(2(1))`

`a = (6 \pm 4)/(2)`

`a_1 = (6 + 4)/(2)`

`a_1 = 5`

`a_2 = (6 - 4)/(2)`

`a_2 = 1`

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.