Question

Find all solutions to the following equation. If there are any extraneous solutions, identify them and explain why they are extraneous.(7/b+3)+(5/b-3)=(10b/b²-9)

Answer 1

b = 3, which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0.

Explanation:

`(7)/(b+3) + (5)/(b-3) = (10b)/(b^2-9)`

The minus common multiply must be (b-3)*(b+3) = b² - 9

`(7*(b-3))/(b^2-9) + (5*(b+3))/(b^2 -9)  = (10b)/(b^2-9)`

7b - 21 + 5b + 15 = 10b

2b = 6

b = 3

Which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0. So this solution is invalid.