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

Answer:

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.

Step-by-step explanation:

$\frac{7}{b+3} + \frac{5}{b-3} = \frac{10b}{b^2-9}$

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

$\frac{7*(b-3)}{b^2-9} + \frac{5*(b+3)}{b^2 -9} = \frac{10b}{b^2-9}$

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.