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:
[tex]\frac{7}{b+3} + \frac{5}{b-3} = \frac{10b}{b^2-9}[/tex]
The minus common multiply must be (b-3)*(b+3) = b² - 9
[tex]\frac{7*(b-3)}{b^2-9} + \frac{5*(b+3)}{b^2 -9} = \frac{10b}{b^2-9}[/tex]
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.
