Question

What is an extraneous solution to the equation b/b-3 -5/b=3/b-3 ?

Answer 1

Answer:B= 3

Step-by-step explanation:

Answer 2

$\frac{b}{b-3} - \frac{5}{b} = \frac{3}{b-3} \\ \frac{ b^{2} -5(b-3)}{b(b-3)} =\frac{b}{b-3} \\ \frac{ b^{2} -5b+15}{b^2-3b} =\frac{b}{b-3} \\ (b-3)(b^{2} -5b+15)=b(b^2-3b) \\ b^3-8b^2+30b-45=b^3-3b^2 \\ 5b^2-30b+45=0 \\ b^2-6b+9=0 \\ (b -3)(b-3)=0 \\ b=3$

To check: 3/3-3 - 5/3 = 3/0 - 5/3 which is not defined.

Therefore, b = 3 is an extraneous solution.