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

b = 5

b = -5

b = -3

b = 3

User MattjeS
by
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2 Answers

6 votes

Answer:B= 3

Explanation:

User Azamat Mahkamov
by
7.9k points
4 votes

(b)/(b-3) - (5)/(b) = (3)/(b-3) \\ ( b^(2) -5(b-3))/(b(b-3)) =(b)/(b-3) \\ ( b^(2) -5b+15)/(b^2-3b) =(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.
User Niklas R
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