How do you solve #1/(6b^2)+1/(6b)=1/(b^2# and check for extraneous solutions?

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Answer 1 · 2016-11-27T19:48:37

#b=5#

Extraneous solution is #b=0#

Write as:# " "1/(6b^2)+b/(6b^2)=1/b^2#

#(1+b)/(6b^2)=1/b^2#

#b^2(1+b)=6b^2#

#1+b=6#

#b=5#
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We are 'not allowed' to have 0 as a denominator. The proper name for this is that the expression is 'undefined'

So the extraneous solution is #b=0#

#1/(6b^2)+1/(6b)=1/b^2#

For #b=0# we have

#1/0+1/0=1/0# which is undefined