How do you solve #1/(6k^2)=1/(3k^2)-1/k# and check for extraneous solutions?

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Answer 1 · 2017-07-31T19:05:34

#k = 1/6#

First, let's multiply everything by the least common multiple of the denominators (which is #6k^2#):

#(6k^2)/(6k^2) = (6k^2)/(3k^2) - (6k^2)/k#

#1 = 2 - 6k#

#6k = 1#

#k = 1/6#

The only domain restriction is #k != 0#, so there are no extraneous solutions:

#color(blue)(ul(k = 1/6#