How do you solve #(5k + 5)/(3k^2) = (k - 4)/(3k^2) + 1/k^2# and find any extraneous solutions?

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Answer 1 · 2016-05-28T22:10:21

#k=-3/2#

Reducing to the same denominator

#(5k+5)/(3k^2)=(k-4)/(3k^2)+3/(3k^2)#

or

#(5k+5)/(3k^2)-(k-4)/(3k^2)-3/(3k^2)=0#

or

#((5k+5)-(k-4)-3)/(3k^2)=0#

or

#(4k+6)/(3k^2)=0#

Solving for #k# we obtain #k=-3/2#