How do you solve #3/(x+1)-1/(x+4)=(x+12)/(x^2+5x+4)# and check for extraneous solutions?

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Answer 1 · 2016-08-29T23:46:11

Solution is #x=1#

#3/(x+1)-1/(x+4)=(x+12)/(x^2+5x+4)#

#hArr(3(x+4)-1(x+1))/(x^2+5x+4)=(x+12)/(x^2+5x+4)# or

#(3x+12-x-1)/(x^2+5x+4)=(x+12)/(x^2+5x+4)# or

#(2x+11)/(x^2+5x+4)=(x+12)/(x^2+5x+4)#

Now assuming #x^2+5x+4!=0# (i.e.#x# is neither equal to #-1# nor equal to #-4#), and multiplying each side by #x^2+5x+4#, we get

#2x+11=x+12# or

#x=1#

As #x# is neither equal to #-1# nor equal to #-4#, there is no extraneous solution.