How do you solve #(x+7)/(x^2-3x-18)+(x-17)/(x^2-x-30)=(x+1)/(x^2+8x+15)# and check for extraneous solutions?

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Answer 1 · 2016-09-11T04:30:20

#x=2# is the Soln., while, #x=-5# is extraneous.

We first factorise all the #Drs.# of the Eqn. Hence,

#(x+7)/((x-6)(x+3))+(x-17)/((x-6)(x+5))=(x+1)/((x+5)(x+3))#.

Now, we multiply the eqn. by #(x-6)(x+3)(x+5)# & get

#(x+7)(x+5)+(x-17)(x+3)=(x+1)(x-6)#

#:. x^2+12x+35+x^2-14x-51=x^2-5x-6#

#:. 2x^2-2x-16-(x^2-5x-6)=0#

#:. x^2+3x-10=0#

#:. (x+5)(x-2)=0#

#:. x=-5, x=2#

Among these, #x=-5# is extraneous soln., since it makes the second and third terms of the eqn. meaningless. #:. xne-5#.

Further, #x=2# satisfies the eqn.

Hence, #x=2# is the Soln., while, #x=-5# is extraneous.

Enjoy Maths.!