How do you solve #(3x-3)/(x^2-13x+40)-(x-7)/(x^2-19x+70)=(x-2)/(x^2-22x+112)# and check for extraneous solutions?

1 Answer
Aug 25, 2016

Answer:

The eqn. has solns. #x=24, x=-1#, and no extraneous soln.

Explanation:

Factorising all the Drs.

#(3(x-1))/((x-8)(x-5))-(x-7)/((x-14)(x-5))=(x-2)/((x-14)(x-8))#.

Multiplying both sides by #(x-5)(x-8)(x-14)#, we get,

#3(x-1)(x-14)-(x-7)(x-8)=(x-2)(x-5)#.

#:. 3x^2-45x+42-x^2+15x-56=x^2-7x+10#.

#:. x^2-23x-24=0#.

#:. ul(x^2-24x)+ul(x-24)=0#.

#:. x(x-24)+1(x-24)=0#.

#:. (x-24)(x+1)=0#.

#:. x=24, x=-1#

These roots satisfy the given eqn.

Hence, the eqn. has solns. #x=24, x=-1#, and no extraneous soln.