How do you solve # [18/(x^2-3x)]-[6/(x-3)]=5/x# and find any extraneous solutions?

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Answer 1 · 2016-08-01T04:54:41

Therefore, the eqn. has No Soln. , but has #x=0, or, x=3# Extraneous Solns.

We rewriting the given eqn. as, #18/(x^2-3x)=6/(x-3)+5/x#

#rArr 18/(x^2-3x)={6x+5(x-3)}/(x(x-3))=(11x-15)/(x^2-3x)#

#rArr 18(x^2-3x)=(11x-15)(x^2-3x)#

#rArr 18(x^2-3x)-(11x-15)(x^2-3x)=0#

#rArr (x^2-3x){18-(11x-15)}=0#

#rArr x(x-3)(33-11x)=0#

#rArr 11x(x-3)(3-x)=0#

#rArr x=0, or, x=3#

But, let us observe that these roots make all the fractions of the eqn. Meaningless.

Therefore, the eqn. has No Soln. , but has #x=0, or, x=3# Extraneous Solns.