How do you solve #(2x)/(x-1) + 3/(x-3) = (x+3)/(x^2-4x+3)#?

2 Answers
Deepak G.
Aug 5, 2016

#x=-1#

Explanation:

#2x/(x-1)+3/(x-3)=(x+3)/(x^2-4x+3)#
or
#(2x(x-3)+3(x-1))/((x-1)(x-3))=(x+3)/(x^2-4x+3)#
or
#(2x^2-6x+3x-3)/cancel(x^2-4x+3)=(x+3)/cancel(x^2-4x+3)#
or
#2x^2-3x-3=x+3#
or
#2x^2-3x-x-3-3=0#
or
#2x^2-4x-6=0#
or
#x^2-2x-3=0#
or
#x^2-3x+x-3=0#
or
#x(x-3)+1(x-3)=0#
or
#(x-3)(x+1)=0#
or
#x-3=0#
or
#x=3# Invalid because it takes deniminator to #0#
or
#x+1=0#
or
#x=-1#

Ratnaker Mehta
Aug 5, 2016

#x=-1# is a soln.

#x=1, 3# are Extraneous solns.

Explanation:

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

#:. {2x(x-3)+3(x-1)}/{(x-1)(x-3)}=(x+3)/((x-3)(x-1))#

#:. (2x^2-6x+3x-3)/((x-1)(x-3))=(x+3)/((x-3)(x-1))#

#:.(2x^2-3x-3)/((x-1)(x-3))=(x+3)/((x-3)(x-1))#

#:. (2x^2-3x-3)(x-1)(x-3)=(x+3)(x-3)(x-1)#, or,

#:. (2x^2-3x-3)(x-1)(x-3)-(x+3)(x-3)(x-1)=0#

#:. (x-1)(x-3)(2x^2-3x-3-x-3)=0#

#:. (x-1)(x-3)(2x^2-4x-6)=0#

#:. 2(x-1)(x-3)(x^2-2x-3)=0#

#;. 2(x-1)(x-3)(x-3)(x+1)=0#

Therefore, the soln. is, #x=1, x=3, or, x=-1#

Among these, #x=1, x=3# make the eqn. undefined , &, hence are extraneous solns.

The root #x=-1# satisfy the given eqn.

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