How do you solve #(3x)/(x-4)=(2x)/(x-3)+6/(x^2-7x+12)# and check for extraneous solutions?

1 Answer
Aug 25, 2016

Answer:

#x = -2#

Explanation:

Start by factoring the denominator.

#(3x)/(x - 4) = (2x)/(x - 3) + 6/((x - 4)(x - 3))#

Put on a common denominator:

#(3x(x - 3))/((x - 4)(x - 3)) = (2x(x - 4))/((x - 3)(x - 4)) + 6/((x - 3)(x - 4))#

#3x^2 - 9x = 2x^2 - 8x + 6#

#x^2 - x - 6 = 0#

#(x - 3)(x + 2) = 0#

#x = 3 and -2#

Let's finally note restrictions on the variable. This is done by setting the denominators to #0# and solving for #x#.

#x - 3 = 0 -> x = 3#

AND

#x - 4 = 0 -> x = 4#

Hence, #x!=3 and x!= 4#

Since #x = 3# contradicts these restrictions, #x = 3# is an extraneous solution. The only actual solution is #x = -2#. Our solution set is therefore #{-2}#.

Hopefully this helps!