How do you solve #(x-2)/(3x+5)=6/(x-3)# and check for extraneous solutions?

1 Answer
Nov 9, 2016

Answer:

#x = 24 " or " x = -1#

Explanation:

Identify the restrictions on #x# before you even solve the equation.
The denominators may not be equal to 0.

#3x +5 != 0" and "x-3!=0#
#3x !=-5color(white)(xxxxxxxxxxx)x!=3#
#x != -5/3#

As there is one fraction on each side of the equal sign you can cross -multuiply.

#(x-2)(x-3)= 6(3x+5)#

#x^2 -3x-2x+6 = 18x+30" "larr# a quadratic, so set =0

#x^2 -5x-18x+6-30= 0#

#x^2 -23x-24=0" "larr# find factors of 24 which differ by 23

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

#x = 24 " or " x = -1#

Both solutions are valid.