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

1 Answer
Sep 11, 2016

Answer:

#x=4#

Explanation:

Because we have an equation which has a variable in the denominators, we need to look at any restrictions for #x#.

#x-2 != 0 rarr x !=2#

Usually we would multiply through by the LCD to get rid of the fractions, but in this case there is another method because both denominators are the same, so we can add the fractions..

#x/(x-2) = 3 color(red)(-2/(x-2))#

#x/(x-2)color(red)(+2/(x-2)) = 3#

#(x+2)/(x-2) = 3" "larr# cross-multiply

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

#x+2 = 3x-6#

#2+6 = 3x-x#

#8 = 2x#

#4 = x#