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

1 Answer
Jun 4, 2017

Answer:

No real solutions.

One extraneous solution: #x=3#

Explanation:

Multiply both sides by #x-3#:

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

#x - 3/2x + 9/2 = 3#

#-1/2x = -3/2#

#x = 3#

Now to check if #x=3# is extraneous, plug it back into the original problem:

#3/(3-3) - 3/2 stackrel(?)(=) 3/(3-3)#

#3/0 - 3/2 stackrel(?)(=) 3/0#

Hmm... we can't divide by 0, so this solution doesn't work. Therefore, #x=3# is an extraneous solution.

Final Answer