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

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Answer 1 · 2017-06-04T01:51:54

No real solutions.

One extraneous solution: #x=3#

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.

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