How do you solve #1+(3y)/(y-2)= 6/(y-2)# and find any extraneous solutions?

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Answer 1 · 2017-11-09T17:53:59

There is only one solution, #y=2#

However this is an extraneous solution because it makes the denominators equal to #0#

Hence there is no solution.

Both the denominators are the same, so re-arrange the fraction terms on the same side:

#1 = 6/(y-2) -(3y)/(y-2)" "larr y!=2#

#1 = (6-3y)/(y-2)" "larr# cross-multiply

#y-2 = 6-3y#

#y+3y =6+2#

#4y = 8#

#y=2#

However, this solution is not valid because it makes the denominators equal to #0#