How do you solve #(2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)#?

1 Answer
Oct 19, 2015

Answer:

Some of these fractions can be simplified.

Explanation:

#(2y)/(2y+6):2/2=y/(y+3)#

#(9y-12)/(3y+9):3/3=(3y-4)/(y+3)#

So now we have:
#y/(y+3)+(3y-4)/(y+3)=(9y+11)/(y+3)#

Since the numerators are all equal we may leave them out, on the condition #y!=-3# for this would make the numerators #=0# and the whole equation would be senseless.

#y+3y-4=9y+11->#

All the #y#'s to one side, all the numbers to the other:

#-5y=15->y=-3#

And this is exactly the solution that was forbidden.

Answer : no sensible solution.