Add the two fractions as they have the same denominator.
# (3n)/(n-1) + (6n-9) /(n-1) = (9n -9)/(n-1) = 9 ne 6#
Now multiply both sides by (n-1)
#{ (n-1) xx ( 9n-9)}/(n-1) = 6 xx ( n-1)# This gives
# (9n-9) = 6(n-1) # the 9 can be factored out giving
# 9(n-1) = 6(n-1) # now subtract 6(n-1) from both sides
# 9(n-1) - 6(n-1) = 6(n-1) - 6(n-1) #
# 3 (n-1) = 0# distribute the 3 across the parenthesis
# 3n -3 = 0 # add 3 to both sides.
# 3n -3 +3 = 0 +3 #
# 3n = 3 # divide both sides by 3
# (3n)/3 = 3/3 # which gives
# n = 1 # But this value for #n# is forbidden, else you divide by zero
