How do you solve #(3n)/(n-1)+(6n-9)/(n-1)=6#?

1 Answer

empty set

Explanation:

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