How do you solve #(n+5)/(n+8)=1+6/(n+1)# and check for extraneous solutions?

1 Answer
Dec 2, 2016

Answer:

#n = -17/3#

Explanation:

Put everything on the least common denominator of #(n + 8)(n + 1)#.

#((n + 5)(n + 1))/((n + 8)(n + 1)) = (1(n + 8)(n + 1))/((n + 8)(n + 1)) + (6(n + 8))/((n + 1)(n + 8))#

We can now solve without the denominators, considering everything is equivalent.

#n^2 + 5n + n + 5 = n^2 + 8n + n + 8 +6n + 48#

#6n - 8n - n - 6n = 56 - 5#

#-9n = 51#

#n = -17/3#

Our original restrictions on the variable are #n!= -8, -1#, and since this solution includes neither of those numbers, the solution is valid.

Hopefully this helps!