How do you solve #(n+5)/(n+8)=1+6/(n+1)# and check for extraneous solutions?
1 Answer
Dec 2, 2016
Explanation:
Put everything on the least common denominator of
#((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
Hopefully this helps!