Name: Randall
Who is asking: Teacher
Question: Hi Randall, I think that the most straightforward way to approach this problem is with a case argument. The cases come from the fact that There are four cases, depending on the sign of n + 4 and 3  2n. Case 1: n + 4 >= 0 and 3  2n >= 0 n + 4 + 3  2n = 16 n =  9 Check:  9 + 4  +  3 2(9)  = 5 + 21 = 26 which is not 16. Hence n = 9 is not a solution. n + 4  3 + 2n = 16 3n = 15 n = 5 Check:  5 + 4  +  3 2(5)  = 9 + 7 = 16. Hence n = 5 is a solution.  n  4 + 3  2n = 16 3 n = 17 n = ^{ 17}/_{3} Check:  ^{ 17}/_{3} + 4  +  3 2( ^{ 17}/_{3})  = ^{5}/_{3} + ^{43}/_{3} = 16. Hence n = ^{ 17}/_{3} is a solution.  n  4  3 + 2n = 16 n = 23 Check:  23 + 4  +  3 2(23)  = 27 + 43 = 70 which is not 16. Hence n = 23 is a solution. Thus there are two solutions, n = 5 and n = ^{ 17}/_{3} Cheers,Penny
