SOLUTION: find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18 I am getting 9, ll, 13, and 15. My math teacher is getting -9, -9, -5, and -3. W

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18 I am getting 9, ll, 13, and 15. My math teacher is getting -9, -9, -5, and -3. W      Log On


   

Question 1142224: find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18
I am getting 9, ll, 13, and 15. My math teacher is getting -9, -9, -5, and -3. Who is correct?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
Numbers n,n+2,n+4,n+6

n%2B%28n%2B2%29%2B%28n%2B4%29-%28n%2B6%29=18
-
Solve for n, not knowing sign initially.
2n%2B6-6=18
2n=18
n=9


The consecutive odd integers: 9, 11, 13, 15.

YOUR answer is correct.

----

Look at your teacher's solution.
-9, -7, -5, -3
check:
%28-9%29%2B%28-7%29%2B%28-5%29-%28-3%29=18
-21%2B3=18
-18=18FALSE

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!

find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18
I am getting 9, ll, 13, and 15. My math teacher is getting -9, -9, -5, and -3. Who is correct?
Is this a REAL situation?

If so, I'd run as fast as I can from this teacher, and encourage my friends and classmates to do the same. 

Also, if so, this reminds of a lot of people on this site who respond to folks who need math help!

BTW, if you don't know my answer by now, here it is: You are CORRECT!!