SOLUTION: To solve integer problems.
Find three consecutive odd integers such that three time the middle integer is seven more than the sum of the first and third integers.
my try:
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Question 192254This question is from textbook Intermediate Algebra
: To solve integer problems.
Find three consecutive odd integers such that three time the middle integer is seven more than the sum of the first and third integers.
my try:
1st integer: n
2nd integer: 3(n+2)
3rd integer: (n+4) + 7
n+3(n+2) = 7+(n+4)
2n + 5 = 11 + n
n + 5 = 11
n = 6
WRONG! what am I doing wrong?!
Thanks!
This question is from textbook Intermediate Algebra
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find three consecutive odd integers such that three time the middle integer is seven more than the sum of the first and third integers.
my try:
1st integer: n
2nd integer: 3(n+2) ---- it's n+2
3rd integer: (n+4) + 7 ----- it's n+4
-----------
3*(n+2) = n + n+4 + 7 ---- 3 times the middle # is the sum of 2 #s + 7
3n+6 = 2n+11
n = 5
5, 7, 9
------------------------
n+3(n+2) = 7+(n+4)
2n + 5 = 11 + n
n + 5 = 11
n = 6
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