SOLUTION: Find three consecutive odd integers such that 3 times the 2nd is 1 less than the 1st and 3rd.

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Question 1099120: Find three consecutive odd integers such that 3 times the 2nd is 1 less than the 1st and 3rd.
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the 3 consecvutive odd integers be
+n+. +n%2B2+, and +n%2B4+
------------------------------------
+3%2A%28+n+%2B+2+%29+=+n+%2B+%28+n%2B4+%29+-+1+
+3n+%2B+6+=+n+%2B+n+%2B+4+-+1+
+3n+-+2n+=+4+-+1+-+6+
+n+=+-3+
and
+n%2B2+=+-1+
and
+n+%2B+4+=+1+
------------------
The consecutive odd integers are:
-3, -1 and 1
----------------
check:
+3%2A%28-1%29+=+-3+%2B+1+-+1+
+-3+=+-3+
OK
get a 2nd opinion if needed

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Find three consecutive odd integers
n, (n+2), (n+4)
such that 3 times the 2nd is 1 less than the 1st and 3rd.
3(n+2) = n + (n+4) - 1
3n + 6 = 2n + 3
3n - 2n = 3 - 6
n = -3
:
The 3 consecutive odd integers: -3, -1, +1