SOLUTION: write (and solve) and inequality to find three consecutive odd integers with a sum between 64 and 74.

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Question 209999: write (and solve) and inequality to find three consecutive odd integers with a sum between 64 and 74.
Found 2 solutions by rapaljer, stanbon:
Answer by rapaljer(4671) (Show Source):
You can put this solution on YOUR website!
Let x = first odd integer
x+ 2 = second odd integer
x+ 4 = third odd integer

The sum of these integers (add them up!) will be 3x + 6.

The inequality will be
64<3x+6 <74

Subtract 6 from each part of the inequality:
64-6<3x+6-6<74-6
58<3x<68

Divide by 3:

19.333. . . < x < 22.666. . .

Since these are INTEGERS you are looking for, the only ones that fit this description would be x= 20, 21, and 22. You must reject the x=20 and x=22 since these are NOT odd integers. The only solution would be

x=21
x+2 = 23
x+4=25

The sum is 69.

R^2

Dr. Robert J. Rapalje
Seminole Community College

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
write (and solve) and inequality to find three consecutive odd integers with a sum between 64 and 74.
------------------------------------
1st: 2x-1
2nd: 2x+1
3rd: 2x+3
-------------------
Inequality:
64 < 2x-1+2x+1+2x+3 < 74
64 < 6x+3 < 74
61 < 6x < 71
61/6 < x < 71/6
10 1/6 < x < 11 5/6
Then x = 11
----------------------
1st = 2*11-1 = 21
2nd = 23
3rd = 25
---
sum = 21+23+25 = 69
==========================
Cheers,
Stan H.
reply to stanbon@comcast.net