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

Algebra ->  Algebra  -> Inequalities -> SOLUTION: write (and solve) and inequality to find three consecutive odd integers with a sum between 64 and 74.      Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Inequalities, trichotomy Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Inequalities 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(4667)   (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(57307)   (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