SOLUTION: Find the three smallest consecutive odd integers such that their sum is at least 40. (Let x = the first odd integer) I've tried x+x+2+x+4 is greater than or equal to 40 3x + 6

Algebra ->  Algebra  -> Inequalities -> SOLUTION: Find the three smallest consecutive odd integers such that their sum is at least 40. (Let x = the first odd integer) I've tried x+x+2+x+4 is greater than or equal to 40 3x + 6       Log On

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Question 78582This question is from textbook Algebra Structure & Method
: Find the three smallest consecutive odd integers such that their sum is at least 40. (Let x = the first odd integer)
I've tried x+x+2+x+4 is greater than or equal to 40
3x + 6 is greater than or equal to 40
3x is greater than or equal to 34
x is greater than or equal to 11.33333This question is from textbook Algebra Structure & Method

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
X+(X+2)+(X+4)>=40
3X+6>=40
3X>=40-6
3X>=34 YOU ARE CORRECT. HOWEVER THE ANSWER REMAINS THE SAME.
X>=34/3
X=11.33333 THE NEXT ODD NUMBER IS THE SMALLEST ODD NUMBER THAT MEETS THE REQUIREMENTS. THUS THE THREE NUMBERS ARE
13,15,17.
PROOF
13+15+17>=40
45>=40
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SORRY ABOUT THAT MY MIND SLIPPED.