algebra 2 word problem question?

Question

Q: Find all ests of three consecutive odd integers whose sum is between 20 and 30.

20<(X)+(x+2)+(x+4)<30
20<3x+6<30
14<3x<24
14/3
I got that far and the answer are {5,7,9}
I dont understand how the answer is 5,7,and 9?

Answer 1

Find all "ests"? What are "ests"? sets? okay.

Your analysis has shown that 14/3
If x=5, you have 5,7,9, okay.

If x=7, you have 7,9,11. That one works too.

So there are two sets.

Answer 2

so, what's the problem now?
u got the value of X between 4.667 and 8. you took the smallest possible odd from the given range and that is 5.
i.e. x = 5, x + 2 = 7, and x + 4 = 9

Now, the next odd in the possible range (4.667 < x < 8) is 7
so, x = 7, x + 2 = 9, and x + 4 = 11, that sums up to 27

{5, 7, 9} and {7, 9, 11} are the possible answers.

Let me know if you meant something else.

Answer 3

you can't put the x's together & 2 and 4 because they are in perentheses. so u would have to find the answers for (X)+(x+2)+(x+4)=21,22,23,24,25,26,27,28,&29 since theyre all in between 20 & 30. i got 5, 5 1/3, 5 2/3, 6, 6 1/3, 6 2/3, 7, etc...9. so you void all the others besides 5, 7, and 9 because those three are consecutive odd numbers (odd numbers one after another). sounds confusing but if you think about it it'll make sense.