odd integers: find three consecutive odd integers such that

H

humakhan

Junior Member
Joined
May 19, 2005
Messages
182
reviewing the concept of consecutive integers, we rememner that we can designate an unspecified integer with the letter N and greater consecutive intergers with N + 1 , N + 2, etc.

Consecutive integers N,N + 1, N + 2, etc

consecutive odd integers are 2 units apart and consecutive even integers are also 2 units apart.

consecutive odd integers N,N + 2, N + 4, N + 6, etc
consecutive even integers N,N + 2, N + 4, N + 6, etc

QUESTION IS
find three consecutive odd integers such that twice the sum of teh first and second is one less than three times the third!

ok with the information that we have i think..
N, N + 2, N + 4, N - 6
????
 
2(2N + 2) = 3N + 12 - 1
4N + 4 = 3N + 13
4N - 3N = 13 - 4
N = 9

but if we check it ..replacing N with 9
it comes out to two different answers on both sides
40 on one side and 38 on the other... ???
 
i messed up i see that
i had to do.

2(N + (N + 2)) = 3(N + 4) - 1
2(2N + 2) = 3(N + 4) - 1
4N + 4 = 3N + 12 - 1
add 1 to both sides
4N + 5 = 3N + 12
subtract 5 from from both sides
4N = 3N + 7
subtract 3N form both sides
N = 7

when you check it both sies are equal to 32
:D :)
 
ok i get how to do this..
but the problem that i generally have is writing out the equation...i mean setting it up. that's what i get trouble in.
 
"setting it up" is the part you are trying to learn.
 
i mean i know how to set up the problem in some cases...but in most problems i have to read it over over again to put the problem together and still i dont seem to get how to put it together
 
humakhan said:
i mean...in most problems...i dont seem to get how to put it together
Click to expand...
Yes. That's the part the tutor was talking about. The set-up process is what you're needing to learn.

Eliz.
 
That's exactly my point. You should know already how to solve set-up problems. That is the material you have been studying to date. The study of mathematics is not the repetition of the same symbol manipulations, over and over. It is learning how to think, organize, plan, and structure (among other things). Don't feel bad if you are struggling with setting up. That IS what you are supposed to be working on. If all you do is struggle through the setup so you can get to the easy stuff, you are missing the point. Pay attention during the struggle. That is where your brain gets retrained. :)

I am not picking on you. Please take these comments to be generally applicable to a vast majority of mathematics students. The sooner the concept of what it is we are working on is understood, the happier and better the student of mathematics. We ALL want good and happy math students, right? :D
 
Yes i guess you are very right. I think if your stuck on a problem try to do it over and over again, read the question over again until you fully understand what they are asking you to do. Math is complicated sometimes but most of the times it is very fun. when your starting to enjoy math and hard problem has to come up , for you to lose your enjoyment. Am i right?
For the setting up of the problems, i am trying to do them over again, many questions come up that ask you to set up the problem your self. i'll understand day by day if i do them everyday. am i right? :)
 
You know what guys, this is my last year of highschool, yes a senior..
i'm just tired of Math . sometimes dont wnat to do it, but then i think i have to do it if i want to finish highschool. well....few more months left. pray taht i do finish highschool with great grades. :)
 
Just for fun, let's make you think about this one jsut a little tiny bit more, just to make sure we emphasize that there is OFTEN more than one way to proceed. Someimtes, one is much simpler.

QUESTION IS
Find three consecutive odd integers such that twice the sum of the first and second is one less than three times the third!

X = the Middle of the three consecutive odd integers.
X-2 = The first of the series
X+2 = The last of the series

"twice the sum of the first and second"

2*((X-2)+X)

"one less than three times the third"

3(X+2) - 1

So

2*((X-2)+X) = 3(X+2) - 1
2*(2X-2) = 3X+6+1
4X-4 = 3X+7
X = 11
And the three values are 9, 11, and 13

Try this one:

y = Not any of the three consecutive odd integers, rather the even integer between the first two.

y-1 = The first of the series
y+1 = The second of the series
y+3 = The last of the series

"twice the sum of the first and second"

2*((y-1)+(y+1))

"one less than three times the third"

3(y+3) - 1

So

2*((y-1)+(y+1)) = 3(y+3) - 1
2*(2y) = 3y+9+1
4y = 3y+10
y = 10
And the three values are 9, 11, and 13

Really, anything that is correct and makes sense to you is what is most important. Don't get caught in one way of thinking.

Now graduate already! :)
 
LOL yeah math always throws some complicating problems........but every subject does that. You will be in my prayers. :)
 
You must log in or register to reply here.
Top