7 times 1st of 2 consecutive odd integers is 5 time 2nd

I

igor_iv837

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Please help|!!!! The problem is this. "Seven times the first of two consecutive odd integers is five times the second. Find the integers.

I wrote it out as 7(n)(n+2)=5(n+2) = 7n+2=5n+10
-5n -5n =2n+2=10
-2 -2
2n=8
2n 2n= 4. but the answer is 5, 7 integers. I dont understand what Have i done wrong?
 
Re: Whats the answer for...

igor_iv837 said:
... Seven times the first of two consecutive odd integers is five times the second. Find the integers.

I wrote it out as 7(n)(n+2)=5(n+2) ... what Have i done wrong?
Click to expand...

Similar to what I posted about your exercise with 3 consecutive integers ...

on this exercise, let the two integers be represented by the symbols A and B.

Here is the equation that models this word problem (before substituting expressions for A and B using the variable n):

7A = 5B

Now do you see where you went wrong in setting up your equation?

~ Mark :)

PS: It's not necessary to send me Private Messages when you post, Igor. If I don't see your post, then somebody else will.
 
Re: Whats the answer for...

igor_iv837 said:
No i still dont understand...
Click to expand...

Heh, heh. Did you try to understand? (Your last post is a fast comeback ...)

Using my notation, the equation you wrote models the following.

7AB = 5B

In other words, the equation that you set up reflects the statement, "Seven times the first integer times the second integer is 5 times the second integer".

Is this clear? You have not set up an equation to model your exercise.

~ Mark :)
 
Re: Whats the answer for...

so basically its 7(n)=5(n+2) and answer is n=5 , 7 THANXXXXXXXX They made it tricky in a question....they asked whats the "FIRST of TWO consecutive" but it doesnt mean its 2 its just saying whats the first of two so its just n ....THANK U
 
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