Hi I'm posting this for my son he is having a hard time with an algebra problem that is due Friday so if someone could please help him out and walk him through it that would be great.

Heres the problem------- Find two consecutive odd integers whose product is 1 less than 6 times their sum. Domain for the smallest -1,1,11

First, to understand the basics of the question:

Odd integers are simply odd (as opposed to even) numbers - whole numbers that aren't divisible by two! You know - 1, 3, 5, 7, 9, 11, 13 etc.

Two consecutive odd integers are two "in a row" - and the difference between two odd integers is always 2. 1 and 3, 3 and 5, 5 and 7, etc. represent pairs of consecutive odd integers.

The domain is the set of numbers you are limited to. In this case, you are limited to either "1" or "11" as your smaller (first) number.

Therefore, the two consecutive odd integers are either 1 and 3 or 11 and 13. You want to see if the product of the two numbers plus one is equal to six times the sum of the numbers. Let's test an equation using both of those sets.

Using 1 and 3:

Does (1 x 3) + 1 = 6 x (1 + 3) ?
If so: 3 + 1 = 6 x 4
or: 4 = 24 <----- not true!

Using 11 and 13:

Does (11 x 13) + 1 = 6 x (11 + 13) ?
If so: 143 + 1 = 6 x 24
or: 144 = 144 <----- TRUE!

so your consecutive odd integers are 11 and 13.

You could also solve this problem without being given the domain. To start this problem, consider the first number as "x" and the second number as "x + 2".

x * (x + 2) + 1 = 6 * (x + x + 2)

However, I don't think your problem demands you go to this point - that's probably for next semester's class and would confuse the issue at hand!

Hope this helps.