find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7 times the second. Please can some one solve this problem so I can see how it's done?

How come no one has aswered my question

The three consecutive odd integers are n, n+2 and n+4. The problem states that "the product of the first and third" - which is n(n+4) - is equal to "4 less than 7 times the second" - which is 7(n+2)-4. So you have:

n(n+4) = 7(n+2)-4

Can you take it from here? You will get two possible answers for n, but only one of which is odd.