Find three consecutive odd positive integers such that 5 times the sum of all three is 38 less than the product of the first and second integers.

Answer 1

The three consecutive odd positive integers can be represented by the following:

x, (x+2), (x+4)

The reason for this is because positive odd integers would be [1,3,5,7,...) and therefore there is an integer between them that we must account for [hence why its not x, (x+1), and (x+2)].

So 5 times the sum would be 5 * (x+(x+2)+(x+4)) =

This would equal 38 less than the product of the first and second integers. The first and second integers are x and (x+2) So the product means we are multiplying the first and second integers. (x*(x+2)). Now remember its 38 less than that.

Now we have

5 * (x+(x+2)+(x+4)) = (x*(x+2))-38

Using this now formed equation, we can solve for x. Our goal is to get x to move to one side completely.

15x+30 = x² +2x-38

-15x -15x

30=x²-13x-38

-30 -30

0=x²-13x-68

We can switch sides if it is easier to read also.

x²-13x-68=0

Now you solve for x.

x=17

Your three integers would then be

x, (x+2), and (x+4) so

17, 19, and 21

We can check this as well.

*5(17+19+21) = 285

(17*19)-38 = 285

285=285 therefore it is correct.