Find three consecutive odd integers such that four times the sum of the first and third integer is five more than the product of three and t

Geno's answer is good....let me see if I can break it down more for you....

Let the first integer = x

Then, the next odd integer is just two more = x + 2

And the 3rd odd integer is just two more than the second = (x + 2) + 2 = x + 4

Now...let's look at what's said (in pieces)

"The sum of the first and the third" is just x  + (x + 4)   = 2x + 4

So....."Four times the sum of the first and the third" is just  4(2x + 4)     =  (A)

And "The product of 3 and the second number means to multiply the second number by 3 =  3(x + 2)  = (B)

Now....here's the tricky part.....

We're told that (A) is 5 more than (B)...this must mean that, if I subtracted 5 from (A), I'd end up with (B) !!

Therefore......(A) - 5 = (B)

And, substituting in for (A) and (B),we have

4(2x + 4) - 5 = 3(x + 2)

Simplify both sides

8x + 16 - 5 = 3x + 6

8x + 11 = 3x + 6

Subtract 6 and 8x from both sides

5 = -5x

Divide both sides by -5

-1 = x

That's the first integer

And the second is just  x + 2 = -1 + 2 = 1

And the third is just x + 4 = -1 + 4 = 3

And that's it !!!

If  x  is the smallest odd integer, then  x + 2  will be the next larger odd integer, and  x + 4  will be the largest of the three odd integers.

4[ (x) + (x + 4) ] = 5 + 3(x + 2)

4( 2x + 4 )  =  5 + 3x + 6

8x + 16  =  3x + 11

5x  =  -5

So the smallest integer is  -1.

Use this to find the other two.

Thanks for your help but I am still totally lost on this one.

Thank you CPhill! I think I understand now.