How do you find four consecutive odd integers such that three times the sum of the first and second is 25 less than the fourth?

First, to set up consecutive odd or even integers you have to add 2 to the previous integer.
Let x be the first integer,
(x + 2), (x + 4), and (x + 6)

Now set up the parameters they give:
3(x + x + 2) = x + 6 - 25
Solving:
Combine like terms and distribute on the left, combine like terms on the right.
3(2x + 2) = x - 19
6x + 6 = x - 19
Subtract x from both sides and subtract 6 from both sides.
5x = -25
Divide both sides by 5
x = -5
x + 2 = -5 + 2 = -3
x + 4 = -5 + 4 = -1
x + 6 = -5 + 6 = 1
Check:
3(-5 + -3) ?=? 1 - 25
-24 ?=? -24 CORRECT

Set square i think so