Math question should be easy but I might be over thinking.

Consider three consecutive odd integers. Two times the first integer minus the third is 6 more than half the second. Find the three even integers.

I have to show the work too. The examples in class had a number or sum that all integers added up to. I don't know if this is a misprint on the handout or not.

Let x be the lowest of the three integers. This makes the other two x + 2 and x + 4.
2 * first integer - third integer = second integer/2 + 6
2x - (x + 4) = (x + 2)/2 + 6
x - 4 = x/2 + 1 + 6
x/2 = 11
x = 22
The numbers are 22, 24, and 26.

It appears to be a misprint. Six more than half an odd integer would end in one half. But twice the first plus the third would be a whole number. I expect three even integers is right.

Let 2y be the middle of the three consecutive even integers. By definition 2y-2 is the first and 2y+2 is the third. By substitution of equals then 2(2y-2) - (2y+2) - 6 = y By subtracting y from both sides and adding 6 to both sides we get 2(2y-2) - (2y+2) - y = 6. You can finish.

We could have just as well called our numbers 2x, 2x+2, and 2x+4, so long as the even numbers have a factor of 2 for our convenience.

Let (x/2) represent an even number.
Then (x/2)+1, (x/2)+3, (x/2)+5 are three consecutive odd numbers.
Can you do it from here?
2*[(x/2)+1] - [(x/2)+5] = .5*[(x/2)+3] + 6
Solving for x yields x=42
Then
(x/2)+1=22
(x/2)+3=24
(x/2)+5=26