Yeah, I would be unable to solve it, given that I'm pretty sure there is no solution if we assume that "consecutive" implies integers (and I'd be curious what 'consecutive' means if we don't).
If instead we made it "2 consecutive numbers and a third number that was five higher than the second number add up to 70" (i.e., 21 + 22 + 27), that's easily solved using 'tricks'. Really, the trick mention (which is what I did in my head, too) is just a rephrasing of the algebra. That is, I would subtract 4 off of the thing that is 5 higher (so that it's now 1 higher; the problem is now the same, get 3 consecutive integers), subtract 4 off of the number I'm trying to get (so 66), so the problem is now 3 consecutive integers that equal 66, and solve the same way (66/3 = 22, so -> 21, 22, 23), and then just add the four back in to the highest (21, 22, 27).
If instead we made it "2 consecutive numbers and a third number that was five higher than the second number add up to 70" (i.e., 21 + 22 + 27), that's easily solved using 'tricks'. Really, the trick mention (which is what I did in my head, too) is just a rephrasing of the algebra. That is, I would subtract 4 off of the thing that is 5 higher (so that it's now 1 higher; the problem is now the same, get 3 consecutive integers), subtract 4 off of the number I'm trying to get (so 66), so the problem is now 3 consecutive integers that equal 66, and solve the same way (66/3 = 22, so -> 21, 22, 23), and then just add the four back in to the highest (21, 22, 27).