so... let us say, we have an integer "a"
a consecutive integer of that can either be a-1 or a+1, since it's before and after "a", anyhow, let us use a+1
so "a" is the smaller one, and a+1 the bigger consecutive integer
twice the smaller, 2*a or 2a
added to
2a +
three times the larger one
the larger is a+1, three times that is 3(a+1)
and that added to 2a +
2a + 3(a+1)
is 96
thus 2a + 3(a+1) = 96
solve for "a", to find the smaller integer
and the bigger, is just a+1