Find two consecutive even integers such that twice the smaller integer added to three times the larger integer is 96.

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

The answer is the first one