four times the lesser of two consecutive even integers is 12 less than twice the greater number. find the integers.

ArcSine

Any even integer can be written in the form 2k, where k is an integer. For example, 32 is 2k, where k is 16; -18 is 2k where k is -9.

Let your "two consecutive even integers" be 2k and 2k + 2. Then your word problem sets up as

4(2k) + 12 = 2(2k + 2)

Solve for k, and then you can easily determine your two integers. Be sure to test your proposed answer back into the problem before you call it a day.

ebaines

I find it easier to think of the smaller number as N and the larger as N+2. Then the problem becomes

4* smaller = 2 * larger -12
4N = 2(N+2)-12
Solve for N.

ArcSine

Roger that...using N instead of 2k for the lesser of the two integers certainly gives an easier set-up, and simpler algebra. I just used 2k so that the original model would be respecting the problem's stated "even" integer requirement right from the get-go. Overkill in this case, sure; just a habit of being overly cautious when I'm getting one teed up, before I take a swing at it.

Adios, gentlemen, and have a nice one...