Find three consecutive even integers such that 3 times the first is 26 less than twice the sum of the last two

Let's say our first integer is "a".

how to get the next consecutive EVEN integer?  well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".

and the next after that, will then be (a + 2) + 2, or "a + 4".

so those are are 3 integers, a           a + 2           a+4

notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.

2 + 2 is 4, or 8 + 2 is 10  some even ones

3 + 2 is 5, or 13 + 2 is 15, some odd ones



what are the other two consecutive integers?  well, a + 2 and a + 4.