Question

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

Answer 1

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

$\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26} \\\\\\ 3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14 \\\\\\ 0=a-14\implies 14=a$

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