Find three consecutive even integers, such that
8 times of first integer is 6 more than the sum
of second and third integers.

2, 4 and 6 are consecutive even integers that fit the criterion.

Step-by-step explanation:

Model the problem as an equation

let a be first integer, b be second, c be third

8a = b+c+6

Since these are consecutive even integers:

b = a + 2

c = b + 2 = a + 4

Substitute these equivalents into the equation

8a = (a+2)+(a+4)+6

Combine like terms

8a = a+2+a+4+6

8a = 2a + 12

Isolate a

6a = 12 (divide both sides by 6)

a = 2

Sub a=2 into b=a+2 and c=a+4

b=a+2 = 2+2 = 4

b=4

c=a+4 = 2+4 = 6

c=6

my guess would be b and d, but i didn’t actually get out the pen and paper. lol