Find three consecutive even integers such that the sum of twice the smallest and three times the largest is 42

6,8,10

Step-by-step explanation:

Two consecutive even integers would have difference of 2

Hence we can assume the three consecutive integers to be of the form

a-2, a and a+2

Smallest number = a-2 and

Largest number = a+2

Twice the smallest number + 3 times largest number = 42 (given)

2 (a-2) + 3 (a+2) = 42

2a-4+3a+6 = 42

5a + 2 = 42

Subtract 2 and then divide by 5

5a = 40 and a = 8

Hence the three numbers are 6,8 and 10