Ding 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

  answer:

  write 548 as 500 + 40 + 8. multiply each addend by 5. then add the three products to get 2,740.