Question

Find three consecutive even integers. Such that 7 times of first integers is 14 more than the sum of second and third integers.

Answer 1

4, 6 and 8

Explanation:

1st number x

2nd number x+2

3rd number x+4

7x = (x+2) + (x+4) +14

7x = 2x+20

5x=20

x=4

7*4=28 (6+8)+14 =14+14 = 28

Answer 2

no clue sorry I couldn't be of more help