find three consecutive even integers such that three times the first equals the sum of the other two

Sagot :

Let 'x' be the first integer
      'x+2' be the 2nd
      'x+4' be the 3rd
Translating the statement into mathematical expression you'll have:
3x = (x+2) + (x+4)
3x = x + 2  + x + 4
3x = 2x + 6
3x - 2x = 6
x = 6  ---1st even integer
x+2 = 6 + 2 
x + 2 = 8  ---2nd even integer
x + 4 = 6 + 4 
x + 4 = 10  --3rd even integer
Therefore the 3 consecutive even integers are 6, 8, and 10