You can put this solution on YOUR website! Three consecutive even integers such that their sum is ten less than twice the smallest
:
Let n = the smallest number
then
(n+2) and (n+4) are the 2nd and 3rd even numbers
:
n + (n+2) + (n+4) = 2n - 10
combine like terms
3n + 6 = 2n - 10
3n - 2n = -10 - 6
n = -16 is the smallest number
and add 2 & 4 to -16:
-14, -12 are 2nd and 3rd integers
:
:
Check:
-16 + (-14) = (-12) = 2(-16) - 10
-42 = -32 - 10
