Question

find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third (only an algebraic solution will be accepted )

Answer 1

X + x + 1 + x + 2

X + 2( x + 1) = 12 + x + 2
X + 2x + 2 = 12 +x + 2
3x + 2 = x + 14
3x + 2-2 = x + 14 -2
3x = x +12
3x – x = x – x + 12
2x = 12
2x/2 = 12/2
X = 6
Plug 6 into the formula to check
6 + 2( 6 + 1) = 12 + 6 + 2
6 + 12 + 2 = 12 + 6 + 2
20 = 20
The numbers are: 20, 21, 22