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 )

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

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