Find three consecutive even integers such that the sum of the smallest integer and twice the second is 12 more than the third

6, 8 , 10

Step-by-step explanation:

smallest = x

medium = x+2

largest = x+4

The smallest integer and twice the second is 12 more than the third:

x + 2(x+2)  = 12+ (x+4)

x + 2x + 4  = x +16

3x  +  4 = x + 16

2x =  12

x = 6

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!