Question 970197
find three consecutive even integers such that three times the first integer is eight more than the sum of the second and third integers
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let x=1st consecutive even integer
x+2=2nd consecutive even integer
x+4=3rd consecutive even integer
..
3x=x+2+x+4+8
x=14
x+2=16
x+4=18
1st consecutive even integer=14
2nd consecutive even integer=16
3rd consecutive even integer=18