SOLUTION: find four consecutive numbers such that the sum of the first three numbers is twelve more than the fourth number

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Question 506974: find four consecutive numbers such that the sum of the first three numbers is twelve more than the fourth number
Answer by Maths68(1474) About Me  (Show Source):
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Let
1st Integer = x
2nd Integer = x+1
3rd Integer = x+2
4th integer = x+3
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Given
x+x+1+x+2=(x+3)+12
3x+3=x+3+12
3x+3=x+15
3x=x+15-3
3x=x+12
3x-x=12
2x=12
x=12/2
x=6
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1st Integer = x = 6
2nd Integer = x+1 = 6+1 = 7
3rd Integer = x+2 =6+2= 8
4th integer = x+3=6+3=9
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Check
x+(x+1)+(x+2)=(x+3)+12
6+7+8=9+12
21=21