SOLUTION: Find three consecutive integers such that three times the first one is eight more than the sum of the last two integers

Question 949590: Find three consecutive integers such that three times the first one is eight more than the sum of the last two integers
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
Find three consecutive integers such that three times the first one is eight more than the sum of the last two integers
First integer=x; second=x+1; third=x+2
3x=((x+1)+(x+2))+8
3x=2x+11 Subtract 2x from each side.
x=11 ANSWER 1: The first number is 11
x+1=12+1=12 ANSWER 2: The second number is 12
x+2=12+2=13 ANSWER 3: The third number is 13
Check:
3 times first= 8 more than sum of last two
3(11)=(12+13)+8
33=25+8
33=33
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