<PRE><B>Question 949590</B>
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</PRE>