SOLUTION: three consective integers are such that the first plus one half the second plus seven less than twice the third is 2101. What are the integers?

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Question 28874: three consective integers are such that the first plus one half the second plus seven less than twice the third is 2101. What are the integers?
Found 2 solutions by Paul, cocakola2003:
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!


Let the three consecutive integers be x, x+1, and x+2.

equation:
x+(x+1)/2+2(x+2)-7=2101
Multiple the whole equation by 2:
2(x)+x+1+2(2(x+2))-2(7)=2(2101)
can you solve it from there?
Paul


Answer by cocakola2003(1) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not sure but I got
2x + x + 1 + 4x + 4 - 14 = 4202
2x + 1 + 4 - 14 = 4202
I don't know what I'm doing
I don't know but somehow I broke it down to the 2x/ 2 = 4202/2 = 2101