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?

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) (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) (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
RELATED QUESTIONS
three consecutive integers are such that the first plus one half the second plus seven... (answered by bonster)
three consecutive integers are such that the first plus one half the second plus seven... (answered by ankor@dixie-net.com)
Three consecutive intergers are such that the first plus one-half the second plus seven... (answered by blubunny01,venugopalramana)
Three consecutive intergers are such that the first plus one-half the second plus seven... (answered by venugopalramana,ikdeep)
Three consecutive intergers are such that the first plus one-half the second plus seven... (answered by venugopalramana)
Three consecutive intergers are such that the first plus one-half the second plus seven... (answered by solver91311)
Please can someone help me with this? Three consecutive integers are such that the first... (answered by ankor@dixie-net.com)
Three consecutive integers are such that the first integer plus twice the second integer... (answered by addingup)
Find 3 consecutive integers such that thrice the first plus twice the second is 16 less... (answered by macston)