Question 246286: Find three consecutive integers such that the sum of twice the first and 4 times the second is equal to 20 more than twice the third Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! let n=first integer
n+1=second
n+2=third
2n+4(n+1)=20+2(n+2)
2n+4n+4=20+2n+4
combine and simplify
6n+4=24+2n
subtract 2n from both sides
4n+4=24
subtract 4 from both sides
4n=20
divide both sides by 4
n=5:n+1=6:n+2=7
integers are 5,6,7
check
2(5)+4(6)=20+2(7)
10+24=20+14
34=34
ok