Question 168829: find 4 consecutive intergers such that the sum of the first three exceeds the 4th by 18. I can not figure out how to write out the problem to solve it.
x+x+2+x+4=x+6-18
is that correct
Found 3 solutions by nerdybill, Mathtut, Alan3354:
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! find 4 consecutive intergers such that the sum of the first three exceeds the 4th by 18.
.
Let x = first of 4 consecutive integers
then
x+1 = second of 4 consecutive integers
x+2 = third of 4 consecutive integers
x+3 = fourth of 4 consecutive integers
.
Then from "sum of the first three exceeds the 4th by 18" we get:
x+(x+1)+(x+2) = (x+3)+18
3x+3 = x+21
2x+3 = 21
2x = 18
x = 9
.
Answer:
9, 10, 11, and 12
Answer by Mathtut(3670)  (Show Source):
You can put this solution on YOUR website! your on the right track but this is 4 consecutive integers not 4 consecutive odd or even integers.....so x,x+1,x+2,x+3
x+x+1+x+2=x+3+18
3x+3=x+21
2x+18
 so integers are 9,10,11,12
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! find 4 consecutive intergers such that the sum of the first three exceeds the 4th by 18. I can not figure out how to write out the problem to solve it.
x+x+2+x+4=x+6-18
is that correct
------------------
It says integers, not even integers. And, the sum exceeds the 4th, so add 18 to the 4th, not subtract.
------------
So x + x+1 + x+2 = x+3 + 18
3x+3 = x+21
2x = 18
x = 9 (the 1st)
Check:
9+10+11 = 30
12 + 18 = 30
|
|
|
