Asked by Crystal
Find 4 consecutive odd integers such that 5 times the sum of the first two was 10 less than 7 times the sum of the second and fourth. What are the four integers?
When I worked the problem, I got -13 as N and then when I finished it, it didn't seem correct. Can anyone help me?
When I worked the problem, I got -13 as N and then when I finished it, it didn't seem correct. Can anyone help me?
Answers
Answered by
Reiny
let the smallest of the odd integers be x
then the next 3 consecutive integers are
x+2,x+4, and x+6
5(x + x+2) = 7(x+2 + x+6) - 10
10x + 10 = 14x + 46
-4x = 36
x = -9
so the 4 integers are -9, -7, -5, and -3
check:
5times(sum of first two) = 5(-9 -7) = -80
7times(sum of 2nd and 4th) = 7(-7 -3) = -70
Is -80 ten less than -70? YES
then the next 3 consecutive integers are
x+2,x+4, and x+6
5(x + x+2) = 7(x+2 + x+6) - 10
10x + 10 = 14x + 46
-4x = 36
x = -9
so the 4 integers are -9, -7, -5, and -3
check:
5times(sum of first two) = 5(-9 -7) = -80
7times(sum of 2nd and 4th) = 7(-7 -3) = -70
Is -80 ten less than -70? YES
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.