Algebra II
posted 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?

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
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