Find four consecutive odd integers so that five times the sum of the first and third integers excceeds four times the sum of the second and last integers by 14.
let the integers be n-3, n-1, n+1, n+3
5(n-3 + n+1) = 4(n-1 + n+3)+14
5(2n-2) = 4(2n+2)+14
10n-10 = 8n+22
2n = 32
n=16
The numbers are 13,15,17,19
Check:
5(13+17) = 150
4(15+19)+14 = 150
To solve this problem, let's assign variables to represent the consecutive odd integers. Let's assume the first odd integer is n.
The four consecutive odd integers can be represented as: n, n + 2, n + 4, n + 6.
According to the problem, "five times the sum of the first and third integers exceeds four times the sum of the second and last integers by 14." We can translate this into an equation:
5(n + n + 4) = 4((n + 2) + (n + 6)) + 14
Now, let's solve this equation step by step:
5(2n + 4) = 4(2n + 8) + 14
10n + 20 = 8n + 32 + 14
10n + 20 = 8n + 46
10n - 8n = 46 - 20
2n = 26
n = 13
Now that we have the value of n, we can substitute it back into our equation to find the four consecutive odd integers:
n = 13
n + 2 = 15
n + 4 = 17
n + 6 = 19
Therefore, the four consecutive odd integers are: 13, 15, 17, 19.