Answer:
Four consecutive odd integers are: 2n+1, 2n+3, 2n+5 and 2n+7
Three times the sum of the first and fourth then are 3(2n+1 + 2n+7)
80 less than 4 times the sum of the second and third then are 4(2n+3 + 2n+5) - 80
Since the two are equal:
3(2n+1 + 2n+7) = 4(2n+3 + 2n+5)-80
expanding the brackets:
6n+3 + 6n + 21 = 8n + 12 + 8n + 20 - 80
combining like terms
3 + 21 - 12 -20 + 80 = 8n + 8n - 6n - 6n
72 = 4n
Explanation: