who can do these problems?

Question

find 3 consecutive odd integers such that 6 times the sum of the first and the third is 28 greater than 8 times the second what is the answer?

find four consecutive inegers such that 4 times the sum of the first and the fourth is 24 greater than 6 tiimes the third what is the answer?

Answer 1

start by defining the numbers as variables, I used x, x+2, and x+4 for the first one, see how that works? Then you just make an equation with them and solve for x:
6*(x+x+4)=8*(x+2)+28
12x+24=8x+44
4x=20
x=5
so if x=5 then the three consecutive odd integers are 5, 7, and 9 (from x, x+2, x+4)

For the second one, do the same thing but this time with variables x, x+1, x+2, x+3 (you won't actually use x+1 but anyways)
4(x+x+3)=6(x+2)+24
8x+12=6x+36
2x=24
x=12
so the consecutive integers are 12, 13, 14, and 15

Answer 2

Let the integers be x, x + 2 and x + 4. This is because for any 3 consecutive odd numbers, there is a difference of 2 between one and the next eg, 1, 3, 5. if 1 = x, then 3 = x + 2 and 5 = x + 4.

If 6 times the sum of the first and the third is 28 greater than 8 times the second then we have
6[x + (x + 4)] - 8(x + 2) = 28

Expanding we have
6[2x + 4] - 8x - 16 = 28
12x + 24 - 8x - 16 = 28

Collect like terms
12x -8x = 28 + 16 - 24
4x = 20
4x/4 = 20/4
x = 5

Substitute this in x, x + 2 and x + 4 we have
5, 5 + 2, 5 + 4
5, 7, 9 are therefore the consecutive odd numbers.

Check!
Substitue these in the equation:
6[x + (x + 4)] - 8(x + 2) = 28
6[5 + 9] - 8(7) = 28
6(14) - 8(7) = 28
84 - 56 = 28
28 = 28!

Answer 3

3 consecutive odd integers n, n+2, n+4:
sum of the first and the third is 28 greater than 8 times the second

n + (n+4) = 28 + 8(n+2)

Answer 4

Let the integers be n, n+2, n+4.

6(n + n+4) = 8(n+2) + 28
12n + 24 = 8n +16 + 28
4n = 20
n=5.

n+2=7
n+4=9