Help me with this hard math problem please to study for math team!?

Question

There are 3 consecutive odd numbers such that the sum of 3 times the smallest number and twice the second smallest number execeeds 3 times the largest number by 26. Find the sum of the 3 consecutive odd numbers.

Answer 1

first number= 2a + 1
second would be 2a + 3
third would be 2a + 5

3(2a+1) + 2(2a+3) - 26 = 3(2a+5)
6a + 3 + 4a + 6 - 26 = 6a + 15
4a = 32
a = 8

thus the integers would be:
1) 2(8)+1 = 17
2) 2(8)+3 = 19
3) 2(8)+5 = 21

the sum of the three numbers would be:
17 + 19 + 21
= 57

Answer 2

Let x be the smallest consecutive odd number. It follows that since they are odd, the next two would be

x + 2
x + 4

"The sum of" means we're using +.
"3 times the smallest number" means 3x.
"Twice the second smallest number" means 2(x + 2)
"exceeds 3 times the largest number..." means 3(x + 4)

The 26 would go with the sum. Therefore

3x + 2(x + 2) + 26 = 3(x + 4)

Let's solve for x. Expanding everything, we get

3x + 2x + 4 + 26 = 3x + 12

Simplifying,

5x + 30 = 3x + 12
2x = 12 - 30
2x = -18
x = -9

Now that we have our first number, we can obtain our next two odd integers: -7, -5.

We're asked to find the sum of the 3 consecutive odd numbers, so we just add them up

(-9) + (-7) + (-5) = -21

Answer 3

Let the numbers be x, x + 2, x + 4.

Then we have

3x + 2(x + 2) = 3(x + 4) + 26
3x + 2x + 4 = 3x + 12 + 26
2x = 34
x = 17

x + (x + 2) + (x + 4) = 17 + 19 + 21 = 57

Answer 4

you have n, n+2, n+4
3n+2(n+2)=3(n+4)+26
3n+2n+4=3n+12+26
3n+2n-3n=38-4
n=34/2
n=17

so the sum is 17+19+21=57

Answer 5

You had a problem here before, it was not even a hard problem. Please show what you have done so far.

This way people can help you and see where you are going wrong. Instead you are just getting answers and not understanding why.

Answer 6

2n+1 / 3 = sin(4x)

Solve for R