Find three consecutive odd numbers such that the sum of the second and third is 1 more than 3 times the first?

Answer 1

n = first odd number

n + 2 = 2nd odd number (which is consecutive ... n + 2 because you add 2 to n to get the 'next' odd number)

n + 4 = 3rd odd number (consecutive to 2nd odd number... adding another 2 to n+2 which was previous 2nd odd number)

(2nd odd #) + (3rd odd #) = 1 + [3 x (1st odd #)]

(n + 2) + (n + 4) = 1 + 3n

combine like terms:

2n + 6 = 1 + 3n

move numbers to left side and n's to right side

6 - 1 = 3n - 2n

5 = n

therefore, the 3 consecutive odd numbers are 5, 7 and 9

7 + 9 = 3 x 5 + 1 ?

16 = 15 + 1 = 16 check... CORRECT!

... so that's how you do this problem...

Answer 2

5, 7, and 9 (7 + 9 = 16, which is 1 more than 3 * 5 = 15).

Answer 3

The answer to this is 5,7, and 9.

the three numbers would be represented as x, (x+2), and (x+4). Since the sum of the 2nd and 3rd is 1 more then 3 times the first that would mean:

(x+2) + (x+4) = 3x+1
2x+6 = 3x+1
-2x -2x
------------------------------
6 = x+1
-1 -1
------------------------------
5 = x

So X=5, which is your first number then (x+2) --the 2nd number-- would be (5+2) which is 7, and then (x+4) -- the 3rd number-- would be (5+4) which is 9.

Answer 4

let a,b,c = 3 consecutive odd numbers
general true equasions of consecutive odds
a+2=b
b+2=c
a+4=c
stated equasion
b+c=3a+1
(a+2)+(a+4)=3a+1
2a+6=3a+1
2a+5=3a
5=a

a+2=b
5+2=b
b=7

b+2=c
7+2=c
c=9

check
b+c=3a+1
7+9=3(5)+1
16=15+1
16=16 true

5,7,9

Answer 5

OKAY i just learned this... You have N+1, N+2, N+3 which are given.
HERE IS HOW YOU WORK IT:
[(N+2) +(N+1)]= 3(N+1)+1
2N+3=3N+4
2N=3N+1
-N=1
ANSWER: N=1

Answer 6

5,7,9

x, x+2,x+4
(x+2)+(x+4)=3x+1
2x+6=3x+1
2x-3x=-6+1
-x=-5
x=5