Question

Find 4 consecutive odd integers , such that 5 times the second is equal to one more than eight times the smalller

Answer 1

Answer:

The numbers are:

3, 5, 7 and 9

Step-by-step explanation:

a = first odd number or "the smaller"

a+2 = second consecutive odd number

a+4 = third consecutive odd number

a+6 = fourth consecutive odd number

Then:

5(a+2) = 1 + 8(a)

(5*a + 5*2) = 1 + 8a

5a + 10 = 1 + 8a

10 - 1 = 8a - 5a

9 = 3a

9/3 = a

a = 3

a+2 = 5

a+4 = 7

a+6 = 9

Check:

5(5) = 1 + 8(3)

25 = 1 + 24

Answer 2

Answer:

• The integers are 3, 5, 7 an d 9

-------------------------------------------------

The difference between consecutive integers is 2.

Let the smallest be n, then we have the following equation:

• 5(n + 2) = 8n +1

Solve it for n:

• 5n + 10 = 8n + 1
• 8n - 5n = 10 - 1
• 3n = 9
• n = 3

The other integers are 5, 7, 9.