math help?

Question

The sum of the reciprocals of two consecutive odd integers is 8/15. Find the integers? How do you do this?

Answer 1

Let the 2 consec. odd integers be called n and n+2. Your equation, then, is

1/n + 1/(n+2) = 8/15

The LCD is 15n(n+2)

Multiply both sides of the eqn. by the LCD and you get:

15(n+2) + 15n = 8n(n+2)

15n+30+15n= 8n^2+16n

8n^2 -14n -30 = 0

2(4n^2 - 7n - 15) = 0

2(n-3)(4n+5) = 0

n = 3 or n = -5/4

But -5/4 is not an odd integer, so n = 3.

The 2 integers were n and n + 2, so 3 and 5 is the answer.

Answer 2

If the first integer is X, the next odd integer is X+2.
Their reciprocals are 1/x and 1/(x+2)

So the formula is 8/15 = 1/x + 1/(x+2).

First, you need to combine fractions -
1/X + 1/(x+2) = (x+2)/(x^2 + 2x) + X/(X^2 + 2x), or (2X + 2)/(X^2 + 2x).

Now you can multiply across, getting
8(x^2 + 2x) = 15(2X + 2), or 8x^2 + 16X = 30X + 30,
or 4X^2 + 8X = 15X + 15, or 4x^2 -7X - 15 = 0
This factors to (4X + 5 ) (x - 3) = 0 so
X = 3 and X = -4/5. Remove 4/5 since it is not an integer.

Checking, 1/3 + 1/5 = 5/15 + 3/15 = 8/15, so we are correct.
Whew!

Answer 3

Always remember that two consecutive odd integers differ by 2. So,

Let n = 1st odd int. and
n + 2 = 2nd odd int.

Then the reciprocals are 1/n and 1/(n + 2). Now your equation is

1/n + 1/(n + 2) = 8/15 I will let you solve the equation.

I get 3 and 5. Do you?

Note: the setup is exactly the same for consecutive EVEN integers. And, the setup for just consecutive integers is

Let n = 1st int
n + 1 = 2nd int
n + 2 = 3rd int. etc.

Answer 4

3,5