algebra problem?

Question

I have a problem in math, can you help me please thank you ^_^ the sum of the reciprocals of two concecutive even integers is 13/84. find these integers.

Answer 1

Let x equal the smaller integer, and then let x+2 equal the larger one.
The reciprocals is 1 divided by a number, so the sum of the reciprocals would look like this:
1/x + 1/(x+2) = 13/84.
Now, multiply the entire equation by (x)(x+2)(84) to get rid of the denominators. This gets you:
(84)(x+2) + (84)(x) = 13(x)(x+2)
Expand the multiplication to get:
84x + 168 + 84x = 13x^2 + 26x.
Combine like terms on one side of the equation to get:
13x^2-142x - 168 = 0.
This factors into:
(x-12)(13x+14) = 0.
This splits into:
x-12 = 0 and 13x+14 = 0.
Because the second equation will not give us an integer, we can ignore it.
The first equation, however, gives us x = 12.
So, 12 is the first number, and we said the second number was x + 2.
12 + 2 = 14, so 14 is our second number.

Final answer: 12, 14

Answer 2

1/x + 1/x+2 = 13/84

x(x + 2)(84)(1/x) + x(x + 2)(84)(1/x + 2)= x(x + 2)(84)(13/84

(x + 2)84 + x(84) = 13(x + 2)(x)

84x + 168 + 84x = 13(x² + 2x)

168x + 168 = 13x² + 26x

168x + 168 - 168x = 13x² + 26x - 168x

168 = 13x² - 142x

168 - 168 = 13x² - 142x - 168

0 = 13x² - 142x - 168

0 = (13x + 14)(x - 12)

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Roots

13x + 14 =

13x - 14 = - 14

13x/13 = - 14/13

x = - 1 1/13

x = - 1.076923077. . .

Or

x - 12

x - 12 + 12 = 12

x = 12

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Answer 3

1/a+1/(a+2) = 13/84 =(a+2+a)/(a*(a+2))

2(a+1)*84=13*a*(a+2)

or 13*a^2 +26*a -168*a-168 =0


try 12 and 14.