Question

the sum of the reciprocals of two consecutive even integers is 9/40. this can be represented by the equation shown.
1/x + 1/x+2 = 9/40

use the rational equation to determine the integers

Answer 1

well I would suggest to use a common denominator

\[x \times (x + 2) \times 40\]

that way the numerators become

\[\frac{40(x+2)}{40x(x+2)} + \frac{40x}{40x(x + 2)} =\frac{9x(x +2)}{40x(x+2)}\]

so you now need to just solve the numerator so its

\[40x + 80 + 40x = 9x^2 + 18x\]

so you have a quadratic... collect the like terms and solve for x.

how it helps

Answer 2

Lets arrange the items here:
\[\frac{ 1 }{ x } + \frac{ 1 }{ x+2 } = \frac{9}{40}\]
lets equalize the denominators:
\[\frac{x+2}{x^2 + 2x} + \frac{x}{x^2 + 2x} = \frac{9}{40}\]
\[\frac{2x + 2}{x^2 + 2x} = \frac{9}{40}\]
Multiply both sides by the denominator to get rid of denominator:
\[(x^2 + 2x) \times \frac{2x + 2}{x^2 + 2x} = \frac{9}{40} \times (x^2 + 2x)\]
After simplification, we will get this:
\[40(2x + 2) = 9(x^2 + 2x)\]
and you will have a quadratic equation. Find 'x's.

Answer 3

im still confused :/

Answer 4

like i dont know how to solve for x

Answer 5

ok... so you could use the general quadratic formula

the equation can be written as

\[9x^2 - 62x - 80 = 0\]

so

a = 9, b = -62 and c = -80

the formula is

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

so when you substitute you get

\[x = \frac{62 \pm \sqrt{(-62)^2 - 4 \times 9 \times (-80)}}{2 \times 9}\]

calculate the values and only use the integer answer