## anonymous one year ago 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

1. campbell_st

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

2. anonymous

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.

3. anonymous

im still confused :/

4. anonymous

like i dont know how to solve for x

5. campbell_st

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