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. Please help!! I have no idea what to do:(

I have waited three years... and seen many things... and yet i still so not know the answer.

I’m still confused

So the answer is just 8? or is it 8 and 10? I need two even integers. This is confusing:(

Always define your variables.

let the smaller even integer be x
let the larger even integer be x+2
then
1/x + 1/(x+2) = 9/40
follow R_scott's steps to get
9x^2 - 62x - 80 = 0
which factors to:
(x - 8)(9x + 10) = 0
x = 9 or x = -10/9, but we want an integer, so x = 8

by my definition, the smaller is 8 and the larger is 10

Yeah, me too Steve.

it’s two consecutive EVEN numbers

meaning we have x and x+2
bc both have to b even

we have to use reciprocals
1 1 9
— + — = ——
x x+2 40

we have to multiply the bottom with everything it is missing (this will cancel out the entire denominator)

1 1 9
———— + ———— = ————-
x(x+2)(40) x+2(x)(40) 40(x)(x+2)

whatever you do to the bottom you do to the top, so multiply the numerators

1(x+2)(40) 1(x)(40) 9(x)(x+2)
———— + ———— = ————-
x(x+2)(40) x+2(x)(40) 40(x)(x+2)

the denominators will all cancel out
you’ll be left will

40(x+2) + 40(x) = 9x(x+2)
distribute

40x +80 +40x =9x^2+18x
combine like terms

80x+80 =9x^2+18x
-80x-80 -80x -80
0=9x^2-62x-80

factor
(9x+10)(x-8)

set both zero
9x+10=0
-10 -10
9x=-10
———
9
x =-10/9 not possible

x-8=0
+8
x=8

we are looking for two consecutive even integers so
8 and 10 is your answer

i hope i helped some of y’all who were confused
good luck :]

the thing messed my spacing up

but the first set up was supposed to be 1/x +1/x+2 =9/40
hope that helps

Can you run that by me again?

so were all just here struggling i see

Its 8 and 10