Asked by Jenn
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:(
Use the rational equation to determine the integers. Please help!! I have no idea what to do:(
Answers
Answered by
R_scott
the common denominator is ... x (x + 2) * 40 = 40 x^2 + 80 x
multiplying ... 40 x + 80 + 40 x = 9 x^2 + 18 x
9 x^2 - 62 x - 80 = 0
use the quadratic formula to find x
multiplying ... 40 x + 80 + 40 x = 9 x^2 + 18 x
9 x^2 - 62 x - 80 = 0
use the quadratic formula to find x
Answered by
Jenn
So the answer is just 8? or is it 8 and 10? I need two even integers. This is confusing:(
Answered by
Reiny
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
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
Answered by
Steve
I’m still confused
Answered by
Steve
Yeah, me too Steve.
Answered by
Steve
Can you run that by me again?
Answered by
Steve
I have waited three years... and seen many things... and yet i still so not know the answer.
Answered by
Janae
Its 8 and 10
Answered by
moony
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 :]
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 :]
Answered by
moony
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
but the first set up was supposed to be 1/x +1/x+2 =9/40
hope that helps
Answered by
Welp
You are a life saver thank you!
Answered by
:0
so were all just here struggling i see
Answered by
meow
stan moony frfr
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