Question

the difference of two numbers is 2. if the sum of their reciprocals is 9/40, find the two numbers

Answer 1

\[x-y=2\\
\frac{1}{x}+\frac{1}{y}=\frac{9}{40}\] is how i would begin it

Answer 2

or maybe
\[\frac{1}{x}+\frac{1}{x+2}=\frac{9}{40}\]

Answer 3

need more help with this?

Answer 4

yeah

Answer 5

do you cross multiply?

Answer 6

OK

Answer 7

we can add them

Answer 8

before you cross multiply you need to simplify the left hand side

Answer 9

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

Answer 10

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

Answer 11

x-y=2
x=y+2

and x-2=y, NOT x+2=y

Answer 12

I think you might have made a mistake

Answer 13

probably

Answer 14

yes, check your work, please

Answer 15

\[\frac{1}{x}+\frac{1}{x-2}=\frac{9}{40}\] might work better

Answer 16

good. Won't interrupt any longer:) Do your work:D

Answer 17

i got 80x+80
-------
9x^2-18x

Answer 18

of course is actually makes no difference at all

Answer 19

lets add
\[\frac{x-2+x}{x(x-2)}=\frac{9}{40}\] or
\[\frac{2x-2}{x(x-2)}=\frac{9}{40}\]

Answer 20

then go ahead and "cross multiply" like you said, get
\[80x-80=9x^2-18x\]

Answer 21

now you have a quadratic equation to solve

Answer 22

\[9x^2-18x-80x+80=0\\
9x^2-98x+80=0\] which you can factor

Answer 23

\[(9x-8)(x-10)=0\] et

Answer 24

you get two different numbers for \(x\) which leads to two different y values

Answer 25

i would like to point out that if you solve
\[\frac{1}{x}+\frac{1}{x+2}=\frac{4}{90}\] you get the same solutions
no one can tell \(x\) from \(y\) in this question

Answer 26

i got x=10 & y=8