Question

Find the addends that are fraction 1/x and 1/y of 3/7. "No negative integer"

Answer 1

For the life of me, I can't figure this one out.
1/x + 1/y = 3/7
(x+y)/xy = 3/7
7x + 7y = 3xy
x and y are not negative integers. If they have to be positive integers, then what values satisfy the above equation? Hmmm....missing something.

Answer 2

,. because my professor told us that the answer must be a fraction that has the numerator of one., this crap is superbly stressful. what about 2/11?

Answer 3

There are no solution where x and y are positive integers. So, they must be fractions. Then, x=7 and y=7/2 will work.

Answer 4

So would probably many other solutions.

Answer 5

,. but it must be a proper fraction with a numerator of one.

Answer 6

1/7 + 1/(7/2) = 3/7

Answer 7

If x and y are fractions less than 1 (numerator of 1), then it is impossible for the sum of their reciprocals to be < 1 as in (3/7).

Answer 8

no it shouldn't be 1/(7/2),.. I tired it but my professor said it was wrong because she said that it must be a proper fraction.
.

Answer 9

I am missing something here man. Sorry.

Answer 10

or maybe the given answer doesn't really fit to the problem.

Answer 11

so than we get the system of :

x+y=3 --- x=3-y
xy=7 so than (3-y)y=7 so 3y-y2=7 so -y2 +3y -7=0 so y2 -3y +7=0
so y_1,_2=(3+/- sqrt(9-28))/2=(3+/- isqrt17)/2
and now for x_1,_2=3-(3+/-isqrt17)/2 =(6-3+/-isqrt17)/2 = (3+/- isqrt17)/2

Answer 12

huh'?..
I can't understand.

It seems really hard to process that ways of solving in my mind.