1. Jul 25, 2017

OcaliptusP

1. The problem statement, all variables and given/known data
What is x ?( Assuming x and y are positive)
2. Relevant equations
1000/x=3600 + (3600(y-x)+1000)/x+y

3. The attempt at a solution

Calculator gives it as x=5/36 but I cannot find a way to get that?

2. Jul 25, 2017

Ray Vickson

Do you mean
$$(1): \; \;\; \frac{1000}{x} = 3600 + \frac{3600(y-x) +1000}{x} + y$$
or do you mean
$$(2): \; \;\; \frac{1000}{x} = 3600 + \frac{3600(y-x) +1000}{x+ y}$$
In either case you have only one equation in two unknowns $x$ and $y$, so there will generally be an infinite number of different solutions. For example, if you put $y=0$ you will get one solution $x$, if you put $y = 1000$ you will get a different solution $x$, and so forth. Are you sure you have not left out some important information?

3. Jul 25, 2017

haruspex

In general, yes.
But here we are given x and y both positive. (1) has no solutions, but (2) has only the given solution for x (y being indeterminate).
Just multiply it out and simplify. Post your working as far as you get.

4. Jul 26, 2017

OcaliptusP

It's the second one. I put it on the calculator and give the result as x=5/36?
I simplified until x/5=x/36y+ 1/36 but cannot get further

Last edited by a moderator: Jul 26, 2017
5. Jul 26, 2017

haruspex

6. Jul 26, 2017

cnh1995

There is no y in the final equation.
Post your working so we can see where you went wrong.

7. Jul 26, 2017

OcaliptusP

Well starting equation was:
1000/x=3600+(3600(y-x)+1000)/(x+y)
Divide all 200
5/x=18+(18(y-x)+5)/(x+y)
Can be written as;
5/x=18+(18y-18x+5)/(x+y)
Multiply 18 with (x+y)/(x+y)
5/x=(18x+18y+18y-18x+5)/(x+y)
5/x=36y+5/(x+y)

8. Jul 26, 2017

OcaliptusP

Yes it was wrong as I noticed, I've sent my final work

9. Jul 26, 2017

haruspex

Continue to simplify. Multiply out.

10. Jul 26, 2017

OcaliptusP

Okay I've missed the detail. Thanks for all your help

11. Jul 26, 2017

OcaliptusP

If we continue
5x+5y= 36xy + 5x
Then x=5/36

12. Jul 26, 2017

OcaliptusP

Can you explain why do we need information of x and y is positive to derivate that?

13. Jul 26, 2017

haruspex

No, my post#3 was a bit misleading there. That comment only appiled to the other interpretation of the equation given, i.e. (1) in Ray's post.

14. Jul 26, 2017

OcaliptusP

Okay. Thanks to everyone who helped.