Problem about equation solving

[OcaliptusP]

Homework Statement

What is x ?( Assuming x and y are positive)

Homework Equations

1000/x=3600 + (3600(y-x)+1000)/x+y

The Attempt at a Solution

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

 

[Ray Vickson]

Homework Statement

What is x ?( Assuming x and y are positive)

Homework Equations

1000/x=3600 + (3600(y-x)+1000)/x+y

The Attempt at a Solution

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

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?

 

[haruspex]

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.

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).

I cannot find a way to get that

Just multiply it out and simplify. Post your working as far as you get.

 

[OcaliptusP]

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?

It's the second one. I put it on the calculator and give the result as x=5/36?

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.

I simplified until x/5=x/36y+ 1/36 but cannot get further

 

[haruspex]

I simplified until x/5=x/36y+ 1/36 but cannot get further

I asked you to post your working. It is wrong somewhere.

 

[cnh1995]

I simplified until x/5=x/36y+ 1/36 but cannot get further

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

 

[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)

 

[OcaliptusP]

I asked you to post your working. It is wrong somewhere.

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

 

[haruspex]

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)

Continue to simplify. Multiply out.

 

[OcaliptusP]

Continue to simplify. Multiply out.

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

 

[OcaliptusP]

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

 

[OcaliptusP]

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

 

[haruspex]

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

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.

 

[OcaliptusP]

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.

Okay. Thanks to everyone who helped.