The discussion revolves around solving an equation involving two variables, x and y, under the assumption that both are positive. The equation presented is 1000/x = 3600 + (3600(y-x)+1000)/(x+y). Participants are exploring the implications of different interpretations of the equation and the conditions under which solutions exist.
The discussion is ongoing, with participants providing guidance on simplification and encouraging the sharing of work to identify mistakes. There is recognition of the need for clarity regarding the assumptions about the positivity of x and y, and how this affects the interpretation of the solutions.
Participants note that the assumption of positivity for x and y is significant in determining the validity of certain solutions, particularly in relation to the different interpretations of the equation.
Do you meanOcaliptusP said:Homework Statement
What is x ?( Assuming x and y are positive)
Homework Equations
1000/x=3600 + (3600(y-x)+1000)/x+yThe Attempt at a Solution
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Calculator gives it as x=5/36 but I cannot find a way to get that?
In general, yes.Ray Vickson said: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.
Just multiply it out and simplify. Post your working as far as you get.OcaliptusP said:I cannot find a way to get that
It's the second one. I put it on the calculator and give the result as x=5/36?Ray Vickson said: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?
I simplified until x/5=x/36y+ 1/36 but cannot get furtherharuspex said: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 asked you to post your working. It is wrong somewhere.OcaliptusP said:I simplified until x/5=x/36y+ 1/36 but cannot get further
There is no y in the final equation.OcaliptusP said:I simplified until x/5=x/36y+ 1/36 but cannot get further
Yes it was wrong as I noticed, I've sent my final workharuspex said:I asked you to post your working. It is wrong somewhere.
Continue to simplify. Multiply out.OcaliptusP said: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)
Okay I've missed the detail. Thanks for all your helpharuspex said:Continue to simplify. Multiply out.
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 said:Can you explain why do we need information of x and y is positive to derivate that?
Okay. Thanks to everyone who helped.haruspex said: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.