Problem about equation solving

  • Thread starter OcaliptusP
  • Start date
In summary: Now I can finally complete my assignment.In summary, after simplifying the given equation, the final equation is 5x+5y=36xy+5x. The solution for x is 5/36, assuming that x and y are both positive. This solution can be obtained by multiplying out the equation and simplifying it further. The information that x and y are positive is necessary for this solution to be derived.
  • #1
OcaliptusP
23
2

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?
 
Physics news on Phys.org
  • #2
OcaliptusP said:

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?
 
  • #3
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.
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).
OcaliptusP said:
I cannot find a way to get that
Just multiply it out and simplify. Post your working as far as you get.
 
  • #4
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?
It's the second one. I put it on the calculator and give the result as x=5/36?
haruspex 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 simplified until x/5=x/36y+ 1/36 but cannot get further
 
Last edited by a moderator:
  • #5
OcaliptusP said:
I simplified until x/5=x/36y+ 1/36 but cannot get further
I asked you to post your working. It is wrong somewhere.
 
  • #6
OcaliptusP said:
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.
 
  • #7
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
haruspex said:
I asked you to post your working. It is wrong somewhere.
Yes it was wrong as I noticed, I've sent my final work
 
  • #9
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)
Continue to simplify. Multiply out.
 
  • #10
haruspex said:
Continue to simplify. Multiply out.
Okay I've missed the detail. Thanks for all your help
 
  • #11
If we continue
5x+5y= 36xy + 5x
Then x=5/36
 
  • #12
Can you explain why do we need information of x and y is positive to derivate that?
 
  • #13
OcaliptusP said:
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.
 
  • #14
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.
Okay. Thanks to everyone who helped.
 

Related to Problem about equation solving

1. How do I solve a complex equation?

Solving a complex equation requires breaking it down into simpler steps. Start by simplifying each side of the equation as much as possible. Then, use inverse operations to isolate the variable on one side of the equation. Finally, check your solution by plugging it back into the original equation.

2. Can I use the same method to solve all equations?

No, there are various methods for solving equations depending on the type of equation. Some common methods include substitution, elimination, and graphing. It's important to understand the type of equation you are dealing with in order to choose the most appropriate method.

3. Why do I need to show my work when solving equations?

Showing your work when solving equations helps to identify any mistakes you may have made. It also allows others to follow your thought process and understand how you arrived at your solution. Additionally, showing your work is important in academic settings as it demonstrates your understanding of the problem.

4. How do I know if my solution is correct?

To check if your solution is correct, you can plug it back into the original equation and see if it satisfies the equation. Another way is to use a graphing calculator to graph both sides of the equation and see if they intersect at the same point. Additionally, you can ask a teacher or tutor to review your solution for any errors.

5. How can I improve my equation solving skills?

Practice is key to improving your equation solving skills. Start by mastering the basic techniques and then move on to more complex equations. You can also challenge yourself by solving equations from different areas of math, such as algebra, geometry, and calculus. Seeking help from a teacher or tutor can also be beneficial in improving your skills.

Similar threads

Replies
8
Views
270
Replies
5
Views
972
Replies
6
Views
1K
Replies
32
Views
2K
Replies
3
Views
2K
Replies
14
Views
2K
Replies
17
Views
837
Back
Top