Finding the single possibility [closed]

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If I have x + y = 4 and I know x and y can either be 1 and 3 or 2 and 2, is there any way to tell which combination it is?




probability




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closed as off-topic by Jendrik Stelzner, zipirovich, Graham Kemp, Key Flex, Xander Henderson Aug 18 at 2:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, zipirovich, Key Flex, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    This isn't clear at all. If you know that $(x,y)$ can be either $(1,3)$ or $(2,2)$ and you have no other information, why do think there is a way of choosing between the two possibilities?
    – saulspatz
    Aug 17 at 3:23






  • 1




    From that information alone, no. There are even more possibilities than that if you allow negative integers such as $x=-50,y=54$ and so on, and even more still if you allow any rational or real results. Two unknowns and one linear equation relating them is never enough information to uniquely determine anything.
    – JMoravitz
    Aug 17 at 3:24










  • You already answered your own question by saying that "I know x and y can either be 1 and 3 or 2 and 2".
    – drhab
    Aug 17 at 7:28















up vote
-3
down vote

favorite












If I have x + y = 4 and I know x and y can either be 1 and 3 or 2 and 2, is there any way to tell which combination it is?




probability




share|cite|improve this question












closed as off-topic by Jendrik Stelzner, zipirovich, Graham Kemp, Key Flex, Xander Henderson Aug 18 at 2:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, zipirovich, Key Flex, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    This isn't clear at all. If you know that $(x,y)$ can be either $(1,3)$ or $(2,2)$ and you have no other information, why do think there is a way of choosing between the two possibilities?
    – saulspatz
    Aug 17 at 3:23






  • 1




    From that information alone, no. There are even more possibilities than that if you allow negative integers such as $x=-50,y=54$ and so on, and even more still if you allow any rational or real results. Two unknowns and one linear equation relating them is never enough information to uniquely determine anything.
    – JMoravitz
    Aug 17 at 3:24










  • You already answered your own question by saying that "I know x and y can either be 1 and 3 or 2 and 2".
    – drhab
    Aug 17 at 7:28













up vote
-3
down vote

favorite









up vote
-3
down vote

favorite











If I have x + y = 4 and I know x and y can either be 1 and 3 or 2 and 2, is there any way to tell which combination it is?




probability




share|cite|improve this question












If I have x + y = 4 and I know x and y can either be 1 and 3 or 2 and 2, is there any way to tell which combination it is?





probability





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 17 at 3:19









SouthwickSolutions

14




14




closed as off-topic by Jendrik Stelzner, zipirovich, Graham Kemp, Key Flex, Xander Henderson Aug 18 at 2:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, zipirovich, Key Flex, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Jendrik Stelzner, zipirovich, Graham Kemp, Key Flex, Xander Henderson Aug 18 at 2:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, zipirovich, Key Flex, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    This isn't clear at all. If you know that $(x,y)$ can be either $(1,3)$ or $(2,2)$ and you have no other information, why do think there is a way of choosing between the two possibilities?
    – saulspatz
    Aug 17 at 3:23






  • 1




    From that information alone, no. There are even more possibilities than that if you allow negative integers such as $x=-50,y=54$ and so on, and even more still if you allow any rational or real results. Two unknowns and one linear equation relating them is never enough information to uniquely determine anything.
    – JMoravitz
    Aug 17 at 3:24










  • You already answered your own question by saying that "I know x and y can either be 1 and 3 or 2 and 2".
    – drhab
    Aug 17 at 7:28













  • 1




    This isn't clear at all. If you know that $(x,y)$ can be either $(1,3)$ or $(2,2)$ and you have no other information, why do think there is a way of choosing between the two possibilities?
    – saulspatz
    Aug 17 at 3:23






  • 1




    From that information alone, no. There are even more possibilities than that if you allow negative integers such as $x=-50,y=54$ and so on, and even more still if you allow any rational or real results. Two unknowns and one linear equation relating them is never enough information to uniquely determine anything.
    – JMoravitz
    Aug 17 at 3:24










  • You already answered your own question by saying that "I know x and y can either be 1 and 3 or 2 and 2".
    – drhab
    Aug 17 at 7:28








1




1




This isn't clear at all. If you know that $(x,y)$ can be either $(1,3)$ or $(2,2)$ and you have no other information, why do think there is a way of choosing between the two possibilities?
– saulspatz
Aug 17 at 3:23




This isn't clear at all. If you know that $(x,y)$ can be either $(1,3)$ or $(2,2)$ and you have no other information, why do think there is a way of choosing between the two possibilities?
– saulspatz
Aug 17 at 3:23




1




1




From that information alone, no. There are even more possibilities than that if you allow negative integers such as $x=-50,y=54$ and so on, and even more still if you allow any rational or real results. Two unknowns and one linear equation relating them is never enough information to uniquely determine anything.
– JMoravitz
Aug 17 at 3:24




From that information alone, no. There are even more possibilities than that if you allow negative integers such as $x=-50,y=54$ and so on, and even more still if you allow any rational or real results. Two unknowns and one linear equation relating them is never enough information to uniquely determine anything.
– JMoravitz
Aug 17 at 3:24












You already answered your own question by saying that "I know x and y can either be 1 and 3 or 2 and 2".
– drhab
Aug 17 at 7:28





You already answered your own question by saying that "I know x and y can either be 1 and 3 or 2 and 2".
– drhab
Aug 17 at 7:28











1 Answer
1






active

oldest

votes

















up vote
2
down vote













No. You require an additional constraint in order to solve for $(x,y)$ uniquely.



In general, systems of linear equations with $n$ degrees of freedom (i.e. variables) require $n$ equations for a unique solution.



If the number of degrees of freedom exceeds the number of equations, as in this case, we call that an underdetermined system. The result is that there either exists no solution or there exists infinitely many solutions. Your example has infinitely many solutions if we include the negative integers.






share|cite|improve this answer


















  • 1




    Replace "polynomial equations" with "linear equations", otherwise you'll end up with things like $x^2+y^2+z^2=0$ which clearly has only one solution over the reals.
    – JMoravitz
    Aug 17 at 4:06










  • @JMoravitz Thanks. Done!
    – zahbaz
    Aug 17 at 4:20

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













No. You require an additional constraint in order to solve for $(x,y)$ uniquely.



In general, systems of linear equations with $n$ degrees of freedom (i.e. variables) require $n$ equations for a unique solution.



If the number of degrees of freedom exceeds the number of equations, as in this case, we call that an underdetermined system. The result is that there either exists no solution or there exists infinitely many solutions. Your example has infinitely many solutions if we include the negative integers.






share|cite|improve this answer


















  • 1




    Replace "polynomial equations" with "linear equations", otherwise you'll end up with things like $x^2+y^2+z^2=0$ which clearly has only one solution over the reals.
    – JMoravitz
    Aug 17 at 4:06










  • @JMoravitz Thanks. Done!
    – zahbaz
    Aug 17 at 4:20














up vote
2
down vote













No. You require an additional constraint in order to solve for $(x,y)$ uniquely.



In general, systems of linear equations with $n$ degrees of freedom (i.e. variables) require $n$ equations for a unique solution.



If the number of degrees of freedom exceeds the number of equations, as in this case, we call that an underdetermined system. The result is that there either exists no solution or there exists infinitely many solutions. Your example has infinitely many solutions if we include the negative integers.






share|cite|improve this answer


















  • 1




    Replace "polynomial equations" with "linear equations", otherwise you'll end up with things like $x^2+y^2+z^2=0$ which clearly has only one solution over the reals.
    – JMoravitz
    Aug 17 at 4:06










  • @JMoravitz Thanks. Done!
    – zahbaz
    Aug 17 at 4:20












up vote
2
down vote










up vote
2
down vote









No. You require an additional constraint in order to solve for $(x,y)$ uniquely.



In general, systems of linear equations with $n$ degrees of freedom (i.e. variables) require $n$ equations for a unique solution.



If the number of degrees of freedom exceeds the number of equations, as in this case, we call that an underdetermined system. The result is that there either exists no solution or there exists infinitely many solutions. Your example has infinitely many solutions if we include the negative integers.






share|cite|improve this answer














No. You require an additional constraint in order to solve for $(x,y)$ uniquely.



In general, systems of linear equations with $n$ degrees of freedom (i.e. variables) require $n$ equations for a unique solution.



If the number of degrees of freedom exceeds the number of equations, as in this case, we call that an underdetermined system. The result is that there either exists no solution or there exists infinitely many solutions. Your example has infinitely many solutions if we include the negative integers.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 17 at 4:19

























answered Aug 17 at 3:34









zahbaz

7,62021636




7,62021636







  • 1




    Replace "polynomial equations" with "linear equations", otherwise you'll end up with things like $x^2+y^2+z^2=0$ which clearly has only one solution over the reals.
    – JMoravitz
    Aug 17 at 4:06










  • @JMoravitz Thanks. Done!
    – zahbaz
    Aug 17 at 4:20












  • 1




    Replace "polynomial equations" with "linear equations", otherwise you'll end up with things like $x^2+y^2+z^2=0$ which clearly has only one solution over the reals.
    – JMoravitz
    Aug 17 at 4:06










  • @JMoravitz Thanks. Done!
    – zahbaz
    Aug 17 at 4:20







1




1




Replace "polynomial equations" with "linear equations", otherwise you'll end up with things like $x^2+y^2+z^2=0$ which clearly has only one solution over the reals.
– JMoravitz
Aug 17 at 4:06




Replace "polynomial equations" with "linear equations", otherwise you'll end up with things like $x^2+y^2+z^2=0$ which clearly has only one solution over the reals.
– JMoravitz
Aug 17 at 4:06












@JMoravitz Thanks. Done!
– zahbaz
Aug 17 at 4:20




@JMoravitz Thanks. Done!
– zahbaz
Aug 17 at 4:20


-

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