Finding the single possibility [closed]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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












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


-

這個網誌中的熱門文章

Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
閱讀完整內容

Carbon dioxide

圖片
"CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
閱讀完整內容