Solving system of equations with $3$ variables: [closed]












-1














I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.



$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$



and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$






algebra-precalculus systems-of-equations







share|cite|improve this question















closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25


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." – Nosrati, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Plug first equation into second and third to get 2 by 2 I'm $x,w$.
    – Zachary Selk
    Nov 22 '18 at 23:13
















-1














I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.



$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$



and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$






algebra-precalculus systems-of-equations







share|cite|improve this question















closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25


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." – Nosrati, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Plug first equation into second and third to get 2 by 2 I'm $x,w$.
    – Zachary Selk
    Nov 22 '18 at 23:13














-1












-1








-1







I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.



$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$



and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$






algebra-precalculus systems-of-equations







share|cite|improve this question















I am looking for some detail about how the following system of equations is solved, I checked using wolfram and got the answer but I am not sure the details of how to get there.



$$x+y=1$$ $$xw+y=frac{1}{2}$$ $$xw^2+y = frac{1}{3}$$



and the solution is $w = frac{1}{3}, x = frac{3}{4}, y = frac{1}{4}$






algebra-precalculus systems-of-equations




algebra-precalculus systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 '18 at 23:17









José Carlos Santos

152k22123226




152k22123226










asked Nov 22 '18 at 23:05









mt12345mt12345

958




958




closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25


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." – Nosrati, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Nosrati, Alexander Gruber♦ Dec 4 '18 at 4:25


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." – Nosrati, Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Plug first equation into second and third to get 2 by 2 I'm $x,w$.
    – Zachary Selk
    Nov 22 '18 at 23:13


















  • Plug first equation into second and third to get 2 by 2 I'm $x,w$.
    – Zachary Selk
    Nov 22 '18 at 23:13
















Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13




Plug first equation into second and third to get 2 by 2 I'm $x,w$.
– Zachary Selk
Nov 22 '18 at 23:13










2 Answers
2






active

oldest

votes


















3














From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…






share|cite|improve this answer





















  • so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
    – mt12345
    Nov 22 '18 at 23:11












  • Yes, that's it.
    – José Carlos Santos
    Nov 22 '18 at 23:13










  • when you say but now dividng the second equality by the first could you show details of that?
    – mt12345
    Nov 22 '18 at 23:13










  • $$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
    – José Carlos Santos
    Nov 22 '18 at 23:15










  • and then after this I can plug x into y = 1 -x to find y?
    – mt12345
    Nov 22 '18 at 23:16



















1














Subtract the second equation from the first equation
begin{eqnarray*}
x(1-w)=frac{1}{2}
end{eqnarray*}
Subtract the third equation from the second equation
begin{eqnarray*}
xw(1-w)=frac{1}{6}
end{eqnarray*}
So $w=?$ ....






share|cite|improve this answer




























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…






    share|cite|improve this answer





















    • so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
      – mt12345
      Nov 22 '18 at 23:11












    • Yes, that's it.
      – José Carlos Santos
      Nov 22 '18 at 23:13










    • when you say but now dividng the second equality by the first could you show details of that?
      – mt12345
      Nov 22 '18 at 23:13










    • $$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
      – José Carlos Santos
      Nov 22 '18 at 23:15










    • and then after this I can plug x into y = 1 -x to find y?
      – mt12345
      Nov 22 '18 at 23:16
















    3














    From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…






    share|cite|improve this answer





















    • so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
      – mt12345
      Nov 22 '18 at 23:11












    • Yes, that's it.
      – José Carlos Santos
      Nov 22 '18 at 23:13










    • when you say but now dividng the second equality by the first could you show details of that?
      – mt12345
      Nov 22 '18 at 23:13










    • $$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
      – José Carlos Santos
      Nov 22 '18 at 23:15










    • and then after this I can plug x into y = 1 -x to find y?
      – mt12345
      Nov 22 '18 at 23:16














    3












    3








    3






    From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…






    share|cite|improve this answer












    From the first equation, you get that $y=1-x$. So, you now have only two equations:$$left{begin{array}{l}xw-x=-frac12\xw^2-x=-frac23,end{array}right.$$which is equivalent to$$left{begin{array}{l}x(w-1)=-frac12\x(w-1)(w+1)=-frac23.end{array}right.$$But now, dividing the second equality by the first one gives you $w+1=frac43$. So…







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 22 '18 at 23:10









    José Carlos SantosJosé Carlos Santos

    152k22123226




    152k22123226












    • so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
      – mt12345
      Nov 22 '18 at 23:11












    • Yes, that's it.
      – José Carlos Santos
      Nov 22 '18 at 23:13










    • when you say but now dividng the second equality by the first could you show details of that?
      – mt12345
      Nov 22 '18 at 23:13










    • $$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
      – José Carlos Santos
      Nov 22 '18 at 23:15










    • and then after this I can plug x into y = 1 -x to find y?
      – mt12345
      Nov 22 '18 at 23:16


















    • so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
      – mt12345
      Nov 22 '18 at 23:11












    • Yes, that's it.
      – José Carlos Santos
      Nov 22 '18 at 23:13










    • when you say but now dividng the second equality by the first could you show details of that?
      – mt12345
      Nov 22 '18 at 23:13










    • $$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
      – José Carlos Santos
      Nov 22 '18 at 23:15










    • and then after this I can plug x into y = 1 -x to find y?
      – mt12345
      Nov 22 '18 at 23:16
















    so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
    – mt12345
    Nov 22 '18 at 23:11






    so then $w = 1/3$ then plug it into the second equation you have there to find $x$? or the first one too
    – mt12345
    Nov 22 '18 at 23:11














    Yes, that's it.
    – José Carlos Santos
    Nov 22 '18 at 23:13




    Yes, that's it.
    – José Carlos Santos
    Nov 22 '18 at 23:13












    when you say but now dividng the second equality by the first could you show details of that?
    – mt12345
    Nov 22 '18 at 23:13




    when you say but now dividng the second equality by the first could you show details of that?
    – mt12345
    Nov 22 '18 at 23:13












    $$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
    – José Carlos Santos
    Nov 22 '18 at 23:15




    $$frac{x(w-1)(w+1)}{x(w-1)}=frac{-2/3}{-1/2}iff w+1=frac43.$$
    – José Carlos Santos
    Nov 22 '18 at 23:15












    and then after this I can plug x into y = 1 -x to find y?
    – mt12345
    Nov 22 '18 at 23:16




    and then after this I can plug x into y = 1 -x to find y?
    – mt12345
    Nov 22 '18 at 23:16











    1














    Subtract the second equation from the first equation
    begin{eqnarray*}
    x(1-w)=frac{1}{2}
    end{eqnarray*}
    Subtract the third equation from the second equation
    begin{eqnarray*}
    xw(1-w)=frac{1}{6}
    end{eqnarray*}
    So $w=?$ ....






    share|cite|improve this answer


























      1














      Subtract the second equation from the first equation
      begin{eqnarray*}
      x(1-w)=frac{1}{2}
      end{eqnarray*}
      Subtract the third equation from the second equation
      begin{eqnarray*}
      xw(1-w)=frac{1}{6}
      end{eqnarray*}
      So $w=?$ ....






      share|cite|improve this answer
























        1












        1








        1






        Subtract the second equation from the first equation
        begin{eqnarray*}
        x(1-w)=frac{1}{2}
        end{eqnarray*}
        Subtract the third equation from the second equation
        begin{eqnarray*}
        xw(1-w)=frac{1}{6}
        end{eqnarray*}
        So $w=?$ ....






        share|cite|improve this answer












        Subtract the second equation from the first equation
        begin{eqnarray*}
        x(1-w)=frac{1}{2}
        end{eqnarray*}
        Subtract the third equation from the second equation
        begin{eqnarray*}
        xw(1-w)=frac{1}{6}
        end{eqnarray*}
        So $w=?$ ....







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 '18 at 23:12









        Donald SplutterwitDonald Splutterwit

        22.3k21344




        22.3k21344















            -

            Popular posts from this blog

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more