Solve the formula for the indicated variable. [closed]

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I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.



Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)



$$x^2 - 5xy + y^2 = 0$$










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closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27


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." – amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
    – Kevin
    Sep 5 at 11:26











  • $(x-frac5y2)^2=frac21y^24$ so...
    – mrs
    Sep 5 at 11:27











  • This is an ordinary quadratic equation in $y$, use the standard formula.
    – Yves Daoust
    Sep 5 at 12:40










  • Treat $x$ as constant and solve the quadratic in $y$.
    – Narasimham
    Sep 5 at 19:30














up vote
0
down vote

favorite












I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.



Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)



$$x^2 - 5xy + y^2 = 0$$










share|cite|improve this question















closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27


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." – amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
    – Kevin
    Sep 5 at 11:26











  • $(x-frac5y2)^2=frac21y^24$ so...
    – mrs
    Sep 5 at 11:27











  • This is an ordinary quadratic equation in $y$, use the standard formula.
    – Yves Daoust
    Sep 5 at 12:40










  • Treat $x$ as constant and solve the quadratic in $y$.
    – Narasimham
    Sep 5 at 19:30












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.



Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)



$$x^2 - 5xy + y^2 = 0$$










share|cite|improve this question















I can't seem to understand how to solve for 'y' in this equation for my college algebra class. Please help.



Solve the formula for the indicated variable. (Enter your answers as a comma-separated list.)



$$x^2 - 5xy + y^2 = 0$$







algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 5 at 11:25









Kevin

5,138722




5,138722










asked Sep 5 at 11:23









Abdullah Mumtaz

111




111




closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27


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." – amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim Sep 5 at 20:27


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." – amWhy, José Carlos Santos, Strants, Ahmad Bazzi, zwim
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
    – Kevin
    Sep 5 at 11:26











  • $(x-frac5y2)^2=frac21y^24$ so...
    – mrs
    Sep 5 at 11:27











  • This is an ordinary quadratic equation in $y$, use the standard formula.
    – Yves Daoust
    Sep 5 at 12:40










  • Treat $x$ as constant and solve the quadratic in $y$.
    – Narasimham
    Sep 5 at 19:30












  • 1




    One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
    – Kevin
    Sep 5 at 11:26











  • $(x-frac5y2)^2=frac21y^24$ so...
    – mrs
    Sep 5 at 11:27











  • This is an ordinary quadratic equation in $y$, use the standard formula.
    – Yves Daoust
    Sep 5 at 12:40










  • Treat $x$ as constant and solve the quadratic in $y$.
    – Narasimham
    Sep 5 at 19:30







1




1




One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
– Kevin
Sep 5 at 11:26





One way would be to treat the above as a quadratic in the variable $y$ and treat $x$ constant, then use the quadratic formula.
– Kevin
Sep 5 at 11:26













$(x-frac5y2)^2=frac21y^24$ so...
– mrs
Sep 5 at 11:27





$(x-frac5y2)^2=frac21y^24$ so...
– mrs
Sep 5 at 11:27













This is an ordinary quadratic equation in $y$, use the standard formula.
– Yves Daoust
Sep 5 at 12:40




This is an ordinary quadratic equation in $y$, use the standard formula.
– Yves Daoust
Sep 5 at 12:40












Treat $x$ as constant and solve the quadratic in $y$.
– Narasimham
Sep 5 at 19:30




Treat $x$ as constant and solve the quadratic in $y$.
– Narasimham
Sep 5 at 19:30










2 Answers
2






active

oldest

votes

















up vote
1
down vote













To solve for y.



I thought of the equation as a quadratic, treating $x$ as a constant:



$$x^2 − 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$



After that, I plugged it into the quadratic formula to solve for $y$:



ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.






share|cite|improve this answer






















  • Thanks to Kevin and Mrs for the help!
    – Abdullah Mumtaz
    Sep 5 at 11:51

















up vote
1
down vote













Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get



$$left(frac yxright)^2-5frac yx+1=0$$



and



$$frac yx=frac5pmsqrt212.$$






share|cite|improve this answer



























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    To solve for y.



    I thought of the equation as a quadratic, treating $x$ as a constant:



    $$x^2 − 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$



    After that, I plugged it into the quadratic formula to solve for $y$:



    ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.






    share|cite|improve this answer






















    • Thanks to Kevin and Mrs for the help!
      – Abdullah Mumtaz
      Sep 5 at 11:51














    up vote
    1
    down vote













    To solve for y.



    I thought of the equation as a quadratic, treating $x$ as a constant:



    $$x^2 − 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$



    After that, I plugged it into the quadratic formula to solve for $y$:



    ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.






    share|cite|improve this answer






















    • Thanks to Kevin and Mrs for the help!
      – Abdullah Mumtaz
      Sep 5 at 11:51












    up vote
    1
    down vote










    up vote
    1
    down vote









    To solve for y.



    I thought of the equation as a quadratic, treating $x$ as a constant:



    $$x^2 − 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$



    After that, I plugged it into the quadratic formula to solve for $y$:



    ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.






    share|cite|improve this answer














    To solve for y.



    I thought of the equation as a quadratic, treating $x$ as a constant:



    $$x^2 − 5xy + y^2 = 0 ; iff ; y^2 - 5xy + x^2 = 0$$



    After that, I plugged it into the quadratic formula to solve for $y$:



    ANSWER: $$ frac5x + x sqrt212 , frac5x - x sqrt212 $$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 5 at 12:30









    amWhy

    190k27221433




    190k27221433










    answered Sep 5 at 11:50









    Abdullah Mumtaz

    111




    111











    • Thanks to Kevin and Mrs for the help!
      – Abdullah Mumtaz
      Sep 5 at 11:51
















    • Thanks to Kevin and Mrs for the help!
      – Abdullah Mumtaz
      Sep 5 at 11:51















    Thanks to Kevin and Mrs for the help!
    – Abdullah Mumtaz
    Sep 5 at 11:51




    Thanks to Kevin and Mrs for the help!
    – Abdullah Mumtaz
    Sep 5 at 11:51










    up vote
    1
    down vote













    Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get



    $$left(frac yxright)^2-5frac yx+1=0$$



    and



    $$frac yx=frac5pmsqrt212.$$






    share|cite|improve this answer
























      up vote
      1
      down vote













      Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get



      $$left(frac yxright)^2-5frac yx+1=0$$



      and



      $$frac yx=frac5pmsqrt212.$$






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get



        $$left(frac yxright)^2-5frac yx+1=0$$



        and



        $$frac yx=frac5pmsqrt212.$$






        share|cite|improve this answer












        Alternatively, noticing that the equation is homogeneous, divide by $x^2$ to get



        $$left(frac yxright)^2-5frac yx+1=0$$



        and



        $$frac yx=frac5pmsqrt212.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 5 at 12:43









        Yves Daoust

        115k666209




        115k666209












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