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How to solve these equation?I try but getting wrong answer

  1. Mar 4, 2012 #1
    Question
    Number of ordered pair (x,y)(for y not equal to zero) solution of this equation.
    (x+y)+(x/y)=1/2
    (x+y)(x/y)=-1/2
    options:
    (a) 0
    (b) 1
    (c) 2
    (d) none of these {this is my answer}

    My answer
    from second equation (x+y)=(-1/2)(y/x)
    putting this in first equation.
    (-1/2)(y/x)+(x/y)=1/2
    let y/x=t;
    -t/2+1/t=1/2
    =>t2+4t-1=0;
    Discriminant of this equation is greater than zero so there will two values for y/x;
    hence there can be infinite sets of values y,x that can satisfy this equation, so answer should (d) none of these..
    But this is not correct where am i doing it wrong.
    thanks!
    please try to help.
     
  2. jcsd
  3. Mar 4, 2012 #2
    Just a quick look and it looks like you've multiplied t by 4 rather then 2..? Does this help?
     
  4. Mar 4, 2012 #3
    You are pointing out it????
    Oh! sorry that's typo error:"
    it will t2+t-2=0
    But even after that same problem is there. It has discriminant >0; so there must infinite solutions..........
     
  5. Mar 4, 2012 #4
    Try solving that quadratic, y/x=t, so you will have two solutions to y/x. Try rearranging these solutions so you can put them in one of the equations.
     
  6. Mar 4, 2012 #5

    spf

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    Now the equation is correct. You will obtain two values for t. You just need to insert them in anyone of the original two equations and substitute x for y/t and then each of the two values for t will give you an ordered pair (x,y). Thus, you will have not have an infinite number of ordered (x,y) pairs but only two.
     
  7. Mar 4, 2012 #6

    Ray Vickson

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    Why not set u = x+y and v = x/y, to get the system
    u + v = 1/2 and u v = -1/2 ? After you know u and v you can get x and y.

    RGV
     
  8. Mar 4, 2012 #7
    solution of the equation is y/x=1 and y/x= -2 so all values where y=x &y= -2x should be satisfied. BUT why these two conditions are not satisfying the main two equations.
    got confused?><?
    did it required to find out the relation in x+y too?
    excellent help but what's wrong in my work...
     
  9. Mar 4, 2012 #8

    Ray Vickson

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    I don't know; I did not check.


    RGV
     
    Last edited: Mar 4, 2012
  10. Mar 4, 2012 #9

    spf

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    Finding the two conditions for y/x is only the first step in solving the original equations. With this step you did not find the solutions to the original equations yet, you just narrowed them down. By now, you only know that the solutions to the original two equations satisfy either y=x or y=-2x (and NOT that ALL the values which satisfy y=x or y=-2x are all solutions to your original equations, most of them aren't). To find the the solutions to the original equations, you have to first insert y=x in anyone of the two original equations and solve it in order to obtain the first ordered (x,y) pair, and then insert y=-2x in anyone of the two original equations and solve it in order to obtain the second ordered (x,y) pair.
     
  11. Mar 4, 2012 #10
    Just return that equation in one of the first equation and you will found all solutions.

    You can also see SOLUTION

    Sorry for bad English.

    EDIT: oh, i didn't see that @spf post
     
  12. Mar 4, 2012 #11
    thanks spf..
    that's genuine mistake i did.
    actually just assembling equation together aways does not give answer.every step may add something new thing or may remove some answers.

    getting answer on wolfram-alpha just a way to check answer..
    Here i am concerned about mistake i did.


    after all i understood my mistake from spf's tiny lecture.
    thanks to all of you for helping me.
     
  13. Mar 4, 2012 #12

    spf

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    Glad that I could help. Thanks to Ray Vickson for the elegant alternative substitution and to Karamata for the nice visual presentation of the solutions with wolfram-alpha.
     
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