Results 1 to 12 of 12

Thread: find the value of k

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    68

    find the value of k

    find the value of k such that the system has infinite number of solutions


    (k-1)x - y = 5
    (k+1)x + (1-k)y = (3k+1)

    plzz solve it wholly

    thnks in advance

    i am aquinted with the process of taking 3 cases and show the common value of k in all is the answer. so it will be helpful if u solve it with this process. i am having problem with factorising so plzz solve it wholly.
    Last edited by saha.subham; May 14th 2010 at 11:01 PM. Reason: extra info.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by saha.subham View Post
    find the value of k such that the system has infinite number of solutions


    (k-1)x - y = 5
    (k+1)x + (1-k)y = (3k+1)

    plzz solve it wholly

    thnks in advance

    i am aquinted with the process of taking 3 cases and show the common value of k in all is the answer. so it will be helpful if u solve it with this process. i am having problem with factorising so plzz solve it wholly.
    Someone will probably post a full solution after me, but I prefer to give a hint.

    Consider the equations

    x = y + 1
    2x = 2y + 2

    Note how this system has an infinite number of solutions.

    So...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    i no the process but just getting stuck in factorising so plzz anybody help me
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by saha.subham View Post
    i no the process but just getting stuck in factorising so plzz anybody help me
    Factorising???

    Write

    $\displaystyle \frac{k-1}{k+1}=\frac{-1}{1-k}=\frac{5}{3k+1}$

    and solve for $\displaystyle k$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    Quote Originally Posted by undefined View Post
    Factorising???

    Write

    $\displaystyle \frac{k-1}{k+1}=\frac{-1}{1-k}=\frac{5}{3k+1}$

    and solve for $\displaystyle k$.
    forget abt it plzz solve it fully
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,776
    Thanks
    3028
    Have you even tried anything yourself? In particular, in the original system,
    (k-1)x - y = 5
    (k+1)x + (1-k)y = (3k+1)
    if you multiply the first equation by 1- k and add to the second equation, what do you get?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    i have solved all this type of sums accept 3 this is one of them plzz help
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by saha.subham View Post
    i have solved all this type of sums accept 3 this is one of them plzz help
    In what I wrote above, we have

    $\displaystyle \frac{-1}{1-k}=\frac{5}{3k+1}$

    Solve for $\displaystyle k$.. cross multiply.. easy stuff... plug the value you get back into the equations and see that it works out very nicely..
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    i understands it but i want to do all the sums by the same process
    my process is
    case1 -
    for terms 1 and 2 i have to show the value of k

    case 2

    for terms 2 and 3 i have to show the value of k

    case 3

    for terms 1 and 3 i have to show the value of k

    now the common value of k in all is the answer.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    i have solved for case 2

    for case 1 i have
    k-1/k+1 = -1/1-k

    for case 3

    k-1/k+1 = 5/3k+1


    plzz solve for k in both the case which i am not being able to do.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    Hi saha.subham,

    If $\displaystyle \frac{a}{b} = \frac{c}{d}$ then

    ad = bc.

    Similarly, can you simplify

    $\displaystyle \frac{k-1}{k+1}$ = $\displaystyle \frac{-1}{1-k}$


    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Ah, it wasn't clear to me that you knew how to write the equations but not solve them. I did not know that the three cases you mentioned were the ones I wrote down. Please try to be a little more specific in the future, and show work; you would have gotten the guidance you needed a lot sooner that way.

    I figured since solving systems of equations is more advanced than cross multiplying that you knew the procedure. Hopefully with sa-ri-ga-ma's post you will now have what you require to complete the problem.

    Best regards.

    Edit: Maybe by "factorising" you meant applying the distributive property? For example if you wanted to solve

    $\displaystyle \frac{k-1}{k+1}=\frac{-1}{1-k}$

    you would need to distribute as follows

    $\displaystyle (k-1)(1-k)=-1(k+1)$

    $\displaystyle k-k^2-1+k=-k-1$

    Or further on, there is a place to factor out a $\displaystyle k$.

    $\displaystyle 3k-k^2=0$

    $\displaystyle k(3-k)=0$

    $\displaystyle k=0 \ \ \text{ OR }\ \ k=3$

    Of course it's much easier to just solve

    $\displaystyle \frac{-1}{1-k}=\frac{5}{3k+1}$

    and plug in the value of $\displaystyle k$ you find to make sure it satisfies

    $\displaystyle \frac{k-1}{k+1}=\frac{-1}{1-k}$

    Anyway, this is a complete solution to one of the three cases you mentioned, so now you really should have all the tools you need to solve the other cases.
    Last edited by undefined; May 17th 2010 at 08:17 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: Mar 22nd 2011, 04:57 PM
  2. Replies: 2
    Last Post: Jul 5th 2010, 08:48 PM
  3. Replies: 1
    Last Post: Feb 17th 2010, 03:58 PM
  4. Replies: 0
    Last Post: Jun 16th 2009, 12:43 PM
  5. Replies: 2
    Last Post: Apr 6th 2009, 08:57 PM

/mathhelpforum @mathhelpforum