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Math Help - find the value of k

  1. #1
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    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.
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  2. #2
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    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...
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  3. #3
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    i no the process but just getting stuck in factorising so plzz anybody help me
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  4. #4
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    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

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

    and solve for k.
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  5. #5
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    Quote Originally Posted by undefined View Post
    Factorising???

    Write

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

    and solve for k.
    forget abt it plzz solve it fully
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  6. #6
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    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?
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  7. #7
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    i have solved all this type of sums accept 3 this is one of them plzz help
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  8. #8
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    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

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

    Solve for k.. cross multiply.. easy stuff... plug the value you get back into the equations and see that it works out very nicely..
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  9. #9
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    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.
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  10. #10
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    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.
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  11. #11
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    Hi saha.subham,

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

    ad = bc.

    Similarly, can you simplify

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


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  12. #12
    MHF Contributor undefined's Avatar
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    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

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

    you would need to distribute as follows

    (k-1)(1-k)=-1(k+1)

    k-k^2-1+k=-k-1

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

    3k-k^2=0

    k(3-k)=0

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

    Of course it's much easier to just solve

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

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

    \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.
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