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Math Help - Simultaneous equations unique, infinite solutions

  1. #1
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    Simultaneous equations unique, infinite solutions

    hello all, frantic rush to get all the h/w done before school starts again lol.

    here it is i have absolutely no idea what i am suppose to do here, the book provides little to no help at all.

    This is the example they give which i need help figuring out as well as the actual problem.

    Consider the simultaneous linear equations
    (m-2)x + y = 2 and mx+2y = k
    Find the values of m and k such that the system of equations has

    a) a unique solution
    b) no solution
    c) infinitely many solutions


    Now for the question......

    Find the value of m for which the simultaneous equations

    (m+3)x +my = 12
    (m-1)x + (m-3)y = 7

    have no solution.

    i no that when they want the equations to have no solutions it means the lines are parallel and when the infinite solutions they are the same line however i have no idea what to do about it.

    Thanks

    Josh
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  2. #2
    MHF Contributor red_dog's Avatar
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    1) From the first equation y=2-(m-2)x

    Replacing y in the second equation we have x(4-m)=k-4
    This equation has:

    a) a unique solution if 4-m\neq 0\Leftrightarrow m\neq 4
    b) infinetely many solutions if 4-m=0, \ k-4=0\Leftrightarrow m=k=4
    c) no solution if 4-m=0, \ k-4\neq 0\Leftrightarrow m=4, \ k\neq 4
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  3. #3
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    hello, thanks for the reply. i dont understand how you interpreted a) b) c) from the two equations?

    thanks

    Josh
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  4. #4
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    Quote Originally Posted by red_dog View Post
    1) From the first equation y=2-(m-2)x

    Replacing y in the second equation we have x(4-m)=k-4
    This equation has:

    a) a unique solution if 4-m\neq 0\Leftrightarrow m\neq 4
    b) infinetely many solutions if 4-m=0, \ k-4=0\Leftrightarrow m=k=4
    c) no solution if 4-m=0, \ k-4\neq 0\Leftrightarrow m=4, \ k\neq 4
    Quote Originally Posted by sinners View Post
    hello, thanks for the reply. i dont understand how you interpreted a) b) c) from the two equations?

    thanks

    Josh
    It should be clear where x(4-m)=k-4 has come from.

    a) It should be clear that from the solution red_dog gives here that you can solve for a unique value of x (and hence y) eg. m = 2 and k = anything (3, say).

    b) What happens when you substitute the solution red_dog gives here ....? Can you get a unique value of x or can x be any value ....?

    c) What happens if m = 4 and k = 2, say ....?
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  5. #5
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    ohh i see thanks guys i understand now
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