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Math Help - Systems of linear equations - no solution, unique solution, infinite solutions

  1. #1
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    Systems of linear equations - no solution, unique solution, infinite solutions

    I'm just gona copy these from my book:

    1. x-y = 6 and 2x-2y = 12 have infinitely solutions, show this using parameters?

    Obviously both equations are the same line, but the answer is x = t+6 where t = real number

    I assume a 3rd variable is involved...

    2. Find the values of b and c for which equations x+5y = 4 and 2x + by = c that have

    a. a unique solution b. infinite set of solutions c. no solution

    3. Solve simultaneous equations 2x-3y = 4 and x + ky = 2 where k is a constant, also, find value of k with no unique solution

    How do I solve these
    Last edited by mr fantastic; March 24th 2010 at 05:31 AM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by NewtoMath View Post
    I'm just gona copy these from my book:

    1. x-y = 6 and 2x-2y = 12 have infinitely solutions, show this using parameters?

    Obviously both equations are the same line, but the answer is x = t+6 where t = real number

    I assume a 3rd variable is involved...
    The set of solutions consists of ordered pairs (x, y). If you want to calculate some elements of this set you need a parameter such that

    (x, y) = (t+6, t) or (x, y) = (t, t - 6)

    2. Find the values of b and c for which equations x+5y = 4 and 2x + by = c that have

    a. a unique solution b. infinite set of solutions c. no solution
    Solve this system of equations considering b, c as constants. You'll get:

    x=\dfrac{5(c-8)}{b-10}+4~\wedge~y=\dfrac{c-8}{b-10}

    You now can determine
    a.) b\neq10~\wedge~c\neq8
    b.) b=10~\wedge~c=8
    c.) b=10~\wedge~c\neq8

    3. Solve simultaneous equations 2x-3y = 4 and x + ky = 2 where k is a constant, also, find value of k with no unique solution

    How do I solve these
    If you solve this system of equation at first for y you'll get y = 0 and x = 2.

    So it seems as if k can have any value. But if you solve the system at first for x you'll get x = \dfrac{4k+6}{2k+3}

    Obviously this fraction is not defined for k = -\frac32
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