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Math Help - linear solutions

  1. #1
    Senior Member DivideBy0's Avatar
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    linear solutions

    Find the value of m for which the simultaneous equations 3x+my=5 and (m+2)x+5y=m have:

    a) an infinite number of solutions
    b) no solutions.

    mostly having trouble with a), please explain how you got your answer,
    thanks
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    Find the value of m for which the simultaneous equations 3x+my=5 and (m+2)x+5y=m have:

    a) an infinite number of solutions
    b) no solutions.

    mostly having trouble with a), please explain how you got your answer,
    thanks
    Hello,

    you know that a linear equation has a straight line as it's graph.

    If you have the 2 linear equations

    l_1:y=m_1x + b_1~\text{ and }l_2:y=m_2x + b_2 then you know that there are infinite many common points if l_1=l_2~\iff~m_1=m_2\ \wedge \ b_1=b_2

    and

    There is no solution (or there aren't any common points) if

    l_1\neq l_2~\iff~m_1=m_2\ \wedge \ b_1\neq b_2

    With your problem:

    1. Rearrange your equations ( y=-\frac3m x+\frac5m , y=-\frac{m+2}{5} x + \frac m5)

    2. Use the general equations which I mentioned above

    3. For confirmation only: Infinite many solutions with m = -5 and no solution with m = 3.
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  3. #3
    Senior Member DivideBy0's Avatar
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  4. #4
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    Quote Originally Posted by DivideBy0 View Post
    Find the value of m for which the simultaneous equations 3x+my=5 and (m+2)x+5y=m have:

    a) an infinite number of solutions
    b) no solutions.

    mostly having trouble with a), please explain how you got your answer,
    thanks
    3x+my = 5
    (m+2)x+5y = m
    The determinant of the coefficients is:
    \left| \begin{array}{cc}3&m\\m+2&5 \end{array} \right| = 15 - m(m+2)
    To have non-unique solution the determinant needs to be zero:
    m(m+2) = 15 \implies m=-5,3.
    Now check which one gives infinite solutions and which one gives no solutions.*


    *)Or you can use determinants again, but there is no point. Because you just need to check two values.
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