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Math Help - Linear Law - Unclear of solving part (b)

  1. #1
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    Linear Law - Unclear of solving part (b)

    Variables x and y are related by the equation y = \frac{ax}{x-b}. The graph of y against \frac{y}{x} has gradient 3 and y-intercept 2.

    (a) Find the value of a and of b.
    (b) Find the values of x for which x = y.

    Attempted Solution

    (a) y = \frac{ax}{x-b}
    yx-yb = ax
    \frac{yx}{x} - \frac{yb}{x} = \frac{ax}{x}
    y - \frac{yb}{x} = a
    y = b\frac{y}{x} + a
    b=3 , a=2

    Unclear of doing part B
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  2. #2
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    Whoops.
    Last edited by Wilmer; January 8th 2010 at 04:43 AM. Reason: none
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  3. #3
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    y = \frac{ax}{x-b} is not equals to
    y = \frac{ax}{xb} therefore, the x cannot be cancelled

    in any case i don't really know what you are talking about, perhaps you meant this
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  4. #4
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    Quote Originally Posted by Punch View Post
    y = \frac{ax}{x-b} is not equals to
    y = \frac{ax}{xb} therefore, the x cannot be cancelled
    YES; BUT you originally showed ax / x + b
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  5. #5
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    False accuse! Your previous post showed ax / x - b, lol main point is solving part (b), thanks for your effort anyway.
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  6. #6
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    Quote Originally Posted by Punch View Post
    Variables x and y are related by the equation y = \frac{ax}{x-b}. The graph of y against \frac{y}{x} has gradient 3 and y-intercept 2.

    (a) Find the value of a and of b.
    (b) Find the values of x for which x = y.

    Attempted Solution

    (a) y = \frac{ax}{x-b}
    yx-yb = ax
    \frac{yx}{x} - \frac{yb}{x} = \frac{ax}{x}
    y - \frac{yb}{x} = a
    y = b\frac{y}{x} + a
    b=3 , a=2

    Unclear of doing part B
    Assuming that what you have so far is correct, that is, that y= \frac{2x}{y- 3}, then "y= x" gives x= \frac{2x}{x- 3}. Multiplying on both sides of the equation gives x(x-3)= x^2- 3x= 2x which is the same as x^2- 5x= x(x- 5)= 0.; There are two values of x that make that true.
    Last edited by HallsofIvy; January 9th 2010 at 04:18 AM.
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  7. #7
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    Lol it was like i suddenly got it, thanks!
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