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Math Help - Point of Intersection (2)

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    Point of Intersection (2)

    Sorry for starting another topic, but I've tried for hours to try and get an answer.

    I need to find the point of intersection (which I know is (4,7)), but I don't know how I'm supposed to get there?

    \displaystyle y=- \frac {4x}{3}+12.33
     \displaystyle y= \frac {3x}{4}+4

     \displaystyle y= \frac {3x}{4}+4  \Rightarrow 4y=3x+16 \Rightarrow 3x=4y-16 \Rightarrow x= \frac {4y-16}{3}

    \displaystyle y=- \frac {4x}{3}+12.33

    \displaystyle y=- \frac {4(\frac {4y-16}{3} )}{3}+12.33

    \displaystyle y=- \frac {\frac {16y-64}{3} }{3} +12.33

    If I move the 4y to the other side? Then do \frac{-16}{3} + 12.33, I get 6.9997, but that's for 4y.

    What am I doing wrong?

    x
     \displaystyle x = \frac {4y-16}{3}
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  2. #2
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    Quote Originally Posted by watp View Post
    Sorry for starting another topic, but I've tried for hours to try and get an answer.

    I need to find the point of intersection (which I know is (4,7)), but I don't know how I'm supposed to get there?

    \displaystyle y=- \frac {4x}{3}+12.33
     \displaystyle y= \frac {3x}{4}+4

     \displaystyle y= \frac {3x}{4}+4  \Rightarrow 4y=3x+16 \Rightarrow 3x=4y-16 \Rightarrow x= \frac {4y-16}{3}

    \displaystyle y=- \frac {4x}{3}+12.33

    \displaystyle y=- \frac {4(\frac {4y-16}{3} )}{3}+12.33

    \displaystyle y=- \frac {\frac {16y-64}{3} }{3} +12.33

    If I move the 4y to the other side? Then do \frac{-16}{3} + 12.33, I get 6.9997, but that's for 4y.

    What am I doing wrong?

    x
     \displaystyle x = \frac {4y-16}{3}
    12.33 should be 37/3 if you want (4,7) to work.

    - \frac {4x}{3}+\frac{37}{3} = \frac {3x}{4}+4

    Multiply through by 12 to remove denominators.

    -16x + 148 = 9x + 48

    25x = 100

    x = 4

    Then go back and find y, easier to use this equation

     \displaystyle y= \frac {3x}{4}+4

    than the other since the 4's cancel.
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    Thanks undefined
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