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Math Help - rearranging equations into y=mx+c

  1. #1
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    rearranging equations into y=mx+c

    Hello, i am new to this site, an i was not sure where to post (as i am not in the US schooling system), i am doing AS level maths in the UK if this helps?

    i need to rearrange equations to y=mx+c

    5x+y=4

    x-2y+6=0

    x+2y=3

    y-6=2(x+4) <-- No idea whatsoever.


    I hope you can help me :S
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  2. #2
    Member courteous's Avatar
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    5x+y=4 \Rightarrow y=-5x+4
    x-2y+6=0 \Rightarrow y=\frac{1}{2}x+3
    x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}
    y-6=2(x+4) \Rightarrow y=2x+14
    Last edited by courteous; September 17th 2008 at 01:04 PM. Reason: Oops!
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  3. #3
    Super Member 11rdc11's Avatar
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    Here is an example

    x-2y+6=0<br />

    -2y=-x-6

    y= \frac{1}{2}x +3
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by courteous View Post
    5x+y=4 \Rightarrow y=-5x+4
    x-2y+6=0 \Rightarrow y=\frac{1}{2}x+6
    x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}
    y-6=2(x+4) \Rightarrow y=2x+14
    Check your 2nd example
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  5. #5
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    Thank-you so much

    Now i shall struggle along with the other questions!!
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  6. #6
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    Extra question

    I got to a later question, which was State whether the following were parallel, perpendicular, or neither, I managed to do the parallel ones, so i need now to know whether these were perpendicular or not

    Is y=3x+4

    perpencicular to

    y=2-\frac{1}{3}x

    and

    x+3y-2=0
    Last edited by vanpopta; September 18th 2008 at 02:23 AM. Reason: Finally managed to get the mathematical format!
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  7. #7
    Member courteous's Avatar
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    Quote Originally Posted by vanpopta View Post
    I got to a later question, which was State whether the following were parallel, perpendicular, or neither, I managed to do the parallel ones, so i need now to know whether these were perpendicular or not

    Is y=3x+4

    perpencicular to

    y=2-\frac{1}{3}x

    and

    x+3y-2=0
    General equation is y=kx+n, where k is the slope of the line and n is the point of line intersecting the y-axis.
    So, the slopes are y=3x+4 \Rightarrow 3,
    y=2-\frac{1}{3}x \Rightarrow -\frac{1}{3},
    x+3y-2=0 (now you know how to rearrange this) \Rightarrow -\frac{1}{3}.

    When k equals the lines are parrarel (or the same), and when k_1*k_2=-1 (where the tan is undefined, that is, denominator is 0; more at here) the lines are perpendicular. Now, you only need to compare those ks.
    Last edited by courteous; September 18th 2008 at 04:34 AM.
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  8. #8
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    For y = 3x + 4, you will notice it has a slope of 3.

    Now, a line is perpendicular to the above line if its slope is the inverse reciprocal - ie. slope of \frac{-1}{3}

    y = 2 - \frac{-1}{3}x does indeed have a slope of \frac{-1}{3} and so, it is perpendicular to the line.

    Also:

    x + 3y - 2 = 0

    x + 3y = 2

    3y = 2 - x

    y = \frac{2}{3} - \frac{1}{3}x

    And its slope is also \frac{-1}{3} and hence, it's perpendicular to the line.
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  9. #9
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    Thanks a lot guys

    That helped me, for the next 5 questions!



    However i got stuck again.


    Line q has the equation y=6x+1

    (a) Give the equation of the line through (1,0) parallel to q
    (b) Give he equation of the line through (3,1) perpendicular to q


    I think (a) may well be y=6x-6

    However, i am not confident of that one bit...
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  10. #10
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by vanpopta View Post
    Thanks a lot guys

    That helped me, for the next 5 questions!



    However i got stuck again.


    Line q has the equation y=6x+1

    (a) Give the equation of the line through (1,0) parallel to q
    (b) Give he equation of the line through (3,1) perpendicular to q


    I think (a) may well be y=6x-6

    However, i am not confident of that one bit...
    You are correct on a)

    Remember

    y_2 - y_1 = m(x_2 - x_1)

    so for a)

    y-0 = 6(x-1)

    y = 6x + 6

    to do part b) remember that the perpendicular line is the negative reciprocal

    y-1= -\frac{1}{6}(x-3)
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