Results 1 to 10 of 10

Thread: rearranging equations into y=mx+c

  1. #1
    Newbie
    Joined
    Sep 2008
    From
    UK
    Posts
    12

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206
    $\displaystyle 5x+y=4 \Rightarrow y=-5x+4$
    $\displaystyle x-2y+6=0 \Rightarrow y=\frac{1}{2}x+3$
    $\displaystyle x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}$
    $\displaystyle y-6=2(x+4) \Rightarrow y=2x+14$
    Last edited by courteous; Sep 17th 2008 at 12:04 PM. Reason: Oops!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Here is an example

    $\displaystyle x-2y+6=0
    $

    $\displaystyle -2y=-x-6$

    $\displaystyle y= \frac{1}{2}x +3$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Quote Originally Posted by courteous View Post
    $\displaystyle 5x+y=4 \Rightarrow y=-5x+4$
    $\displaystyle x-2y+6=0 \Rightarrow y=\frac{1}{2}x+6$
    $\displaystyle x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}$
    $\displaystyle y-6=2(x+4) \Rightarrow y=2x+14$
    Check your 2nd example
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2008
    From
    UK
    Posts
    12
    Thank-you so much

    Now i shall struggle along with the other questions!!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2008
    From
    UK
    Posts
    12

    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 $\displaystyle y=3x+4$

    perpencicular to

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

    and

    $\displaystyle x+3y-2=0$
    Last edited by vanpopta; Sep 18th 2008 at 01:23 AM. Reason: Finally managed to get the mathematical format!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2008
    From
    nowhere
    Posts
    206
    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 $\displaystyle y=3x+4$

    perpencicular to

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

    and

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

    When $\displaystyle k$ equals the lines are parrarel (or the same), and when $\displaystyle k_1*k_2=-1$ (where the $\displaystyle tan$ 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; Sep 18th 2008 at 03:34 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Feb 2008
    Posts
    60
    For $\displaystyle y = 3x + 4$, 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 $\displaystyle \frac{-1}{3}$

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

    Also:

    $\displaystyle x + 3y - 2 = 0$

    $\displaystyle x + 3y = 2$

    $\displaystyle 3y = 2 - x$

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

    And its slope is also $\displaystyle \frac{-1}{3}$ and hence, it's perpendicular to the line.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2008
    From
    UK
    Posts
    12
    Thanks a lot guys

    That helped me, for the next 5 questions!



    However i got stuck again.


    Line q has the equation $\displaystyle y=6x+1$

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


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

    However, i am not confident of that one bit...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    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 $\displaystyle y=6x+1$

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


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

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

    Remember

    $\displaystyle y_2 - y_1 = m(x_2 - x_1)$

    so for a)

    $\displaystyle y-0 = 6(x-1)$

    $\displaystyle y = 6x + 6$

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

    $\displaystyle y-1= -\frac{1}{6}(x-3)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rearranging Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 15th 2011, 10:52 AM
  2. Rearranging Equations?
    Posted in the Algebra Forum
    Replies: 13
    Last Post: Apr 7th 2011, 02:22 PM
  3. rearranging equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Feb 8th 2010, 08:31 PM
  4. Rearranging Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Dec 8th 2008, 01:38 PM
  5. REARRANGING EQUATIONS
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 7th 2008, 12:06 PM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum