Results 1 to 10 of 10

Math Help - Linear Equations - I am stuck!

  1. #1
    Newbie Frosty's Avatar
    Joined
    May 2012
    From
    Adelaide, Australia
    Posts
    5

    Linear Equations - I am stuck!

    Calling all math geniuses

    I am really stuck with Linear Equations, I do not get it, at all....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2010
    Posts
    151
    Thanks
    3

    Re: Linear Equations - I am stuck!

    y= mx + c
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Linear Equations - I am stuck!

    you'll have to tell us a bit more than that. give us an example (not just your homework. try to think of a "similar problem" and we'll walk you through that).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie Frosty's Avatar
    Joined
    May 2012
    From
    Adelaide, Australia
    Posts
    5

    Re: Linear Equations - I am stuck!

    Deveno, I am no good at math, but what I saw on the board, was what linalg123 said, and other ones, of course...

    So, it was something like... y = -3x + 1 idk something like that...
    Last edited by Frosty; May 28th 2012 at 12:38 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Linear Equations - I am stuck!

    well, that is not a "linear equation", but rather "the equation of a line" (the difference may not seem that big a deal to you, but it is important later on).

    as you may be aware, a line can be completely decided if we know two points that lie on it (imagine lining up an "infinite ruler" on the two points, and then running one's pencil alongside the edge of the ruler). another way of describing a line is to say it is a "curve" that is straight (so an "uncurvy" curve, sort of bizarre, hmm?), with a constant "tilt".

    the technical term for "tilt" is "slope". the usual way to measure slope is:

    rise/run (how far up, divided by how far over).

    well, if our two points are (x1,y1) and (x2,y2) the "rise" is the difference of the y's (we usually draw the y-axis runnning "up and down"), and the "run" is the difference of the x's.

    for example, let's say we have a line going through (3,3) and (4,4). then the difference of the y's is 1 (4 -3 = 1) and the difference of the x's is also 1 (4 - 3 = 1, again), so the slope is 1/1 = 1 (we go "one over, and one up").

    similarly, a line that goes through (1,1) and (4,2) has a slope of 1/3 = (2-1)/(4 -1) (we are only rising 1 for every 3 we go over). this line isn't as "steep" as the first one (the slope is smaller).

    but naming other points besides the two we know (the ones we drew the line from) seems quite complicated. the slope of a line seems like one good way to tell different lines apart, but what about two parallel lines? they will both have the same "steepness", but they will obviously go through "different points".

    what we need, is some way of telling WHICH line with a given slope we have. the easiest way to do this, is to measure "how high the line is" when x = 0. often this is called the "slope-intercept form" of a line:

    y = mx + b

    where m is the slope, and b is the "y-intercept" (which is just a fancy name for the height of the line at x = 0).

    let's see if this new-fangled way of looking at slope matches our old one.

    suppose (x1,y1) and (x2,y2) are both on our line. this means:

    y1 = mx1 + b and
    y2 = mx2 + b

    so our two points are (x1,mx1+b) and (x2,mx2+b).

    the difference of the y's is: (mx2+b) - (mx1+b) = mx2 + b - mx1 - b = mx2 - mx1 = m(x2 - x1).

    the difference of the x's is: x2 - x1.

    so slope = rise/run = m(x2 - x1)/(x2 - x1) = m. hey! it works.

    let's look at your example:

    y = -3x + 1

    here, m = -3, and b = 1. this tells us the line has a slope of -3 (it goes down 3 as we go over (right) 1), and when x = 0, it is "1 high".

    let's check this: suppose x = 0.

    then y = -3(0) + 1 = 0 + 1 = 1, so the point on the y-axis that the line crosses is (0,1).

    now if our slope is -3, we should get a new point also on our line if we go over to the right by 1, and down by 3.

    well 1 over from 0 is 0+1 = 1, and 3 down from 1 is 1 - 3 = -2, so (1,-2) should be a point on our line.

    so let's check, we set x = 1:

    y = -3(1) + 1 = -3 + 1 = -2.

    that's really all there is to it.
    Last edited by Deveno; May 28th 2012 at 03:08 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,160
    Thanks
    70

    Re: Linear Equations - I am stuck!

    Quote Originally Posted by Frosty View Post
    Deveno, I am no good at math, but what I saw on the board, was what linalg123 said, and other ones, of course...
    So, it was something like... y = -3x + 1 idk something like that...
    If you can't even ask a proper question, you better talk to your teacher...

    2x + 1 = x + 3 ; can you solve that for x ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie Frosty's Avatar
    Joined
    May 2012
    From
    Adelaide, Australia
    Posts
    5

    Re: Linear Equations - I am stuck!

    I asked my teacher to help me and I think I have a little bit of a better understanding of it... But I still have to get the full grasp of it.

    I think it'd be better if I had a step by step talk through with you guys.
    I small example in my book is...

    Find the gradient of:
    P(3,1), Q(7,3)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2010
    Posts
    151
    Thanks
    3

    Re: Linear Equations - I am stuck!

    gradient = rise/run = \frac{y_2-y_1}{x_2-x_1}

    So, if you have the points P(x1,y1) and Q(x2,y2), you should be able to work it out from there.

    If you don't do well remembering formulas, just think of it logically. As x increases from 3 to 7, y increases from 1 to 3. So an increase of 4 in x leads to an increase of 2 in y.
    so, all the question is really asking is, if x increases by ONE, how much does y increase by?

    As a final point, if you are still struggling, it will help you if you make a quick graph of the two points and connect them with a line! From this you can tell if your gradient should be positive ( y increases as x moves to the right) or negative (y decreases as x moves to the right) and should also give you an idea of the steepness or magnitude of your gradient.
    Last edited by linalg123; May 30th 2012 at 09:09 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,160
    Thanks
    70

    Re: Linear Equations - I am stuck!

    Frosty, at least show SOME initiative; try googling.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie Frosty's Avatar
    Joined
    May 2012
    From
    Adelaide, Australia
    Posts
    5

    Re: Linear Equations - I am stuck!

    Quote Originally Posted by linalg123 View Post
    gradient = rise/run = \frac{y_2-y_1}{x_2-x_1}

    So, if you have the points P(x1,y1) and Q(x2,y2), you should be able to work it out from there.

    If you don't do well remembering formulas, just think of it logically. As x increases from 3 to 7, y increases from 1 to 3. So an increase of 4 in x leads to an increase of 2 in y.
    so, all the question is really asking is, if x increases by ONE, how much does y increase by?

    As a final point, if you are still struggling, it will help you if you make a quick graph of the two points and connect them with a line! From this you can tell if your gradient should be positive ( y increases as x moves to the right) or negative (y decreases as x moves to the right) and should also give you an idea of the steepness or magnitude of your gradient.
    So, the answer would be 1/2?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Algebra - I'm Stuck ona Question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 13th 2010, 09:29 PM
  2. Stuck solving set of linear equations [not solved]
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: October 3rd 2009, 08:06 AM
  3. Replies: 3
    Last Post: February 27th 2009, 08:05 PM
  4. Linear Algebra Question I'm Stuck On
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 20th 2008, 10:34 AM
  5. Linear Algebra question I'm stuck on
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 20th 2008, 03:23 AM

Search Tags


/mathhelpforum @mathhelpforum