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X/a + y/b, y=mx+c question

  1. Mar 10, 2013 #1
    {{Really sorry for messing up your boards, I won't use latex in the title again}}


    1. The problem statement, all variables and given/known data
    Find the x and y intercepts of [tex]3x+4y=8[/tex]

    2. Relevant equations
    [tex]\frac{x}{a}+\frac{y}{b}[/tex]

    3. The attempt at a solution
    I'm reading from a book but I can't really see why their coming to the answers using this method.

    The book tells me to make to denominators the same as the constant which I've done.
    [tex]\frac{3x}{8}+\frac{4y}{8} = \frac{8}{8} = 1[/tex]
    Then I proceed onwards with making the [itex]x[/itex] intercept [itex]8[/itex] and the [itex]y[/itex] intercept the reciprocal of [itex]\frac{4y}{8}[/itex] and making it [itex]\frac{8}{4}[/itex]


    I'm not sure if this is correct and the book only gives me one example and I've followed the steps they used but I'm still really unconfident.

    I've graphed the x and y intercepts and then checked them against the original equation and they line up, a happy coincidence though I think.
     
  2. jcsd
  3. Mar 10, 2013 #2

    tiny-tim

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    hi uperkurk! :smile:
    no, i think the book means you to use the reciprocal in both cases,

    ie the intercepts are 8/3 and 8/4

    (but it's much simpler just to put x = 0, or y = 0, in the original equation, giving the same result :confused:)
     
  4. Mar 10, 2013 #3
    The problem is when I'm plotting these graphs they're not matching up with what wolfram is telling me, neither are they matching up when I just go about solving them the standard way. Also when the C is something like [tex]\frac{5}{3}[/tex] is that still graphed like:

    5 on the y axis then 3 on the x axis?
     
    Last edited: Mar 10, 2013
  5. Mar 10, 2013 #4

    SammyS

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    You're getting [itex]\displaystyle \ \ \frac{5}{3}\ \ [/itex] in what context?

    By itself, [itex]\displaystyle \ \ \frac{5}{3}\ \ [/itex] is just a number .
     
  6. Mar 10, 2013 #5
    for example let's say the equation I'm given is [tex]y=2x+\frac{5}{3}[/tex] The first thing I usually do is plot the constant, the y-intercept. How would I go about plotting the y-intercept in this case?
     
  7. Mar 10, 2013 #6

    Mentallic

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    If you're given an equation of the form
    [tex]ax+by=c[/tex]
    Then the x intercept is when y=0, hence
    [tex]ax=c[/tex]
    [tex]x=\frac{c}{a}[/tex]

    And the y intercept similarly is when x=0, hence
    [tex]by=c[/tex]
    [tex]y=\frac{c}{b}[/tex]

    Now, what the book has done is played around with the equation a bit, and it turns out that you can get the info of where the x and y intercepts are without plugging in y=0 and x=0 respectively,

    Going from [itex]ax+by=c[/itex] and dividing through by c to make the RHS 1, we get

    [tex]\frac{ax}{c}+\frac{by}{c}=1[/tex]

    And this is equivalent to

    [tex]\frac{x}{c/a}+\frac{y}{c/b}=1[/tex]

    And as you can see, the denominator of the x term is the x-intercept, and similarly for the y term. So this is WHY it works, as for the gradient intercept form [itex]y=mx+k[/itex] (I'll use k instead of c to avoid confusion with the other c)

    We can convert the previous equation into this form:

    [tex]ax+by=c[/tex]

    We are looking to get y on its own

    [tex]by=-ax+c[/tex]

    [tex]y=\frac{-ax}{b}+\frac{c}{b}[/tex]

    So as you already know, the constant is c/b which is the y-intercept, but the x-intercept isn't immediately obvious, because we know it's c/a and that info isn't really popping out at us.
    From this point, we'd take the usual route of letting y=0 and solving for x to find the x-intercept.
     
  8. Mar 10, 2013 #7
    So I'm not correct? I hate the way they word this type of equation. I'd much prefer to just graph every equation in the form y=mx+b and just plot the points and connect the dots.
     
  9. Mar 10, 2013 #8

    Mentallic

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    3x+4y=8 has an x-intercept of 8/3 and a y-intercept of 8/4=2, just plot those points on your graph and draw a line through them, it's that easy.

    y=2x+5/3 would be a graph with a y-intercept of 5/3 and gradient 2, so you'd need to plot the point (1,11/3) or some other point to be able to draw the graph.

    The second form is easier to visualize, but since we aren't being asked to quickly visualize the graph but instead to graph it on paper, the first form is arguably just as easy or even easier to work with.

    So when you say that you'd prefer to graph the equations in the point-gradient form and "just plot the points and connect the dots", which points are you plotting exactly? The y-intercept is easy to plot, but the others require some work to find.
     
  10. Mar 10, 2013 #9

    SteamKing

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    Don't you know what a fraction is?
     
  11. Mar 10, 2013 #10
    Sorry this may seem very very stupid and I am having a bit of a brain collapse but when you say 8/3 do you literally mean 8/3 expressed as an integer? 2.6 in this case?
     
  12. Mar 10, 2013 #11

    LCKurtz

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    @uperkirk: Just to add my two cents worth to this thread. The reason for the various forms like the slope-intercept form, the intercept form, or the two point form is to make certain properties of the line visible in the equation. This makes it easy to identify the properties from the equation and to generate the equation from the properties. For example, if you have the intercept form$$
    \frac x a + \frac y b = 1$$you can read the intercepts x=a and y = b directly. Actually (a,0) and (0,b) of course. Similarly if you have$$
    y =mx + b$$you can read the slope and y intercept. And if you have the point-slope form$$
    y-y_1 = m(x-x_1)$$you immediately know it is a line through ##(x_1,y_1)## with slope ##m##. Working the other way, depending on what you are given, you choose the appropriate form. For example if you want the equation of the line with x intercept (3,0) and y intercept (0,-2) you an immediately write$$
    \frac x 3 + \frac y {-2} = 1$$and you can multiply both sides by 6 to put it in the general form. If you only have the ##y=mx+b## form to work with you first have to figure out ##m## before you can use it. The point is that some forms are a bit easier than others, depending on what you are given or what you want to know.
     
  13. Mar 10, 2013 #12

    Mentallic

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    Yep!

    [tex]\frac{8}{3}=2\frac{2}{3}=2.666...[/tex]
     
  14. Mar 10, 2013 #13

    SteamKing

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    You've got to brush up on arithmetic.

    An integer is a number like 1, 2, 3, etc. otherwise known as a 'whole number'.

    a fraction, like 8/3, is the ratio of two integers. The value of the fraction may also be expressed in terms of a decimal number, so that 8/3 = 2.666..., where the 6 keeps repeating.
     
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