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Algebra question

  1. Mar 22, 2004 #1
    why does y equal mx+b? I know what this equations is for, how its used. But I don't understand the conceptual meaning of it....
  2. jcsd
  3. Mar 22, 2004 #2


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    y = mx + b

    is the equation for a line. It is different for different shapes (curves, etc. which you'll see eventually).

    We'll start with a simple y = x

    if you grab some graph paper and for each value of x plot its corresponding value of y (x=1, y=1; x=2, y=2; ...), you'll see that you have a slanted line going from the bottom left corner to the top right corner.

    If you then put the 'm' in there... set it equal to two and replot. You'll see that the slope of the line (the "slantiness") has changed. In this case it has changed by a factor of two (x=1, y=2; x=2, y=4; ...)

    If you then put the b in, and replot, you'll see that you have shifted the line up or down.

    That's all there is to it!
  4. Mar 22, 2004 #3
    so that's why y equals mx+b?
  5. Mar 22, 2004 #4


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    For a line, yes.
  6. Mar 22, 2004 #5
    I'm not sure i get it.

    [b(] :frown:
  7. Mar 22, 2004 #6
    if y = mx + b
    then m is the "steepness" of the line and "b" is where the line
    cuts the y axis

    Stating this more formally the y-intercept is b and the slope is m

    so y =2x + 3
    cuts the y= axis at the point (0,3) and has gradient (or slope) of m=2.
  8. Mar 22, 2004 #7
    My inquiry concerns the reason WHY Y equals mx+b. As I stated on my first post, I understand what y=mx+b is used for.
  9. Mar 22, 2004 #8
    The "why" is almost the same as the "how." It is that way because that's what works.

    On the y-axis, the x coordinate is zero. So plugging into y = mx + b we get

    y = 0x + b = b

    so it's easy to see why the b is where it intersects the y-axis.

    For the m part, you need to understand slope. There isn't much "why" other than "this is the slope required to connect points."

  10. Mar 22, 2004 #9


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    One basic answer would be to say that lots of times y does NOT equal mx + b. You get the graph of a parabola instead of the graph of a staight line if you set y equal to ax^2 + bx + c, where a is a nonzero constant and b and c are constants. Does this make sense to you?
  11. Mar 22, 2004 #10
    Ok the easy part to explain that y-intercept = b
    to find the y-intercept you notice that at the y intercept x=0
    so if you sub in x= 0 into y = mx + b
    you get y=m*0 + b
    so y=b

    As for the why slope =m you should notice if you increase x by one the formula y=mx+b means that y must increase by m so the slope is m
  12. Mar 23, 2004 #11


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    I did once this on a classroom, and it may help you "get it".

    Imagine people sitting in lines and rows on a big classroom. Assign the ceter of the room the row number 0 and line number 0, and have them memorize their coordinates.

    Then, ask all people for which row = line to stand up. What do you get? (i.e., what is the shape formed by the people standing up?)

    Have them sit down, and now ask people with
    row = twice the line number
    to stand up. Again, you get a *line*.

    It turns out that all combinations of the type:

    (row) = [ (something) times (line) ] plus (a number)

    produce a straight line.

    When you say that "y=mx+b", you are basically restricting your points to only those that have that kind of relationship (as opposed to having squares, square roots, trig functions, etc. on the relation between x and y).

    IOW, any strainght line you choose in a plane will have coordinates such that, for some m and some b, the relation "y=mx+b" holds for each and all points on your selected line.

    or, to say it in a different way,

    The reason y is equal to mx+b in a straight line is that any further complication on the right hand side (the "mx+b" part) would NOT "produce" a straight line.
  13. Mar 23, 2004 #12


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    Think of 'y = mx + b' as a processor. You stick values in and they become changed. That's what a function is. This one happens to be f(x) = mx+b. That means you stick x in, and it comes out as mx+b.

    The graph is the set of all inputs and their corresponding outputs. For that function, this looks like a line. The reason 'y' is equal to 'mx+b' is because you are plotting the value of that function on the y-axis.
  14. Mar 23, 2004 #13
    The y-intercept has already been shown by some other people, so ill just forget that.

    For simplicity reasons, lets pretend that a line pass through the origin (b=0). Then you have the equation y=mx. With out the B in there it is easy to see that “m” is the ratio of x to y. So if there is more “y” than “x” in a linear function, then m would be a small number over a big number (i.e. less x per y). Draw this graph down, y=3x, Then draw a three by three square starting at the origin, and ending at (3,3). Notice there is three times as much Y as there is X in that box. You can draw a square box in the same manner, in any direction (as long as that direction includes the line), for any distance, and as long as one of the corners of the box is touching the line the box will have the same ratio of X to Y. That’s why M tells us the slope. Does this help at all?

    P.S. my graph attached is pretty bad, all I have on this computer is mspaint.
  15. Mar 23, 2004 #14
    Also this may help. The “y =” part of y=m+b is only there because we have isolated (solved for) y. A lot of times it’s easier to write a line like this: [tex] (Y - Y_1) = M(X - X_1) [/tex]
    Last edited: Mar 23, 2004
  16. Mar 24, 2004 #15


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    SquareItSalamander, realise that we could plot this function f(x) = mx+b on different axes, let's say the p-axis and the q-axis. Then the equation would become p = mq+b.

    I think this, along with my previous post, answers your original question.
  17. Mar 25, 2004 #16
    Thank you guys!
    Arhkron: your explanation is very good! thank you!

    JonF: GROOVY! I believe that's a wonderful example.

    yes vertigo, that answers my question just fine.

    I have another though:

    When finding the intersection of two(or more) lines on a graph, you could do it multiple ways: graphing the equations of the lines and finding their intersection that way, substitution method, and addition method. These seem to be completely distinct ways of calculation. Yet, you arrive at the same answer. WHY? What is the logic behind that?
  18. Mar 25, 2004 #17
    They all get the same answers because they are actually all doing the same things.

    When you solve multivariable equations, what you are really doing is asking: When are these two equations both true. The answer graphically is where they cross. Algebraically it is where they both can have the same Y and X values. (which are really just different ways of saying the same thing)

    Let’s say you have two equations y= x and y= -x.

    You could graph these, and see where both equations are true, that’s where they cross. The both have a possible X and Y values there.

    You could do “addition” as you put it, but what you are really doing is adding equal constants to an equation. Think of it like this. Our first equation is y=x. That means y and x are the same. Also we know that you can add things that are the same to both sides of an equation, and that equation will still be true. For example, when you have x - 4=10, you add 4 to each side giving you x – 4 + 4= 10 + 4. This is really the same thing you do when you do the addition you are talking about, since y = x you can add y to one side of the equation and x to other side of the equation y = -x.

    Substitution works on the same principles! When you substitute, what you are doing is figuring what x equals in relation to y (or you could figure out what y equals in relation to x). Then you plug that into the other equation to see where that ratio of x to y holds true for it also.

    You say, “These seem to be completely distinct ways of calculation”. You are right, they are distinct but they are really all the same thing just dressed up in different ways.
  19. Mar 26, 2004 #18


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    This is pretty much a repeat of JonF post, but the different wording may help.

    The difference between algebraic methods (substitution, addition) is basically a matter of order. If you "follow" carefully what you do to each variable, you can see that, in the end, the same operations were performed (actually, once you finish solving for each, what remains is precisely the "trace" of the operations that were needed).

    On the other hand, the agreement between the algebraic and the geometrical methods has a more interesting root: the map we do between (x,y) pairs and points in the cartesian plane is a one-to-one correspondence, and this implies another correspondance between equations and curves or lines.

    Does the pair of numbers (x,y) satisfy this equation?
    is the same as asking
    does the point in position (x,y) belong to such curve?

    pairs of numbers --> points
    equations --> curves and lines
    does pair (x,y) satisfy such eqn? --> is this point in that curve?
    does pair (x,y) satisfy both eqns simultaneously? --> is this point in those two curves simultaneously?
  20. Mar 27, 2004 #19
    I see. I'll work out a few problems myself and consider your explanations. Thanks again.

    Another question:
    When you are calculating a linear equation say in the substitution method, sometimes you can get a "no solution" or "infinite solution" answer when your answer reads as follows:
    3=0 3=3
    (no solution) (infinite solutions)

    Exactly what do these numerical values represent? the slope? b? y? x?

    My guess is that it represents the equivalence of y and the slope or x.
  21. Mar 27, 2004 #20
    When you solve linear equations for two lines with substitution and get something like

    3=3, that simply means: Graphically that are both the same line, or algebraically that for any X both equations will have the same Y value.

    3=0 would mean: Graphically that they are parallel lines, or algebraically that there is no x that will give the same y for both equations.
  22. Mar 27, 2004 #21
    ah, does that not have anything to do with slope?
  23. Mar 28, 2004 #22
    For straight lines (i.e. of the form y = mx+b), the two lines must have the same slope in order to have no solution or infinitely many solutions. If the y-intercept (b) is the same, then there will be infinitely many solutions because the lines are exactly on top of each other. If the y-intercept is different, then there will be no solutions because they are parallel and will never intersect.

    Two lines with the same slope will be parallel.

    Plot some lines with the same slope and various y-intercepts to see.

    Last edited: Mar 28, 2004
  24. Mar 28, 2004 #23
    True to my words,

    I have worked out (well almost) a "linear" equation of sorts. I have a math poster project that I'm working on. I'm supposed to choose a problem and work it out all three different ways (substitution, graphing, addition). I chose a problem that's supposed to be challenging. I got the hang of it for a bit, but got totally lost.
    Here it is:

    The perimeter of a rectangle is 48 meters. The width of the rectangle is 2 more than half the length. Find the length and the width.

    my work:
    2L + 2w = 48
    1/2L + 2 = w


    2L + 2w = 48, w = 1/2L + 2
    2L + 2(1/2L + 2)=48
    2L + L + 4 = 48
    3L + 4 = 48
    L= 14 2/3

    2L + 2w = 48
    2(14 2/3) + 2w = 48
    28 2/3 + 2w = 48
    2w=20 2/3 ---> do I convert 48 - 28 2/3 into mixed fractions before subtracting?

    Anyway, I have to do the substitution method even when I do the addition method, right?
  25. Mar 28, 2004 #24
    At the steps

    3L = 44
    L = 14 2/3

    most of us would just leave L as

    L = 44/3

    which answers your later question about mixed fractions.

    As for doing the substitution method even when doing the addition method, you have to do it in the sense that it's the same thing, but the process is different. I think that point has been covered already, though.


    2l + 2w = 48

    equation (1) and

    l/2 + 2 = w

    equation (2).

    Multiply (2) by 4 to get

    2l + 8 = 4w

    and then subtract equation (1) from equation (2) in order to get rid of the l variable. Notice how the 2l cancels out the 2l and all you have left are w's. Once you solve for w, multiply equation (1) by 2 and subtract it from equation (1) in order to eliminate the w's. Then solve for l.

  26. Mar 28, 2004 #25
    funky graphing problem here...

    I'm graphing a problem for that algebra poster. (now I'm doing another problem)

    the slope is 1/-1. The run can't go into the left quadrant, can it? I'm sure it could, but I think my algebra teacher says otherwise....she may have meant something else. She's too smart to make an error like that. So could you please explain it to me? can it go into the left quadrant or not?
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