- #1
Imparcticle
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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...
Understanding the Concept of "y=mx+b
- Thread starter Imparcticle
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In summary, y = mx + b is the equation for a line, with m representing the slope and b representing the y-intercept. The equation works because it produces a straight line when graphed, and any further complication on the right hand side would not result in a straight line. The y-intercept is found by plugging in x=0, and the slope is the ratio of x to y. Overall, the equation is a way to represent and manipulate the relationship between x and y values in a linear function.
- #2
enigma
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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!
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!
- #3
Imparcticle
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so that's why y equals mx+b?
- #4
enigma
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For a line, yes.
- #5
Imparcticle
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I'm not sure i get it.
[b(]
[b(]
- #6
Damned charming :)
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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.
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.
- #7
Imparcticle
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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.
- #8
cookiemonster
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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."
cookiemonster
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."
cookiemonster
- #9
Janitor
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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?
- #10
Damned charming :)
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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
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
- #11
ahrkron
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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.
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.
- #12
verty
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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.
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.
- #13
JonF
- 621
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The y-intercept has already been shown by some other people, so ill just forget that.
For simplicity reasons, let's 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.
For simplicity reasons, let's 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.
- #14
JonF
- 621
- 1
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:
- #15
verty
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SquareItSalamander, realize 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.
I think this, along with my previous post, answers your original question.
- #16
Imparcticle
- 573
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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?
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?
- #17
JonF
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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.
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.
- #18
ahrkron
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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?
Basically
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?
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?
Basically
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?
- #19
Imparcticle
- 573
- 4
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.
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.
- #20
JonF
- 621
- 1
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.
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.
- #21
Imparcticle
- 573
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ah, does that not have anything to do with slope?
- #22
cookiemonster
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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.
cookiemonster
Two lines with the same slope will be parallel.
Plot some lines with the same slope and various y-intercepts to see.
cookiemonster
Last edited:
- #23
Imparcticle
- 573
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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
SUBSTITUTION:
2L + 2w = 48, w = 1/2L + 2
2L + 2(1/2L + 2)=48
2L + L + 4 = 48
3L + 4 = 48
3L=44
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?
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
SUBSTITUTION:
2L + 2w = 48, w = 1/2L + 2
2L + 2(1/2L + 2)=48
2L + L + 4 = 48
3L + 4 = 48
3L=44
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?
- #24
cookiemonster
- 978
- 0
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.
Call
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.
cookiemonster
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.
Call
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.
cookiemonster
- #25
Imparcticle
- 573
- 4
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?
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?
- #26
cookiemonster
- 978
- 0
Hold on here. Let's get some terminology straight.
Quadrant means splitting the graph into four different regions based on the coordinate axes (the x- and y-axes). The positive x and y quadrant (I), the negative x and positive y quadrant (II), the negative x and negative y quadrant (III), and the positive x and negative y quadrant (IV).
Which do you mean by "left?" Are you asking if the x value can be negative? And is the slope 1 divided by -1 (i.e. -1), or is that the slope of two different lines, one with slope 1 and the other with slope -1?
cookiemonster
Quadrant means splitting the graph into four different regions based on the coordinate axes (the x- and y-axes). The positive x and y quadrant (I), the negative x and positive y quadrant (II), the negative x and negative y quadrant (III), and the positive x and negative y quadrant (IV).
Which do you mean by "left?" Are you asking if the x value can be negative? And is the slope 1 divided by -1 (i.e. -1), or is that the slope of two different lines, one with slope 1 and the other with slope -1?
cookiemonster
- #27
Imparcticle
- 573
- 4
OH! This is confusing! When I graph x+y=26.4 and 5y=x, the intersection point doesn't match my substitution answer. the substitution answer I got is 22,4.4
did i do something wrong?
did i do something wrong?
- #28
Imparcticle
- 573
- 4
cookiemonster said:
here is the equation:
y=1/-1x + 26.4
the slope--run---is negative. sorry about not being specific.
- #29
cookiemonster
- 978
- 0
(22,4.4) is the correct intersection point. You probably made a mistake in graphing them.
cookiemonster
cookiemonster
- #30
cookiemonster
- 978
- 0
y = 1/(-1)*x + 26.4
simplifies to
y = (-1)x + 26.4
which is in the form y = mx+b, so treat it like you treated any others.
I'm not really sure what you're asking by "the slope can't go into the left quadrant."
cookiemonster
simplifies to
y = (-1)x + 26.4
which is in the form y = mx+b, so treat it like you treated any others.
I'm not really sure what you're asking by "the slope can't go into the left quadrant."
cookiemonster
- #31
Imparcticle
- 573
- 4
Okay...
What I meant was, the line the slope was describing cannot go into the 2nd quadrant.
true?
What I meant was, the line the slope was describing cannot go into the 2nd quadrant.
true?
- #32
cookiemonster
- 978
- 0
Sure it can.
x = -5, y = 21.4
(-5,21.4) is in the second quadrant.
Here's a question, though. Can this line ever go into the third quadrant?
cookiemonster
x = -5, y = 21.4
(-5,21.4) is in the second quadrant.
Here's a question, though. Can this line ever go into the third quadrant?
cookiemonster
- #33
Imparcticle
- 573
- 4
okay, I got the graph right. I sort of cheated though.
Thank you cookiemonster.
Thank you cookiemonster.
- #34
Imparcticle
- 573
- 4
cookiemonster said:
according to my graph, it can only go into the 4th quad. Did you graph it? I could be wrong u know.
- #35
Imparcticle
- 573
- 4
Now I'm doing the addition method. (The last part of the poster)
x+y=26.4
5y=x
Should you isolate the y intercept in 5y=x into the form y-mx=b?
x+y=26.4
5y=x
Should you isolate the y intercept in 5y=x into the form y-mx=b?
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