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Hi, class.
Today we're going to be solving systems of equations by graphing.
And before we get into that,
I want to have a short review of graphing lines
to make sure that everyone remembers how to graph a line.
There's two forms of the line that you will mostly likely see.
One form is this slope-intercept form, [ y = mx + b ]
which you have a y by itself
and then you have the slope times x
plus b, and b is always your y-intercept.
The other form which you will graph in a different manner,
is the standard form. [ Ax + By = C ]
And for this, you have some number times x
plus some number times y equals some other number.
Now, first thing you want to do is recognize,
do I have slope-intercept form, or do I have standard form,
or can I manipulate the equation
to make it into one of these forms?
And then, once you determine which form you have,
you can determine the easiest method for graphing.
We're going to start by reviewing
this slope-intercept form. [ y = mx + b ]
With the slope-intercept form, the b,
which in this equation here [ y = -2/3x + 1] happens to be 1,
is your y-intercept.
So if you look at this b,
that's giving you a point on your line
which is (0,b).
And that's telling you where you're going to start
finding your points for your graphing.
So if we look at this particular equation, [ y = -2/3x + 1 ]
the b is 1, so that's telling me that I have a point of (0,1).
You can think of this as this is where you begin.
So we're going to go to the graph over here on the right
and we're going to plot the point (0,1).
And always make sure you label your points.
That's where we're going to start everything.
Now, the slope is the number that's being multiplied by x,
[ -2/3 is slope] right here.
And, if you'll remember,
slope is rise over run. [ Slope = rise /run ]
This is going to tell you how to move from your beginning point
to the next point that's going to be on your line.
If we look at this particular slope in this equation,
it's a negative 2/3. [ y = -2/3x + 1 ]
So, the rise is negative 2 [ -2 ] and the run is 3.
Since the rise is negative,
that tells you you're going to go down from your beginning point
And if it was positive, you would go up.
Since it's negative, we're going to go down.
So we're going to go over to our point
that's our starting point of (0,1) and from there go down 1, 2.
Then we're going to use our run of 3 to move forward 1, 2, 3.
And right here is where
another point on that line happens to land.
And that point, if we label the coordinates,
which you should always do, is the point (3,-1).
Now, if we want to find another point,
we could continue from this point
and go down 1, 2 more and forward 1, 2, 3 more.
And that ends up landing us right over here at the point,
Oops, I'm running out of room there. I'll write it underneath.
(6,-3)
Once you get at least two to three points,
you can go ahead and just connect those points
to create your line, as we did right here.
And that is all there is to graphing
when it's in slope-intercept form.
You first find your beginning point, which in this case was (0,1),
and then you use your slope
to move from that point to other points that you can label.
And then connect your points.
Now, let's look at standard form. [ Ax + By = C ]
When your equation is in this form
as our example right here
of 4x plus 5y equals 20, [ 4x + 5y = 20 ]
it's easiest to graph it by using the x and y-intercepts.
And the x-intercept is when you cross this x-axis.
When you cross the x-axis here, the y-value is zero.
So notice I made a table and I filled in a y-value of zero.
This particular point that we're going to label
across this row is going to be our x-intercept.
When you cross the y-axis which is the y-intercept,
you're crossing this axis right here.
And on that line, the x is zero.
So notice our other point here has a zero filled in for the x.
And this row here is going to represent our y-intercept.
Now, the reason why we choose the x and y-intercepts
is because zero is the easiest value to plug into an equation,
and normally makes things just kind of disappear.
So, what we're going to do is start with this one here,
and we're going to plug zero in for this x right here.
And we're going to see what we get.
So we get 4 times zero plus 5y equals 20. [ 4(0) + 5y = 20 ]
And 4 times zero is just zero, [ 4(0) = 0 ]
so we can just make that disappear
and then solve what's left over here.
And what's left over here is 5y equals 20, [ 5y = 20 ]
so you can just divide both sides by 5, [ 5y/5 = 20/5 ]
and y is 4. [ y = 4 ]
That allows us to fill in a 4 here [ the y column of the table ]
which gives us the point (0,4) to plot on our graph.
So, we're going to go up here and plot the point (0,4).
Now, one thing that you might want to do
when you're doing these x and y-intercepts is,
you might have noticed that really
since this right here [ 4x ] just disappears,
you could have gone to this equation
and just covered this term right here up with your finger
and looked at what's left.
And most people can solve what's left here in their head.
You just divide the 20 by the 5 and you say y is 4.
I call that my 'covering up method.'
You just take your finger and you 'cover up' that [ x-term ],
and divide, in order to solve what is remaining.
Now, let's go ahead and find our x-intercept
by plugging in our zero for our y.
We're going to replace this with zero, [ y = 0 ]
so we have 4x plus 5 times zero equals 20. [ 4x + 5(0) = 20 ]
Again, 5 times zero just cancels and disappears
and what's left is 4x = 20. [ 4x = 20 ]
So we can divide both sides by 4, [ 4x/4 = 20/4 ]
and x is 5. [ x = 5 ]
So we can fill a 5 in, right here [ the x-column of the table ],
and that gives us the point (5,0).
Again, you may have noticed
we could have just covered this up [ 5y ] with our finger
and looked at what was left.
And gone ahead and divided this [ 4x ] by 4
and that [ 20 ] by 4 to get our 5 also.
Now, let's plot our second point. It's over here at (5,0).
You really only need two points
in order to determine where a line goes.
It's fine to choose any other x you would like to plug into this
and solve in order to find a third point.
The third point is nice as a check point
just to make sure your first two points are in the right spot.
But, if you have two points,
that's really going to determine where the line goes.
We just want to make sure they're accurate. And there's your line.
So that's the two different ways for graphing.
Now, let's see what a system of linear equations is
and how we solve those by graphing.
We'll need to recognize whether we're dealing with equations
that are in slope-intercept form or standard form,
because that will determine
whether we're going to graph like we did in this example
or graph like we did in this example.
Now, when we're graphing, really all you're going to do is,
a system is two equations.
You're going to graph them both on the same coordinate plane
and see where they cross.
Now, you need to keep in mind one of three things will happen.
And so those three things you're going to have to recognize
in order to know what kind of an answer you're going to get.
So, you may have a graph of a coordinate plane.
And you may actually graph both lines.
Say, a line like this and a line like that.
And they end up crossing like you see right here.
If that happens, all you're going to want to do
is read off what the point is for where they cross
and give me the ordered pair (x, y).
However, there's two other situations that could happen.
You could have your coordinate plane
and you graph your first line and it lands here.
And then you go to graph your second line,
and it lands right here.
And you notice that they're parallel.
If they're parallel, they're never, ever going to touch.
Which means you'll never have a spot where they cross.
And what we're looking for in these problems
is where do the two lines cross.
So if something like this happens, your answer is "no solutions."
Now, the other situation that may happen is
you may have a coordinate plane, we're going to put it over here,
And you graph your first line, which I'll do in red,
and it lands right here.
And you go to graph the second line, which I'll do in blue,
and it lands right here.
And you'll notice that both lines are in the same exact spot.
When both lines are in the same exact spot,
what that means is
that the two equations were really the same line.
They just may have looked different.
Maybe one was in slope-intercept form
and one was in standard form.
But, essentially, they're the same exact line.
When that happens,
your answer is that there's "infinite solutions."
And that's because every point that works in one line
is going to also work in the other line.
Now, when you're dealing with these,
there are some fancy names for these kinds of answers.
And you may or may not be required to give those types of answers.
These types of situations here
where you actually have a point for your answer
are what we call "independent."
The type of situation over here
where the lines were the same line
and there's an infinite number of solutions, we call "dependent."
And then this other situation here where they're parallel,
we also call "independent,"
but they are "inconsistent,"
because they never cross.
Now, a way to remember these particular situations,
there's a little story,
some of my students actually came up with that you might enjoy.
This situation right here where the two lines are independent,
you could think of as two people had a blind date,
they didn't like each other, so they went their separate ways.
And, so, they're independent of each other
for the rest of their life.
Over here, you have two people that met,
fell in love, and were married forever.
So, they were dependent on each other for the rest of their life.
And then the other situation which is inconsistent,
is the two people that never ever crossed paths, never met,
so they don't even know if they like each other.
So that's one little story that might help you
remember these things that my students made up
and they kind of enjoyed them.
So, now, let's take a look at a system of equations
and how we are going to go about solving them.
We're going to first do a system where we have the two equations,
y equals 1/3x minus 2 [ y = 1/3x - 2 ]
and then the other equation is going to be
y equals negative 2/3x plus 1. [ y = -2/3x + 1 ]
Okay, now,
For this first situation,
actually, let's get a coordinate plane on here
and then we'll go ahead and graph it.
We want to go ahead and graph the first one.
And then we're going to graph the second one,
and then we're going to read off the coordinate plane
as to where they actually cross.
Now, keep in mind that since we don't have graph paper on here,
that we may have a little bit of human error.
It's very important to have graph paper
when you do these particular problems
because it's very easy to make a mistake.
You can use the slope
since these are both in slope-intercept form,
to try and make yourself a little bit more accurate,
but you'll want to be very careful
when you're on a coordinate plane with your two equations
to make sure that they're very accurate.
Graphing is actually
not the most preferred method of doing these equations
because there's the higher probability of some human error
when you're putting them on the coordinate plane.
These are both in slope-intercept form
because they both have y by themselves.
So we're going to be graphing both of them
using the methods that we used for slope-intercept form equations.
Now, your system will typically have two equations.
Often you see a set bracket off to the side. [ { ]
Not always, but some books, you will see that.
What we're going to do is I'm going to color code this
so that this first equation is going to be graphed
over on a coordinate plane in blue.
The second one is going to be red, and that's just
so that you can keep them separated and know which one is which.
It's very important to completely graph the first one
and have it on your coordinate plane
before you ever even look at the second one.
So, on this first one,
I first notice that my b is negative 2. [ y = 1/3x - 2, b = -2 ]
That means that I have a starting point
where I'm going to begin of (0,-2).
So let's plot that on our coordinate plane. This is (0,-2).
Now, we're going to use our slope, which is right here, [ 1/3 ]
in order to find another point,
so that we can connect it with our first point
and see where our line goes.
And the rise here is 1 and the run is 3.
That means we're going to go up 1 and forward 3.
So let's go to our starting point
where we're going to begin at (0,-2)
and go up 1 and over 1, 2, 3.
We land right here which is the point (3,1).
So, let's go ahead and label our point (3,-1)
because that's (3,-1).
Now, it's nice to have a third point just to check
and make sure that everything is lined up,
so we're going to go from here up 1 more and forward 1, 2, 3 more
And we land right here which is at the point (6,0).
We can now connect those and we get our line.
Notice I completely finished off the blue one
before I've even looked at my one that I'm going to do in red.
The blue one is now on the graph and completely done.
So, looking at our red one, we have a plus 1 here.
That means our beginning point is going to be (0,1).
[ y = -2/3x + 1 ] So let's go ahead
and plot that as (0,1), like that.
Now, if we look at our slope here [-2/3], our rise is negative 2.
So we're going to go down 2, and our run is 3,
so we're going to go forward 3.
So, we're going to start at our beginning one, spot,
and go down 1, 2, and forward 1, 2, 3.
And we land right there.
And that is the point (3,-1).
I can find another point by going down 1, 2, more
and forward 1, 2, 3 more.
And I land right here at the point (6,-3).
Connecting those, I get a picture of my red line.
And notice they cross each other at this point (3,-1).
So, that is going to be the solution to our system.
And we will say that "The solution is (3,-1)."
Now, because there's so much error in graphing,
one thing to keep in mind when you're doing systems by graphing,
is that it's very important to check your answer
in order to make sure that it's correct.
Just in case you had some error in your graphing.
So, we're going to go ahead and do the check.
If your directions say, "Is the point a solution to the system?"
the check is all that the problem is asking you to do.
Let's do our check right here.
When you're checking your point, which in this case is (3,-1),
you essentially just plug it into both equations
to see if it works.
So we're going to start with
y equals 1/3x minus 2. [ y = 1/3x - 2 ]
We're going to plug in the negative 1 for the y
and the 3 for the x and see what happens. [ (-1) = 1/3(3) -2 ]
So, first thing I notice is these 3s cancel. [ 1/3(3) = 1 ]
So this turns into negative 1 equals 1 minus 2, [ -1 = 1 - 2 ]
or negative 1 equals negative 1. [ -1 = -1 ]
And that checks out and is true,
so it does work in the first equation.
Let's go to our second equation
of y equals negative 2/3x plus 1. [ y = -2/3x + 1 ]
Replacing the y with negative 1 and the x with 3,
we can calculate. [ -1 = -2/3(3) + 1 ]
The 3s here cancel, and so we end up with
negative 1 equals negative 2 plus 1, [ -1 = -2 + 1 ]
which means negative 1 equals negative 1. [ -1 = -1 ]
And that also checks out.
So we know that the solution we came up with, is correct.
Let's do one more of these and see what happens.
Our next system is going to be
y equals negative 4/5x plus 4. [ y = -4/5x + 4 ]
And the other equation is going to be
4x plus 5y equals 20. [ 4x + 5y = 20 ]
Now, we first are going to make a coordinate plane over here.
And label it.
We have our y-axis and our x-axis.
And we'll put some values on here so we know our scale.
Now, I'm going to again color code this.
This one will be in blue and this one will be in red.
If we first look at our blue one here,
it's in slope-intercept form.
So we can graph it the same way
we graphed the problems in the last example.
[ y = -4/5x + 4, b = 4 ] So, this 4 is our y-intercept
which is going to be (0,4).
So we're going to go ahead and plot that right here, at (0,4).
Then this slope here [ -4/5 ] is telling us how to move
from the starting point to the next point.
And our rise is -4, so we're going to go down 4.
And our run is 5, so we're going to go forward 5.
So, starting here, we're going to go down 1, 2, 3, 4.
And then forward 1, 2, 3, 4, 5 and we get a point right there.
And that point happens to be the point (5,0).
Now, we can actually just go ahead
and connect those,
and we have our line.
Let's go to the red one. [ 4x + 5y = 20 ]
Now, the red one, I notice is in standard form. [ Ax + By = C ]
So I'm going to need to use my x and y-intercepts.
So, let's make a table
for recording these in.
I'm going to first plug in zero for x,
and then I'm going to plug in zero for y and get two points
that will help me find where this line is located.
So, first, plugging in zero for x, [ 4(0) + 5y = 20 ]
notice that the x term disappears. [ 4(0) = 0 ]
And I have 5y equals 20. [ 5y = 20 ]
So I'm going to go ahead
and divide both sides by 5. [ 5y/5 = 20/5 ]
And y is 4. [ y = 4 ]
So now I have the point (0,4).
So, I'm going to locate that up here and label it.
Notice that I'm completely ignoring the blue one
while I'm doing the red one.
You'll want to do that so that you don't get the two confused.
Now, I'm going to go ahead and take a look
at plugging in zero for the y.
That gives me 4x plus 5 times zero equals 20. [ 4x + 5(0) = 20 ]
Notice that this disappears. [ 5(0) = 0 ]
I divide both sides by 4 [ 4x/4 = 20/4 ]
and x ends up being 5. [ x = 5 ]
So that gives me the point (5,0).
And then I'm going to go ahead and plot that (5,0), right here.
And connect the two red ones
and notice that the red one and the blue one
are in the same exact spot.
That means that these two equations were really the same line.
They just looked different.
One was in slope-intercept form and one was in standard form.
That means that we have infinite solutions.
And that is how you go about
solving systems of equations by graphing.