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>> Instructor: Hi there, this is Professor Shannon Gracie starting with Chapter 4, which
introduces us to solving systems of linear equations. So, before in our last chapter
remember we're using the Blitzer combo book for introductory and intermediate algebra
for college students and the last chapter we learned all about lines; how to make a
picture of a line also known as graph a line, how to make the equation what we call naming
the line so the equation that goes with the line given certain information and now what's
going to happen is we are going to talk about how we can have two lines on the same coordinate
plane and what it means for their solutions. So, here we go. Why don't you guys go ahead
and warm up. Pause the movie and warm up and then we'll see how you did. On your mark,
get set, go. Okay, let's see how the warm up went. We've got the first part we needed
to determine if the given number ordered pair is a solution to the given equation. So remember
on the first one we just have a single variable. So we would plug that 18/5 into the, in for
the x so we would have 5 times 18/5 plus 3 is that equal to 21? So, our x value took
on 18/5 and then 5 goes into itself once so you see we get 18 plus 3 is that equal to
21? Twenty-one sure equals 21. So, we're good. Yes. All right on to Part B. This time we
have a linear equation and two variables. So we have an ordered pair that we need to
plug in. So for the x we're going to plug in 4 and for the y coordinate we'll plug in
1. So this will be negative 4 plus 2 times 1 is that equal to 0? This is what we plugged
in. So negative 4 plus 2 is that equal to 0? Negative 2 is not equal to 0. So, we would
say no. Now let's think about what this means. If an ordered pair is not a solution to a
linear equation and two variables, that means that ordered pair is good. It's not located
on that particular line. Now, let's take a look. So keep that in mind. We'll be coming
back to that thought when we start learning about systems of linear equations. All right.
Let's see how you did on the graphing portion. Graph the line which passes through the points
0, 1 and negative 5, 3. Let's name our axes. I'll just have a square scale and so 0, 1
is here and negative 5, 3 is here and then we just connect the dots. Okay. How did you
guys do? Awesome. All right. So, next up systems of linear equations and their solutions. We
have seen that all equations in the form ax plus by equals c and that is the standard
form of the line and the equations of this type when x is raised to the first power and
y is raised to the first power are straight lines when graphed. Two such equations are
called a system of linear equations or a linear system. A solution to a system of 2 linear
equations in 2 variables is an ordered pair. That satisfies both equations in the system. So going back to our warm up,
the second warm up actually 1b, we found that the ordered pair wasn't a solution to that
line and that meant that the ordered pair at that point was not located on that line.
Now, for a system of equations you've got two lines. So, the ordered pair would have
to be located on both lines and what'll happen is that point will represent the point of
interception. So here we go. Let's check it out. Now we have a negative 2, which we'll
plug in for each x coordinate so we've got to do this twice, and negative 5, which will
go in for each y coordinate. So, first up we do the first line. We say 6 times negative
2 minus 2 times negative 5 is that equal to negative 2? Again, plug that in for x and
that in for y so we'll get negative 12, minus a minus is plus, is that equation to negative
2? Negative 2 is equal to negative 2. So we've got one part of it done. So, we have determined
that that point is located on the first line. Now, we have to do the same thing all over
again except we do it on that second line. So 3 times negative 2 plus negative 5 is that
equal to negative 11? So, now we're working on that second line in that first example.
So we get negative 6 plus a negative is minus 5 is that equal to negative 11? Negative 11
equals negative 11, check. So, what have we found? Yes, and let's write out a little more
so you guys know what the heck you found. Negative 2, the ordered pair negative 2, negative
5 is a solution of the system. All right. Go ahead and pause the movie and you guys
give Part B a chance. All right? Let's see how you did for Part B. First up, again, in
for the x's we'll be putting 10 and in for the y's we'll be putting 7. So, the first
one 6 times 10 minus 5 times 7 is that equal to 25? So this is what was plugged in. Now
we need to see if it's true. Sixty minus 35 is that equal to 25? Well, 25 equals 25 so
the point is located on the first line. Now, we just need to check the second one. So then
4 times 10 plus 15 times 7 is that equal to 13? All right so 40 plus let's see that'll
be 105 I believe is that equal to 13? So, 145 is not equal to 13. So that means that
the ordered pair 10, 7 is not located on the second line. So what have we found out? Ten,
seven is a solution to the first line. So it's located on the first line but not the
second line. So, as far as the whole system goes, we would say good, no. Ten, seven is
not a solution of the system. Good. All right. So moving right along solving linear systems by graphing.
The solution of a system of two linear equations and two variables can be found by graphing
both of the equations in the same rectangular coordinate system. For a system with one solution
the coordinates of the plane of intersection give the systems solution. All right. So,
your answer will be if you get one answer it will be in [inaudible] notation and it
will be an ordered pair instead of just a single value. Okay, steps for solving systems
of two linear equations and two variables, x and y, by graphing. Graph the first equation.
So you'll be using all your Chapter 3 skills here. Then graph the second equation on the
same set of axes, so the same grid. If the lines representing the two graphs intersect
at a point, determine the coordinates of this point of intersection. The ordered pair is the solution of the system and then, of course, and we won't be doing
this in the lecture but it's always good for you to check the solution in both equations
to see if you get true statements. All right. Now the thing, this is just one method. We're
going to learn three different types of methods for solving systems of linear equations. Solving
by graphing is actually what your calculator uses. If you have ever used a graphing calculator,
which we won't be using in this course, the graphing calculator basically plots a bazillion
ordered pairs in order to make the picture of the graphs and then gets a very good approximation
of the intersection point. Now, your solution is only going to be as good as the graph that
you make or the two graphs that you make. So it's very important in this section for
you to use straight edges to draw your lines and graph paper so that you get a really good
grid going. So here we go. Use the graph below to find the solution of the system of linear
equations. So these steps have all been done, steps 1 and 2 graphing the equations, that's
been done for you. Now, step 3 do these lines intersect? Yeah. They intersect. So, let's
check it out. They seem to intersect right here. Now to me that seems to be the ordered
pair negative 1 because it's 1 to the left and then negative 5. So, guess what? The solution
to the system is the ordered pair negative 1, negative 5 and [inaudible] notation because
that is the only ordered pair that will work. How are we doing so far? You hanging in there?
Okay, good. Now, next step you need to do the solving. Now, you're going to have to
decide how you like to graph these lines. You could either use intercepts, maybe Part
A, I might use intercepts Part B. It's already in which format? Good. Slope intercept form.
So, you'll have to make that decision. So I'd like you to pause the movie and at the
very least you can graph the lines. So you're going to end up graphing 6 lines, 2 per grid,
2 per coordinate plane. If you feel confident, you can go ahead and find the solution to
each one of them. Okay? Good. On your mark, get set, go. You can do it. All right let's
see how you guys did. As I mentioned before for these, I think I would like to use intercepts
for Part A. So, I'll do, I'll graph this line in blue and we'll do the other line in green.
So, x and y. If x is 0, do you see that y is going to be 2? And if y is 0, x will be
good, 2. So, we get, I'm going to use a regular grid. So here's x and y, 10, negative 10,
10, negative 10. So we'll have 0, 2 as the sorted pair, 2, 0 is this one and then again
we just need to be very careful when we're graphing. Okay, now the green line, again,
I'm going to use intercepts so we'll have x and y. If x is 0, do you see that y is going
to be negative 4? And if y is 0, x will be positive 4. So, 0 negative 4, 1, 2, 3, 4,
and then 4, 0, 1, 2, 3, 4. I don't like that second part. There we go. That's much more
better. Okay. Other now we've graphed them. This is the line x minus y equals 4 and this
is the line x plus y equals 2. All right? So, where do they intersect? Good. They seem
to intersect here which it looks to me is the ordered pair 1, 2, 3, negative 1. So,
what have we found out? We found out that the answer will be the ordered pair 3, negative
1 and we're all done. How did you do? Did you at least get the pictures of the lines?
Well, good. That means that you've learned your Chapter 3 stuff. Now you just need to
take that extra little step to connect the Chapter 4 concept with the Chapter 3 concepts.
All right. So let's see how you did on the next one. Now, these lines are already in
slope intercept form. So, the blue, I'll do this one in blue and this one in green like
before. So for the blue line, do you see that M is 3 and B is negative 4. So the y intercept
is other negative 4. Again, I'll do a regular grid. This is x and this is y. Okay, so starting
with the intercept 0, negative 4 for the y intercept I'm going to use the slope, which
is 3. So remember we can do, I'll go ahead and do positive 3 over positive 1, which translates
to be 3 up and 1 right. You know what just so I stay on track I'm going to make a couple
more points using the slope. So I'll go 3 up, 1, 2, 3, and 1 right from that point and
I'll go 1 left and 3 down to get this one. Okay, we've got that line going on and then
now the green we've got m is equal to negative 2 and b is equal to positive 1 so the y intercept
must be good, 0, 1. So, in order to graph that negative 2, I'll go ahead and do negative
2, which is 2 down and then 1 right. So here we go, the y intercept was 0, 1 and then going
2 down and 1 to the right. One, two down and one to the right. One, two down and one to
the right. Okay. All right. So this is the line y equals negative 2x plus 1 and this
here is the line y equals 3x minus 4. So, our solution do you see they crossed here
and that seems to be the ordered pair 1, negative 1. So, what's the answer to our system? Good.
The answer to our system is ordered pair 1, negative 1. All right. So, next example. We'll
do our grid and here's the blue line. I'll go ahead and graph using intercept. So, x
and y. So, if x is 0, do you see y will be 6? If y is 0, x will be 6. So, graphing 0,
6, 6, 0, I'm a little off. Okay. So there's the blue line and then the green line is what
type of line? Good, horizontal line where y is equal to negative 3. This is a horizontal
line and remember the y intercept will be 0, negative 3. One, 2, 3, okay, and they seem
to cross here and that's at 9 and then 1, 2, 3 down. So 9 negative 3 and that's our
answer. The ordered pair 9, negative 3. How's it going? You feeling okay? Great. So now
we get into different situations, different situations for solutions of systems of equations.
So, linear systems having no solution or infinitely many solutions. We have seen that a system
of linear equations in two variables represents a pair of lines. The lines either intersect
at one point, which is what we were doing in the last three examples, are parallel or
are identical. Thus, there are three possibilities for the number of solutions to a system of
two linear equations. Let's check it out. If you have exactly one ordered pair of solutions,
you end up with a situation where you have 2 lines that cross at exactly one point and
the solution is going to be whatever the x coordinate is and whatever the y coordinate
is in [inaudible] notation. They intersect at one point. So let's go over here the two
lines intersect at one point. This is a consistent system. Now, think about when you're driving.
It would be kind of scary if the road randomly crossed, right? If the oncoming traffic just
randomly crossed into the incoming traffic. That's why we have traffic lights to control
all that. Normally the two lanes that are going, you know, against each other are, good,
they're parallel. Right? So, parallel lines will never intersect, right. So, that means
that there's no ordered pair where they intersect. So you have a situation where you've got,
you know, online here and one line here and they never cross. So, the answer to this equation
is the empty set because there are no ordered pairs which will make them, there's no ordered
pairs where these two lines will cross. So let's fill this in. The two lines are parallel
and this is considered an inconsistent system. If your boyfriend is always late to pick you
up or doesn't show, you'd consider him to be inconsistent, right? If he shows up on
time every time, you consider him to be consistent. So, now what if you have a situation so let's
say this is 1, 9 and then what if you have another line that has the same picture that's
right on top of that one. How many ordered pairs would I cross in? A bazillion, right?
Or in math infinitely many ordered pairs is where they cross. So there's infinitely many
solutions. Now the answer looks weird, right? So these intersect at infinitely many points.
So, what the answer looks like is you do set building notations. So this time you have
an ordered pair x and y such that then you pick one of the equations of the line. So,
I'm going to put it in general form ax plus by equals c but what you would do is you would
just pick one of the equations of the line that was in the system. It can be either one
because they end up being the same line. So the two lines are identical. This is a system
with dependent equations and this is at this end this is a consistent system. All righty. So now three more examples
and then we've got an application and we'll be done. So, solve each system by graphing
and if there's no solution or infinitely many so state and use that notation to express
solution sets. So, why don't you guys go ahead and as I said at the very least you can graph
the lines. On your mark, get set, go. All right. Let's see how you guys did. Here's
x and y. I'll just do the regular scale, let's square scale, and here we go. I'm going to
go ahead and graph these using intercepts. So we have x and y. If x is 0, do you see
that y will be 4? If y is 0, x is going to be 4. So, the blue line I have 0, 4, 1, 2,
3, 4, 4, 0, and so our line will look something like this and this is the line x plus y equals 4. Now,
this guy, again using intercepts we've got x and y. If x is 0, you see you'll get 2y
equals 8 so y will be 4 and if y is 0, similarly x will end up being 4. So you've got your
same stuff going on here. This is very important, you know, this shows that so 2x plus 2y is
equal to 8. They're the same line. So, this is something that's very important for the
next sections. Do you see that everything from the blue line if you multiplied everything
by 2 say you get 2 x's plus 2 y's equals 8, so, lines that are multiplies of each other
have the same graph. They're the same line. So, what do we say about this solution? They're
identical lines so that means there's infinitely many solutions. So, our answer will be ordered
pair xy such that the ordered pair has to be on the line. So, let's just go ahead and
choose the first guy, x plus y equals 4 and we're set. So any ordered pair that falls
on the line is a solution to the system. Okay, so next step these are already in slope intercept
form. So for the blue line m is 3, b is negative 1 so the y intercept is going to be 0, negative
1 and we'll have m be 3 up over 1 left. Oh, 3 up over 1 right. Sorry. Remember to stop
me when I make up my own math there. So, here's x and y, 10 negative 10, 10 negative 10. So
this one we're at 0, negative 1 for the intercept. My second point I'll get going 1, 2, 3 up
and 1 right. I'll get a couple more 1, 2, 3 up, 1 right. So, this here is the line y
equals 3x minus 1. Okay. So, the green line m equals 3b equals 2 so the y intercept is
going to be 0, 2 and, again, we have b be 3 up over 1 right. So, here we are at 0, 2
for the intercept and then 1, 2, 3, 1, 2, 3. So all these other points are using slope.
So, what do you think about these two lines? Are they ever going to cross? Good, they're
never, ever going to cross. So, they are parallel lines so that means there will be no solution.
How do we write no solution in math ease? Empty set brackets and you're done. All right
so Part C. So the blue line I'll go ahead I guess and do intercepts. So if x is 0, y
is 0, and if, so we've got to pick a different ordered pair. So now I'll plug in, because
we found both x and y intercept. So here if x is, I don't know we could do 2, 2 times
2 is 4. So, y will be 4. Here's x and y. So, we have 0, 0 and then 1, 2, 1, 2, 3, 4. So,
this is the line 2x minus y equals 0 and then the green line y equals 2x I'll do using slope
intercept so the slope is 2, b is 0, so the y intercept is 0, 0, and we can use m as positive
2 over positive 1. So 2 up and 1 right. So, 0, 0 I think we might have found our point
and then 1, 2 up and 1 right, 1, 2 up and 1, right, so it looks to me like these might
end up being the same line. So, y equals 2x. Again, identical lines. So that means infinitely
many solutions. So our answer will be the ordered pair x, y such that it's on one of
the lines. So what do you want to do this time? We could just say y equals 2x. If you
pick the other line, you're good. Okay, now we're onto our application. Our application
will be the next step. A band plans to record a demo. Studio A rents for $100 plus $50 per
hour. Studio B rents for $50 plus $75 per hour. The total cost, y in dollars, of renting
the studios for x hours can be modeled by the linear system y equals 50x plus 100. So
that's that first one, 50 per hour, that's 50x, and then 100 that's a base cost. So if
you ended up leaving right away, you'd still pay $100. All right so then the second one
75 per hour and then $50 fixed cost. So use graphing to solve the system. So, we're going
to extend the x axis from 0 to 4 and let each tick mark represent 1 unit. So 1 hour in a
recording studio and we're going to extend the y axis from 0 to 400 and let each tick
mark represent 100 units. That's rental cost of $100. So here x and y so, you know, I'm
just going to go ahead and do, I'll have this be 5 just by the way our tick marks are so
this would be, you know, 1, 2, 3, 4 and 5. We don't need negative for X because we don't
have negative hours and then here 0 to 400, well, the way it was recommended this would
end up being we'll call this 1,000 because that would be 10, 100s, so then we'd have
1, 2, 3, 4, 5 this would be 500. Here and here. Okay, so graphing the lines I don't
know how you guys wanted to do it. We certainly have our intercept. So if this, we'll call
this the blue one. So, you see that the slope is equal to 50 and the y intercept is 0, 100.
Now, you know, just to make our lives simpler this is up to you so don't get all freaked
out, isn't 50 if I multiplied 50 by 2 over 2 isn't that equivalent to a slope of 100
over 2? And that will fit with our scale so we can go corner to corner. So 100 up and
2 right. So, let's do our intercept. So we had an intercept at 0, 100 and then we got
our next point by going up 100 and then to the right by 2. So, 1, 2. Okay. All right
and then the green line, m, is equal to 75 and the y intercept is 0, 50 and, again, if
I were to multiply this by 2 over 2, it's going to make it easier to do the graph. So
this would be, m would be, well, let's see, well, I don't know, 150 over 2 a little bit
better, but not perfect, but that's okay. We can work with it. So, 0, 50 would be about
here. Now 150 up would take us to here and then 1, 2 over it looks like we land on the
same spot. So, this one would end up being something like that. So, if you see where
they cross, they cross at the ordered pair, 2, and that was at 200, right? Yes. Two, 200.
So, now let's take a look at, let's put our solution. So our solution is the ordered pair
2, 200 and now what the heck does that mean? So, what that means is if the band uses either
studio the cost is the same, which was, of course, $200. But what happens, you know, if they
use it for less than 2 hours? If the band uses a studio for less than 2 hours, which one
is the better bargain? Good, the green line is below, right, in cost. So the green one
was studio B. Studio B is the better bargain
and you know finally if the band uses studio for more than 2 hours Studio A is the better
bargain. Because if you see after 2 hours the blue line is below the green line so that's
better in terms of cost and not this one, of course, was Studio A and the green one
is Studio B. So there you go. We are all finished for this lesson. Make sure and go do your
homework like I always tell you and have a fabulous day.