Hey guys, welcome to Algebra 1.
Today's lesson's going to focus
on determining
if a direct variation exists.
Now, your knowledge
of independent
and dependent variables
and patterns
and the coordinate plane
are really gonna help you
get through this lesson.
You ready to get going?
Let's start.

(Describer) She stands at a screen.

To warm you up a little bit
about direct variation,
I want to first start with
this particular word problem.
Let's take a look.
"Meagan earns $10 per hour
"working
at a local electronics store.
How much will she earn
after working four hours?"
There's a couple of different
ways you could attack this
and get to the answer.
Maybe you would want to use
a pattern
and just develop
a few situations
until you got to the point where
Meagan had worked four hours.
This is what I mean.
After the first hour,

(Describer) She writes.

Meagan will have earned $10
if she's making $10
for every hour she works.
At 2 hours, that would be $20
that she's earned.
After 3 hours,
that would be $30.
Let's scroll down a little bit.
After 4 hours,
she will have earned $40.
Maybe you could generate,
which really is
a table of values,
and be able to tell
that after 4 hours,
Meagan will have learned $40.
That's one strategy.
Something else you may have done
in order to figure out how much
Meagan earned after four hours
is maybe intuitively
you figured out an equation
to represent this situation.
If I wrote an equation,
y=10x, let's say,
and I want to say
that y is going to be
the amount of money
Meagan earns,
and x is going to be
the number of hours she works.
If she's working for 4 hours,
that would be 10x4,
and I'd see there that y is 40.
I could get to my answer there
and be able to tell
that after 4 hours working,
Meagan will have earned $40.
What you did right here
without even knowing
is you were working with
a direct variation situation.
When you have a relationship
that's a direct variation,
all that means is y changes
directly as x changes.
It varies directly.
As x increases, y increases
at some constant rate.
That's all
a direct variation is.
Let's keep going with this.
I want to show you
the equation formally
because there's actually
a couple of different equations
you'll work with
for direct variation.
Let's get that out of our way.
I could write a direct
variation equation as y=kx,
where k is
that constant of variation.
Like in our previous example--
let's get that up here
just to remember--
we said that we could work
with the equation y=10x.
We would call ten
the constant of variation.
It's that constant rate
at which
you're seeing that change.
When you want to figure out
what your constant rate is,
say you've already been given
a table of values,
and you want to know what is
that constant change,
you could figure out what k is
by using the equation k=yx.
All that is is that we've taken
y=kx and we've solved for k,
like a literal equation.
Let me show you what I mean
just right underneath here.
If I have y=kx
and I want to solve for k,
I could divide both sides by x
from what we know about solving
literal equations.
The xs would cancel out,
and I can see right here
if I just flipped it around,
law of symmetry,
that k=y/x.
You're going to be working with
these two different equations
as we go through these
direct variation problems.
Let's keep going, let's get
some practice with this now.

(Describer) The next example has three tables. Each has an x column and a y column, and each column has three numbers.

(female describer)
The next example
has three tables.
Each has an x column
and a y column,
and each column
has three numbers.
Let's take a look at this.
"Which of the following tables
shows that y varies directly
as x?"
What we want to do
in order to solve this problem
is we want to figure out
what is k.
Which one of these tables
has a constant of variation,
where my y divided by x is
the same for every ordered pair.
We're just gonna run through
these three tables and see
which one actually gives us
what we're looking for.
Let's see here.
Working with table A,
I'm just gonna put my work
right underneath this,
because I know this is the first
time we've tackled this style.
If we see four divided by one.
We want to know y divided by x,
what is that?
We're trying to figure out
what k is.
In this case, it's four.
Let's take that second
ordered pair in that table,
I'm going to scroll down
a little bit.
Now I'm going to do
six divided by two.
Let's see what that is.
Six divided by two,
that's three.
I really don't need to go
any further
because I can tell right now
that I don't have
a constant of variation.
In my first ordered pair,
y was obtained
by multiplying x by 4
because 1x4 would give me 4.
In that second pair,
my constant was three.
I got y by multiplying 2x3.
I'm not getting
a constant rate of change
for each of these ordered pairs.
I can tell right now,
A is not gonna be the table
that I'm looking for.
Let's cross that one out.
Now let's tackle B.
Again, what we're going to do
is just divide.
Y divided by x.
We're going to do that
for every ordered pair.
We want to see
which is the table
that gives us the same value
every time we do y divided by x.
We want that constant.
Let's try this one out.
The first ordered pair
we have three divided by one.
Let's close that up a bit,
there we go.
That's three.
If this next ordered pair is
three when we do y divided by x,
then we're on a good roll here.
As I'm looking at it,
eight divided by two,
that's four.
Again, I don't have
a constant rate of change
for each set of ordered pairs.
B does not represent
a direct variation.
Process of elimination,
it has to be C,
but you know the math teacher
in me has to work through these
to verify that.
Let me show you.
For that first set,
five divided by one,
that's five.
Now, ten divided by two
is also five.
Let's scroll, get a little more
work space up here.
Scroll up a bit,
catch that table.
The last set of ordered pairs
is 15 divided by 3 which is 5.
You see, we got five
every single time.
We found that quotient,
that y divided by x.
Table C, that is the table
that's representing
a direct variation.
That's the one you want.
Let's try another one here.
Actually, I do believe
it is your turn.
So go ahead and press pause.
Take a few minutes
and I want you to figure out
which one of these tables
represents a direct variation.
Remember from our previous
example, to check that,
just y divided by x
for every set
of ordered pairs in your table.
You're looking
for that constant.
After you've gotten
your answer,
go ahead and press play
so I can see what you got
and compare our answers.

(Describer) Title: Which of the following is a direct variation?
Each of the three tables has three numbers in an x-column on the left and a y-column on the right.
In Table A, the ordered pairs are 1 and 6, 2 and 10, and 3 and 9.
In Table B, the ordered pairs are 1 and 8, 2 and 16, and 3 and 24.
In Table C, the ordered pairs are 1 and 10, 2 and 24 and 3 and 18.

(describer)
Title: Which of the following
is a direct variation?
Each of the three tables
has three numbers
in an x column on the left
and a y column on the right.
In Table A, the ordered pairs
are one and six,
two and ten,
and three and nine.
In Table B, the ordered pairs
are one and eight,
two and 16,
and three and 24.
In Table C, the ordered pairs
are one and ten,
two and 24, and three and 18.
Let's see how you did.
Gonna work through these,
and let's figure out
which one of these tables
represents a direct variation.
For table A, we know we're doing
y divided by x.
This time,
I'm just gonna write my answers
just right off to the side here
since I know that you know
that we're dividing
y divided by x.
Six divided by one, that's six.
Ten divided by two, that's five.
If you did keep going,
nine divided by three,
that's three.
It can't be A because
I didn't have a constant.
It wasn't the same answer
every time.
Let's try B.
Eight divided by one,
that's eight.
Sixteen divided by two,
that's eight.
Looking good so far.
Twenty-four divided by three,
also eight.
The correct answer for this one,
it was B.
I do want to run through C
just to make sure
that you didn't make
any mistakes there
and maybe picked C
for the answer.
If we find these quotients,
ten divided by one, that's ten.
Twenty-four divided by two,
that's 12.
Eighteen divided by three,
that's six.
We can see, C wasn't the answer.
Our correct answer for this one
was indeed B.
Eight was our k.
Eight was that constant
of variation.
Let's keep going with these.
We're gonna do
a few word problems here.
I'm gonna read through it
and then I'm gonna show you how
you attack these.
"Determine if the situation
represents a direct variation.
"Yesterday, Juliet ran
two miles in ten minutes.
Today she ran three miles
in 15 minutes."
Basically, what we want
to figure out here is,
is Juliet running
at a constant rate?
What I do want to start out
by doing is I want to write out
those direct variation
equations.
I want to show you
which one you'll use here
to figure out what's going on.
We know we have y=kx,
and then we also know
that we can use k=y/x.
Now, because we're trying
to determine
if there's
a constant rate of change,
what is that constant
of variation,
we're going to use this equation
here to figure out
if this does indeed represent
a direct variation.

(Describer) Y over X.

Now what we have to figure out
is which one of these values
is y and which one is x.
We've got to throw back
to your pre-algebra skills here.
If you remember,
y is your dependent variable.
Let me write it out this time.
I don't want to abbreviate it
the first time
because I want to make sure that
you know what I mean by that.
Let me just scoot it in,
I'll write it in the middle.
Y is our dependent variable
and x is
our independent variable.
What that means is,
we can control x.
We can control
the independent variable.
The value of y depends on
what the value of x was.
For example, in this situation,
Juliet ran two miles
in ten minutes.
The number of miles she runs
is going to depend on how many
minutes she spends running.
In this situation,
I could say that y is the miles,
the distance that she's running,
and x is the minutes,
or the amount of time
she's spending running.
She can control
how long she runs.
That's her independent variable.
The distance she runs,
the number of miles,
is the dependent variable.
Now that we've identified that,
what we want to do is figure out
for the first situation,
the yesterday,
what's our k there, what's
our constant of variation,
and we want to determine
for the second part,
for the part she ran today,
what's our k there,
what's our constant
of variation.
If we get the same thing
each time,
then we know that this is
a direct variation situation.
Let's do that math,
let's see what we get.
If our independent variable
is the minutes,
our dependent variable
is the miles.
We're trying to figure out
what k is.
K=y/x.
For the first situation,
let's put a little line here
to separate this.
Yesterday, she ran two miles
in ten minutes.
The miles, that's my y.
The minutes is my x.
I'll say, okay,
two divided by ten.
Let's see what that fraction
simplifies to.
If you throw back
to your fraction skills,
I know I can simplify
this fraction
if I just divide the numerator
and the denominator by two.
Two divided by two,
that would be one.
Ten divided by two is five.
For that first situation,
k is 1/5.
When I figure out what's going
on for the today part,
for what she ran today,
the three miles
and the 15 minutes,
if I also get 1/5,
then I have a direct variation
situation here.
Let's see what we get there.
I think I can fit that
right beside.
Let's get it right underneath,
actually.
Just get it a little spaced out
for us.
For today, she ran three miles.
We know the miles is our y.
In 15 minutes, so that's our x.
Again, we end up with a fraction
and we can simplify this.
If we divide our numerator
and our denominator by three,
three divided by three is one,
and 15 divided by three is five.
We see, we did have
a constant of variation here.
We did get 1/5 for both times.
Does the situation
represent a direct variation?
Yes it does.
We did have
a constant rate of change.
Let's try another one together.
Then I'm gonna let you try one
on your own and see how you do.
Here we go.
"James prepared six sandwiches
in three minutes
and then prepared 12 sandwiches
in four minutes."
We're assuming here
James must work
at some fast food restaurant
or something like that
because he's working
pretty fast here.
First, let's write
our direct variation equations.
What we saw from the last one,
this is actually the one
that we need here.
We need the k=y/x.
We'll go ahead and we'll start
by writing that one down.
Now we need to figure out,
we know that y
is our dependent variable.
This time I'll abbreviate it
because I know that you know
what I mean.
X is our independent variable.
In this situation,
the number of sandwiches
that James prepares
depends on how much time
he's spending working.
He can control the time.
The number of sandwiches
depends on the length of time.
For this situation, we'll say
our dependent variable,
that's our sandwiches,
and our independent variable,
that's the minutes,
that's the time.
We've got that part sorted out,
so now we can go ahead
and figure out
what exactly is the answer
to this question.
Is it a direct variation
situation or not.
Let's see here.
James prepared six sandwiches
in three minutes.
That's the first situation.
Six sandwiches, so that's the y.
In three minutes,
so that's the x.
We have six divided by three,
that's two.
If, when I figure out the second
part of this, if I also get two,
then I know that this is
a direct variation situation.
Let's test that out.
This time,
then he prepared 12 sandwiches.
That's the y.
In four minutes, that's the x.
Twelve divided by four,
that's three.
Because I did not get
the same k, the same constant,
for both situations,
this is a no.
This does not represent
a direct variation situation.
My constant is not constant.
I'm not getting the same thing
every time here.
Good job with that.
Now it's time for you
to try one.
Go ahead and press pause,
take a few minutes,
work your way
through this problem,
and when you're ready to check
your answer against mine,
press play.

(describer)
Titles:

(Describer) Titles: Determine if the situation represents a direct variation.
Yesterday, Alan washed five cars in 20 minutes.
Today, he washed 8 cars in 32 minutes.

Determine if the situation
represents a direct variation.
Yesterday, Alan washed
five cars in 20 minutes.
Today, he washed
eight cars in 32 minutes.
Let's work this one out.
Let's see how you did.
"Yesterday Alan washed
five cars in 20 minutes.
Today he washed eight cars
in 32 minutes."
You know we've been starting out
by first writing
the direct variation equation
that we're going to be using,
then we go ahead and we identify
what's our independent variable,
what's our dependent variable.
We know that y
is the dependent variable.
We know that x is
the independent variable,
the one we can control.
In this situation, the number
of cars that Alan washes
depends on how much time
he's washing the cars.
How much time he's working.
Our dependent variable,
that's the number of cars
he gets washed.
Our independent variable,
that's the minutes,
that's the time.
I abbreviated minutes there.
What we need to do is,
we need to see,
for yesterday
and then for today,
do we have
a constant of variation here.
Let me scroll a little bit
and let's check this out.
For the first situation,
y is the cars.
X is the minutes.
Five cars in twenty minutes.
Five divided by twenty.
I can simplify this fraction
if I divide the numerator
and the denominator by five.
Five divided by five,
that's one.
Twenty divided by five,
that's four.
We're good there.
Now let's check out
the second part of this.
Today he washed eight cars
in 32 minutes.
The number of cars is y,
the minutes is x.
Eight divided by thirty-two.
Let's move that
out of our way there.
I think it can simplify this
by dividing the numerator
and the denominator by eight.
Eight divided by eight is one.
Thirty-two divided
by eight is four.
What do you know; we do have
a constant of variation here.
We got 1/4 for each part
of this problem.
Does this represent
a direct variation situation?
Yes, it does.
Good job on that one.
Let's do a little more with
these direct variations here.
This time, we're gonna discuss
the graph of a direct variation.
To understand this,
we actually use
the other direct variation
equation, the y=kx.
It probably looks
a little familiar to you
because it's not much unlike
slope intercept form.

(Describer) Y equals mx plus b.

In the case of direct variation,
your y intercept, or your b,
is always zero.
K is kind of like your slope.
You're thinking about it
in the same kind of way.
Thinking back
to our first example,
when Meagan was working at
an electronics store
earning $10 per hour,
that could be represented
by the equation
y=10x.
If we take a peek at the graph
of that equation,
let's get this
out of our way here,

(Describer) She uncovers it.

then you can see
that that actually does go
through the origin.
Let's erase that point.
Don't think it's part
of the graph.
We see that the graph of y=10x
does pass through the origin.
That's true for every single
direct variation equation.
Regardless of what your k is,
what your constant is,
the graph of a direct variation
will always pass
through the origin.
It's always a line that passes
right through the origin.
It may have a positive slope
or a positive rate of change,
it may have a negative one,
but it will always go
through the origin.
Let's look at some graphs.
"Select the graph in which y
varies directly as x."
We have two up top,
then I'm gonna scroll down,
then we have two
right underneath.
Let's analyze these, keeping in
mind that a direct variation,
the graph of that
will always be a line
that passes through the origin.
If we take a look at A,
you don't have a line there,
you've got some curves.
You can tell that A is not
what we're looking for here.
This is not
the direct variation.
B, you see I have
a V-shaped graph.
That's actually the graph
of an absolute value.
You'll get more into that
in Algebra 2.
I know that what I'm looking for
when y varies directly as x,
when I have a direct variation,
I'm looking for a line
that passes through the origin.
I don't have a line here.
It's not B.
Let's look at these two
underneath here.
For C, I'm getting better here
because I do have a line,
but this line doesn't pass
through the origin.
It's not C.
Let's check out D.
I've got a line and my line
passes through the origin.
D is actually the graph
of a direct variation.
You see how we analyze
all those graphs
to be able to determine
which one
was a line that passed
through the origin.
That's the graph that's going
to be your direct variation.
Let's move on to the next one.
It is your turn,
so press pause.
Take some time
and analyze these graphs.
See if you can determine
which one
represents a direct variation.
When you're ready to check
your answer, press play.

(Describer) Title: Select the graph in which y varies directly as x.
Three graphs are displayed.
In Graph A, a straight line crosses the y-axis, It is above the x-axis and is parallel to it.
In Graph B, a straight diagonal line sloping downward crosses the positive sections of the x-axis and the y-axis.
In Graph C, a straight diagonal line sloping downward goes through the origin, where the axes cross.

(describer)
Title: Select the graph
in which y varies
directly as x.
Three graphs are displayed.
In Graph A, a straight line
crosses the y axis.
It is above the x axis
and is parallel to it.
In Graph B, a straight diagonal
line sloping downward
crosses the positive sections
of the x axis and the y axis.
In Graph C, a straight diagonal
line sloping downward
goes through the origin
where the axes cross.
Let's see how you did here.
For A, I did have a line, but it
didn't pass through the origin.
It wasn't A.
For B, I did have a line, but it
didn't pass through the origin.
It's not B.
Now, for C, I've got a line
and that line passes
through the origin.
C is the graph that represents
a direct variation.
Great job, guys, working with
those direct variation problems.
I hope you saw
how your knowledge
of independent
and dependent variables
and the coordinate plane
and patterns
helped you get through
this lesson.
Hope to see you soon
for more Algebra 1, bye.

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