Hey guys.
Welcome to Algebra I.
Today's lesson focuses on
determining
if an inverse
variation exists.
Your knowledge of independent
and dependent variables
and patterns
in the coordinate plane
will help you get through
this lesson.
You ready to get started?
Let's go.
Okay, so before we start talking
about inverse variation,
I want to do
a right variation
so you understand
the difference between the two.
Okay?
"Megan earns $10 per hour
working at an electronic store.
How much money will she earn
after working 4 hours?"
We could represent
this situation
a couple of ways
to get the answer.
We could derive
a table of values
and look for a pattern,
or you may be able
to see an equation that
represents the situation.
If we generated a table
of values to figure out
how much Megan
will earn after 4 hours--
need a little space here--
I could say, after 1 hour,
she will have earned $10.
After 2 hours, it would be $40.
I need more room for our work.
After 3 hours--oop--
back this up a little bit.
After 2 hours, she would have
only earned $20
because she is making
$10 an hour.
I tried to give her
a raise.
After 2 hours, $20.
After 3 hours, $30--
extend that.
After 4 hours, $40.
If she's making $10 per hour,
we could tell here
that after 4 hours,
she'll make $40.
If you didn't attack it
by generating a table,
maybe you understood that
I could represent the situation
by the equation y equals 10x.
If I let y represent the amount
of money she's earning
after her work shift,
and x represents
the number of hours she works,
and I want to determine
after 4 hours
how much money
she will have earned,
then I could say,
10 times 4, that's 40.
After 4 hours, Megan earns $40.
This is a direct
variation situation.
We've seen these before.
In a direct variation situation,
as x increases, y increases
at some constant rate, right?
Or as x decreases, y decreases
at some constant rate,
but they are changing
in the same direction.
Now, how inverse variation
is different is,
as x increases, y decreases
at a rate proportional,
so that the product
of x and y is the same.
You really need to see that
so it makes sense.
Let me first show you this.

(female describer)
Inverse Variation.
y equals k over x
or k equals xy;
k is the constant
of variation.

(teacher)
Get that out of the way.
This is our equation
for an inverse variation,
so it differs from that
of a direct variation.
We have,
y equals k divided by x,
or we can say
that k equals x times y.
This is what I meant
when I said that y changes
in a rate proportional to x
so the product of x and y
is always the same.
That second equation represents
what I was saying right there.
Let's get an example
so you can see what I mean.
"In which of these tables
does y vary inversely as x?"
That's what we're
figuring out here.
To figure out
which one of these
is an inverse variation,
we find the product
of x and y.
For each ordered pair,
we multiply x times y,
and we look for the table
that gives us the same answer
every time.
The table where the product
of x and y is constant
will be the table that
represents an inverse variation.
Let's see what we've got here.
I put the work underneath
because it is
our first time doing it.
This time we're multiplying,
so 1 times 2--
Let's get our pen back.
So 1 times 2, that's 2.
Now, 5 times 3/5.
So we throw back a little bit
to remember how to work with
those fractions, but we got it.
5 times 3/5;
I can make that 5.
I can represent it
as a fraction
by putting it over 1, right?
Let's line that up better
so it looks
more in sync here.
Get that out of the way.
Bring that up.
There we go.
I have 5 over 1 times 3/5.
We'll multiply straight across
to get to our answer.
So 5 times 3, that's 15.
1 times 5, that's 5.
And 15 divided by 5, that's 3.
I can tell now,
k is not constant.
I don't have a constant product
for each ordered pair.
I don't really need to bother
to check that last one
because I can tell
the products are different.
So I know that table A
is not the table in which
y varies inversely as x, okay?
Let's check out B here.
Let's see if we get
the same answer
for each product
of x and y, okay?
Let's get a little
space underneath here, okay.
For the first pair,
negative 10 times 3--
that's negative 30.
1 times 5, that's 5.
So, I can tell right there,
k isn't constant.
I'm not getting the same thing
for every product,
my products of "x"s and "y"s.
I can tell right now,
B is not the answer.
So process
of elimination--
it should be C,
but let's run through these
and see that we get the same
product each time, right?
Move that out of our way so
we can work right under there.
For the first pair,
negative 1 times negative 5--
get that pen back.
So, negative 1
times negative 5,
that's a positive 5, okay.
For the second pair,
5 times 1,
that's a positive 5.
Looking good so far.
For the last pair,
we have 10 times 1/2.
Like we saw in the first table,
I put that 10 over 1
to represent it as a fraction,
and multiply straight across.
So, 10 times 1, that's 10.
Going to need
some more room.
1 times 2, that's 2.
I am going--
Let's scoot this over a bit
so I could fit my answer
right beside that.
There we go.
Get that pen back here.
Now I'm ready
to do 10 divided by 2,
and that is 5.
So, I do get that constant
variation every single time.
The product in the third table,
the product of each x and y,
each ordered pair,
was indeed 5.
So, C is the table
that represents
an inverse variation.
Okay? All right.
Now it is time for you
to try one of these, okay?
I'll move this so you can see
that table there.
Press pause and take
a few minutes.
Work your way
through these problems.
Look for the table that is
actually an inverse variation.
To find that out,
check those products.
When you're ready to compare
your answers, press play.

(female describer)
There are three tables,
each with a column x
and column y.
In Table A, the ordered pairs
are negative 5, negative 10;
2, 4; and 6, 12.
In Table B, the ordered pairs
are negative 1/2, negative 4;
6, 1/3; and 8, 1/4.
In Table C, the ordered pairs
are 2, 3; 3, 2; and 4, 1.
Let's see how you did here.
Let's scroll down here;
we can get room to work.
Let's check this first table.
Now, remember, this time
we're checking the products.
So, negative 5
times negative 10.
I write that right here.
Make sure I've got the pen.
That's 50.
2 times 4, that's 8.
So, I can tell,
it can't be A.
Let's check out B.
We've got some fractions.
So, let me work underneath here
to handle that.
Negative 1/2
times negative 4.
So, negative 1/2
times negative 4--
I put that negative 4 over 1
so I can visualize it
as a fraction.
Negative 1 times
negative 4 is positive 4;
2 times 1 is 2;
4 divided by 2 is 2.
For the first product,
I got 2.
I've got another fraction.
I need more room to work
so I can do
that multiplication.
Put these in a calculator
if you want.
It's completely up to you.
This time I have 6 times 1/3,
so 6 times 1/3.
I'm going to put
that 6 over 1.
6 times 1 is 6,
1 times 3 is 3,
6 divided by 3 is 2.
So, B is looking good now.
Let's see what happens
at the last product.
So 8 times 1/4--
going to scroll down here
so I can do that work--
8 times 1/4.
Going to need
a little more space here, okay.
I put that 8 over 1, so I can
visualize it as a fraction.
8 times 1 is 8,
1 times 4 is 4,
8 divided by 4 is 2.
I see I did have a constant
variation in this table.
It was the same
for each ordered pair.
It was 2, so that tells me,
it has to be B.
But I have to show you C
because the math teacher in me
has to check to make sure
that it's not C.
Let's get those
products quickly for C.
So 2 times 3, that's 6.
I'll squeeze it there
and circle it
so you know that was
the answer for that product.
3 times 2, that's 6.
But 4 times 1, that was 4.
So, it couldn't have been C
because you didn't get that
same product every time.
So we were right;
the answer to this one was B.
Let's keep going
with these here.
Got a word problem for you.
"Determine if this
situation represents
"an inverse variation:
"3 people painted
the room in 6 hours,
and 9 people painted
the room in 2 hours," okay?
I want to discuss inverse
variation a little bit.
Earlier, I said that
in an inverse variation,
as x increases, y decreases.
See what I mean
by that in this problem.
Make sure
it's a constant change,
but you do see how 3 people
painted the room--6 hours.
If I increase the number
of people to 9,
it took 2 hours--
a fewer number of hours.
As the more people
were working,
the less time
it took to paint.
We have to make sure that time
was actually constant,
that we saw a constant
variation with that.
For this situation,
we want to use
our inverse variation equation
that is k equals y times x
because we're doing the same
thing as in that table.
We're checking the products.
They presented the values
in a different way here
because we have a word problem.
We start after we have
the equation written down.
We identify our independent
and our dependent variables.
We can control the number
of people working.
Right?
So x,
that's our number of people.
And that is
our independent variable.
The time it
takes to paint the rom
depends on the number
of people,
so that's our y--
the number of hours.
And that's
our dependent variable here.
We make sure
that we have the same product
for each situation here
to know if we have the inverse
variation situation.
Let's get more work space.
For the first part, 3 people
painted the room in 6 hours.
So, k, I already
have y times x--
You can multiply in any order;
it doesn't matter.
I'll follow the y and x
to be consistent.
If y is the number of hours--
for the first situation,
it was 6--
and x is the number
of people, so that's 3.
So, I'd have 6 times 3;
that's 18.
If I also get 18
when I find that product,
then I do indeed have
an inverse variation situation.
Okay? Let's see here.
Get a little more space.
We have 9 people
painted the room in 2 hours.
Okay?
This time, y is our number
of hours.
So, that's 2.
And x is the number of people,
so that's 9.
2 times 9 is 18.
So, we do have
a constant variation.
It's the same thing--
the same constant for each part.
So, we do indeed
have an inverse
variation situation here.
Okay? All right.
Good job on that one.
Let's do another one together.
We want to see, does this
represent an inverse variation?
We've got 6 landscapers
that earn $40 each
and then 3 landscapers
that earned $120 each.
Let's see what's going on.
We're testing out k to see
if it's the same thing--
k equals y times x.
So, here we can control
the number of landscapers.
And how much money each earns?
It depends on how many
landscapers we have.
If you have more people,
they divide that money.
It's a smaller amount
that each person gets.
Because it is divided
amongst more people there.
Here, because we can control
the number of landscapers,
that's our dependent variable--
I'm sorry--
that's
our independent variable,
the number of landscapers.
This is independent because
that's what we can control.
The money they earn,
that's our dependent variable.
We'll say the money earned,
that's our dependent variable
because the money earned
depends on
the number of landscapers
working, okay?
Let's see here;
let's test out these products.
For the first part,
we have 6 landscapers, $40 each.
So, k equals--y is
the money earned, so $40.
x is the number of landscapers,
so that's 6.
Either mental math or your
calculator to get this answer--
40 times 6, that's 240.
So, we've got
the first value for k.
Let's figure out what we get
for the second part.
Now we have 3 landscapers
that earned $120 each, okay?
Let's get a little more space.
Here y is the amount
of money earned,
so this time it was 120.
x is the number of landscapers,
so this time, that's 3.
Either the calculator
or mental math--
120 times 3, that's 360.
You see
that it's not constant.
The first time I got 240;
the second time I got 360.
This does not represent
an inverse variation situation.
I didn't get that same product
for each part of this problem.
All right.
Okay, let's keep going here.
It is your turn
to try one on your own.
Press pause
and take a few minutes.
Work your way
through these problems.
When you're ready, press play.

(female describer)
Determine if the situation
represents
an inverse variation--
4 campers assembled
a tent in 15 minutes,
and 5 campers assembled
a tent in 12 minutes.

(teacher)
Let's see how you did here.
We want to determine if this is
an inverse variation situation,
so the equation we're working
with is k equals y times x.
We see we have 4 campers
assembled a tent in 15 minutes;
5 campers assembled
a tent in 12 minutes.
We can control
the number of campers.
And the time it takes
to assemble this tent
depends on the number of campers
we've got working on it.
Let's get that part down--
x is what we control,
and we said we can control
the number of campers,
so that is
our independent variable.
y, that's going to be the time,
as the time it takes the campers
to assemble the tent
depends on the number
of campers working, right?
So, that's our
dependent variable.
Let's test these values
and see if we get
the same product each time.
Let's get a little
more work space here.
For the first part,
4 campers
assembled the tent
in 15 minutes.
So, k equals--
y is the time, so 15 minutes.
x is the number of
campers, so that's 4.
In your calculator
or mental math--
15 times 4, that's 60.
Okay, let's test that second
part and see if we get 60.
This time we have 5 campers that
assembled a tent in 12 minutes.
So, k equals--y is the time,
which, this time,
is 12 minutes.
And x is the number of campers,
which this time is 5.
In a calculator or mental math,
12 times 5 is 60.
So, we did get
that same product.
We have the same k
for each part of this problem,
so the answer is yes.
This is an inverse
variation situation here.
Okay, good job on that one.
Let's move on and discuss the
graphs of inverse variations.
Earlier, we said
that y equals k divided by x
is also an inverse
variation equation.
It's just solve for y.
We use that one a lot
when we're dealing with
another type of word problem
and the graphs
of inverse variations.
Let's get that out of our way;
there you go.
This is actually the graph
of y equals--
to show an example--
of y equals 6 divided by x.
And you see
for inverse variation,
you don't have a line.
You actually have two curves
for a graph
of an inverse variation.
So, if you see this
for your graph,
you know that you have
an inverse situation--
or an inverse variation
situation.
A little tongue-tied there.
There're two curves,
and they're reflections
of each other.
Remembering
that a reflection is a flip,
if you envision yourself
folding this graph diagonally,
one graph would lay right
on top of the other.
Two curves are reflections
of each other
is what you have for a graph
of an inverse variation.
Let's do a little
practice with the graphs.
"Select the graph in which y
varies inversely as x."
Which one of these graphs
represents
an inverse variation?
We're looking for two curves
that are reflections
of each other.
We've got two graphs up top,
and two we're going
to look at underneath.
If you look at A,
I have curves,
but don't have 2 curves
that are reflections
of each other.
This graph doesn't have that
same style or that same set-up
of the example
we saw earlier.
So, it's not A, okay?
For B, I have
a V-shaped graph.
I don't have curves
at all for this one,
so I can immediately
tell here, it's not B.
Let's scroll down; look
at the two underneath here.
For C, we have a line that
passes through the origin.
This is actually a direct
variation situation,
not an inverse
variation situation.
So, it's not C.
Let's take a look at D.
You've got two curves.
They are reflections
of each other.
So, this is the graph that
represents an inverse variation.
D is the one that you're
looking for, all right?
Keep that in mind, and I want
you to look at this one.
Press pause and work your way
through this problem.
See if you can identify
which graph represents
an inverse variation.
When you're ready to compare
your answer, press play.

(female describer)
Select the graph in which
y varies inversely as x.
Graph A has a straight line
above the x axis,
and it crosses the y axis.
Graph B has two curves
opposite each other
in the lower-left quadrant
and the upper-right quadrant.
They reflect each other.
Graph C has a straight line
which passes diagonally
through the origin.
All right,
let's see how you did.
For A, we have
a horizontal line.
For inverse variation,
I'm looking for two curves that
are reflections of each other,
so it's not A.
Does appear to be B.
I've got my two curves
reflecting each other.
So, the inverse
variation here was B.
To check,
if you take a look at C,
you have a line that passes
through the origin.
That's actually
y varying directly as x,
not inversely as x,
so it's not C.
Our right answer
here was B, okay?
Good job on that.
You've reached the end
of your lesson
on inverse variation.
I hope you saw how knowledge
of the coordinate plane
and independent and
dependent variables
and even direct
variation knowledge
helped you
through this lesson.
Hope to see you back
here for more Algebra I. Bye.