Hey, guys.
Welcome to Algebra 1.
Today's lesson's going to focus
on solving
real-world problems
involving systems
of linear equation.
All the practice
that you've had
solving systems of equations
and real-world problems
will come in handy
in this lesson.
You ready? Let's go.

(Describer) She uses a stylus on a screen.

(Describer) She uses a stylus on a screen.

(Describer) She uses a stylus on a screen.

Okay.
Before we begin solving
systems of equations
that deal
with real-world problems,
let's just jump back
and review
real-world problems
in general.
If you look
at this problem,
I won't solve it,
but I'll review the process
to solve it,
just to jog your memory.
"Mateo's monthly cable
subscription is $80.
"As part of his service,
"he can order newly released
movies to view at home
"at a cost of $6 per movie.
"Mateo was charged $104
for his first month of service.
How many movies did he order?"
If you remember how we've
solved this kind of equation,
we would read through it,
and then go back
and highlight anything
that sounded like math--
anything that helps us focus
on the information we need
to set up this equation
and solve for the unknown.
For example, "His cable
subscription's $80."
Sounds like math,
it's a number.
"As part of his service,
"he can order
newly released movies
to view at home at a cost
of $6 per movie."
Let's get that.
"Mateo was charged $104 for
his first month of service."
Then "how many movies
did he order?"
The question
we're trying to answer.
We'd get this information,
we'd set up an equation.
I'll switch to my pen.
So $80,
plus $6 per movie,
if I let X represent
the number of movies.

(Describer) She writes 6x.

(Describer) She writes 6x.

(Describer) She writes 6x.

And then first bill
was $104.
We'd set up that equation
and then take the necessary
steps to solve for X.
At the end, we'd figure out
how many movies
Mateo ordered.
If you need
a memory jog on that,
go look at that video
and see how we worked out
those problems.
Now, to jog your memory
on systems of equations.
A system, in Algebra 1,
we'd stick with 2 equations.
In Algebra 2, we add
more equations to a system.
But for Algebra 1,
you'd have two equations.
There were three
strategies we could take
to solve a system
of equations.
We could use
the elimination method,
where we add
equations together
and try to cancel out
a variable.
For help with that, go back
and check out that video
on solving systems of equations
using elimination.
Another method
was substitution,
where we would take--
for example, in this system,
it's set up well
for substitution because
I see my second equation
is already solved for Y.
So I'd insert--I'd substitute
that 2X plus 10
back into the first equation.
That would help me get through
and figure out my solution.
To see that worked out,
watch that solving systems
using the substitution
method video;
that would really help.
The third way
to solve systems
of equations is graphically.
We'd solve for this
first equation,
for Y, in this case--
put them both in the calculator,
graph them, and look
for the point of intersection
to find the solution
to our system.
What we're going to do here
is combine these two things
that we know.
We'll combine our knowledge
of solving real-world problems,
and of solving
systems of equations,
and we'll get through
this lesson
about solving
real-world problems
involving systems
of equations, okay?
Let's get started.
Well attack this
and see how we do it.
"Tickets at the local
movie theater
"are $10 for adults
and $8 for children.
"Yesterday the manager
"reported $1,360
in admission sales
"for yesterday's 11 a.m. show.
"If 150 tickets
were purchased
"for yesterday's
11 a.m. show,
"how many
were adult tickets?
How many were tickets
purchased for children?"
I know is seems
like a lot going on there,
but we'll use our steps
to solve it.
We've read through it;
now we'll go back
and highlight everything
that seems like it's math,
anything that helps us
solve the problem.
Pull out your highlighter
and I'll switch
to mine up here,
and let's get going.
"Tickets at the local
movie theater
are $10 for adults--"
seems like something
I'll need to know--
"and $8 for children."
More things
I'll need to know.
"Yesterday the manager reported
$1,360 in admission sales--"
Okay.
Ooh, here we go,
my highlight didn't want
to stay there.
"--for yesterday's
11:00 am show.
"If 150 tickets were purchased
"for yesterday's
11:00 a.m. show,
"how many were adult tickets?
How many were tickets
purchased for children?"
Okay, so now that we've
highlighted the information,
let's attack this and see
how we'll set up this system.
When you're solving
real-world problems
involving systems of equations,
there will be more than
one unknown you're solving for.
If we go back and look at
the information we highlighted,
we see we're dealing with
$10 for adults, $8 for children,
$1,360 in admission sales--
so we're dealing
with money, in one case.
Then we see, we have
"150 tickets were purchased.
How many adult tickets,
how many children's tickets?"
So we're also dealing
with tickets.
We want to figure out
how many adult tickets were sold
and how many
children's tickets were sold.
To find that information,
we need to involve
the amount of sales,
the money for the sales,
and the number of tickets sold.
That will help
with this problem.
Let's start getting
some things down.
That'll help make
more sense to you.
Since I'm trying to figure out
how many adult tickets
were sold
and how many
children's tickets were sold,
those are my unknowns
I'm solving.
I'm going to come underneath
my problem a little bit
and start writing notes here.
I'm going to let X
represent the adult tickets,
and let the Y represent
the children's tickets.
For one equation
I want to involve the money,
I want to involve
the $10 for adults,
the $8 for children
and then the $1,360
in admission sales.
That's going to get me
one equation
for my system.
So $10 for adults.
And I'm letting X represent
the number
of adult tickets.
So 10X.
$8 for children,
and I'm letting Y represent
the number
of children's tickets.
So plus 8Y equals...
Then I had $1,360
in admission sales.
So equals 1,360.
So that's one equation
from my system.
Let's keep going.
"150 tickets were purchased.
"How many
were adult tickets?
How many
were children's tickets?"
So I had a certain number
of adult tickets,
I don't know how many.
I had a certain number
of children's tickets,
don't know how many.
But I know in total
I had 150 tickets purchased.

(Describer) X plus y equals 150.

(Describer) X plus y equals 150.

(Describer) X plus y equals 150.

So there we go.
That is my system
of equations.
Scroll down a bit.
The first equation's
representing
the amount of money
in sales.
And the second equation's
representing the number
of tickets that were sold.
Those two equations
are going
to help me
solve this system.
Remember we had different
methods for solving systems:
elimination,
substitution, graphing?
Here, the method I'm choosing
is substitution
because my numbers are large.
I don't want to start
multiplying
by even larger numbers
to get even larger numbers
in my problem.
So I'll solve
that second equation.
I'll solve it for Y---
I could also solve it for X--
but I'll solve
this one equation for Y
because I can solve for Y
in one step, and from there,
I'll use
the substitution method
to solve this system.
Okay?
Let's get a little more
workspace.
Because if you remember,
substitution took a little more
work than elimination.
But in this case,
it's the easier route to take.
Let's just move that
to the center
and let's get a little
more room here for work.
Okay.
The first thing
I'm going to do
is solve that second equation
for Y.
I'll number my steps
so you can keep them straight.
Get the pen back.
I'm going to call this
"Step 1."
I'm going to take
that X plus Y equals 150,
and I'm going to solve
this equation for Y.
I'm going to subtract X
from both sides,
cancel out over there.
And I have Y equals
negative X plus 150.
Now, maybe you wrote
150 minus X,
it doesn't really matter.
You should
get into the habit
of writing your X term first
because it'll help you
as we go along in the course,
when we solve
for Y in an equation
with two variables.
So you want that X term
written first.
For now, if you reversed it,
it doesn't matter,
but try to get into habit
of writing it first.
Now that I know
that second equation
when I solve for Y,
that it's negative X plus 150--
what it equals--
I'm going to take that value
and substitute it back
into my first equation
to see what I get
when I can solve it for X.
Let me show you
what I mean.
Let's get some more room here,
because remember I said
these do take a bit.
Okay.
My second equation
is 10X plus 8Y equals 1360.
Let me just write that down.
10X plus 8Y...
equals 1360.
I know that
the first equation,
when I solve it for Y,
that Y equals
negative X plus 150,
what I'll do is substitute
that negative X plus 150
into that equation for Y.
Watch me do it.
I'll have 10X plus 8.
Open up parentheses.
Negative X plus 150,
close 'em.
Equals 1360, okay?
Did you see what I did?
I literally just replaced Y
with that negative X plus 150.
If you haven't already,
you can use a calculator
because I'm not
going to trust myself
to do mental math
with this problem
because these numbers
are large.
That's going to come up next.
If we start
cleaning this up,
simplifying this equation,
we've got 10X.
All right, 8 times negative X,
that's negative 8X.
So minus 8X.
Then I have 8 times 150,
which for that,
I am going to pull out
my calculator
for 8 times 150.
And I already cleared
the memory on this one.
So 8 times 150 is 1200.
So back to my problem.
So that is plus 1200...
equals 1360.
Okay.
Let's move things
out of our way,
keep solving for X.
I have like terms
to combine right here,
this 10X and this minus 8X.
If I combine those,
I'll get 2X
plus 1200 equals 1360.
Going to keep going--
we want to isolate the 2X,
get it by itself.
Let's subtract 1200
from both sides.
Going to need more room here,
see it happening.
All right.
So my 1200s, or my 1200
minus 1200, cancels out,
bring down my 2X.
1360 minus 1200.
Kind of large, but I trust
myself--that's 160.
If not, use the calculator
and see what it is.
I have 2X equals 160.
All right, so I'll divide
both sides by 2.
X is 80. Okay.
We have found out the solution
to one of our variables.
X is 80.
Remember, we're dealing
with a word problem.
We want to go back
and determine
what that means
as far as our word problem.
We're scrolling.
If X is 80,
then that means
that we sold 80 adult tickets.
Right now, we know the answer
to one of our unknowns.
Let me go ahead and write,
"80 adult tickets."
Now we want determine how many
children's tickets were sold.
You don't have to go back to
your equations in your system;
you can go back to the problem.
You sold 150 tickets,
and 80 of them were adults.
Let's figure out
how many of them were children.
We can just do 180 minus 50--
sorry, 150 minus 80.
I'm going to move down
so I can write down the work.
This is our third step.
Remember what we said:
that 80 represented
80 adult tickets.
And we know that we sold
150 tickets in total.
So 150 minus 80.
That leaves 70.
That means that we sold
70 children's tickets.
You've solved your system.
I believe it was
a local movie theater
we were dealing with here.
Let's scroll--
the 11 a.m. show, mm-hmm.
There were 80
adult tickets sold
and there were 70
children's tickets sold.
Let's try another one
just to make sure you got it.
It's a lot of steps,
but if you break it down,
you'll have it in no time.
Check this one out.
"Tickets to the local
high school sporting events
"are $7 for students
and $12 for adults.
"During last week's event,
107 tickets were sold
"for a total of $949.
"How many student tickets
were sold?
How many adult tickets
were sold?"
We've read through
the problem;
now let's go back
and let's highlight
all of the key information.
If you find you're able
to read the problem
and highlight the information
all at one time,
that's fine.
Go ahead and do that.
But you may want to
take one time
to read through it
with no pressure,
no extra work to do,
then go back and attack
the problem, pull it apart,
highlighting what you need.
If you find you can do that
all at once
as you do more problems
and get more comfortable,
that's okay.
This time, since I've
read through it once,
I'll highlight the information
I need to know.
Just highlight anything
that sounds like math.
"Tickets to the local
high school sporting events
are $7 for students and $12
for adults--" I got that.
Highlighter
didn't want to stay.
"During last week's event,
107 tickets were sold
for a total of $949--"
I got that.
"How many student tickets
were sold?"
And "how many adult tickets
were sold?"
Okay.
Didn't want to highlight.
Okay, there we go.
I'll pass through it
one more time.
$7 for students,
$12 for adults.
107 tickets were sold
for a total of $949.
How many student tickets?
How many adult tickets?
I have a general idea
of what I'm trying to solve
and what information
will help me
get to that solution.
Let me switch to my pen
to get to work.
Since I'm dealing with student
tickets and adult tickets
as far as what I'm trying
to solve in this problem,
I'll let X represent
the number of student tickets.
I'll make myself a note here.
And I'll let Y represent
the number of adult tickets.
I've got that to the side.
Then I see I'm dealing
with the cost of the tickets,
and the total.
So I'll keep all the money
things together
in one equation;
so $7 for students,
so 7X,
plus $12 for adults,
so plus 12Y, equals $949.
So there is one.
The other situation
in my problem
is just the number of tickets
that were sold.
I don't know how many
student tickets were sold,
but I'll let X represent that.
Y is the adult tickets,
so X plus Y equals 107,
because it said that I sold
a total of 107 tickets.
There lies my system.
Now that I've got it,
it's time to start solving it.
I'm going to do a little
rearranging here,
just to get us
some extra workspace.
I'm going to move--
ooh--do want that.
Going to move this down
a little bit to the center,
and I'm going to scroll down,
and just make this page
a little longer for us.
Because I know it's going
to take multiple steps,
I'll label this as "Step 1."
get the pen back--
so Step 1.
I'm going to solve this system
using substitution.
I'll take that second equation
and go ahead
and solve it for Y,
like on the first one.
So X plus Y equals 107.
I'm going to solve it for Y.
Maybe you chose X, that's fine,
but I'm going to choose Y.
I'll subtract X
from both sides.
Cancel out over there,
bring down my:
Y equals
negative X plus 107.
Now that I have that second
equation solved for Y--
I'm using
the substitution method
to substitute that expression,
that negative X plus 107,
into my first equation for Y.
All right, watch me do that.
I need a little more
space here.
Make the page longer again.
That equation,
that first one,
is 7X plus 12Y equals 949.
So 7X plus 12Y equals 949.
So I'm going to substitute
that negative X plus 107
into this equation for Y.
So 7X plus 12...
times negative X
plus 107...
equals 949.
Again, it just took
the place of Y.
Let's get some more space here.
Those numbers are large.
We'll pull out a calculator.
But to simplify this first,
let's bring down that 7X.
We've got 12
times negative X.
So that's negative 12X.
12 times 107--
let's pull out
the calculator for that.
12 times 107.
Okay, so 12 times 107.
That is 1,284.
Let's write that down: 1284.
So plus 1284 equals 949.
Let's get some more
space here to work.
So we can combine those
two like terms in the front,
that 7X minus 12X--
that's negative 5X.
Bring down that plus 1284.
Equals 949, okay?
So I'm trying to solve
for negative 5X,
so I'm going to subtract
1284 from both sides.
So cancel out right here.
Let's bring down
that negative 5X
and pull out
the calculator again.
949 minus 1284.
Okay, 949 minus 1284.
Okay, that is negative 335.
So let's go back
to our paper.
So that equals negative 335.
And we need
more space again.
So now we can divide
both sides by negative 5.
So X equals--
back to the calculator--
negative 335
divided by negative 5.
Okay, negative 335
divided by negative 5.
And we've got 67.
Back to our work.
All righty, so X equals 67.
So if we bring that back
to our word problem
to get the meaning
of that answer...
Let's go back and remember
what X represented.
Scrolling.
So X represented the number
of student tickets sold.
Right now we know
that there were 67
student tickets sold.
And since we know that 107
tickets were sold in total,
we can make a shortcut;
we don't have to go back
to our equation.
Let's just subtract,
let's do 107 minus 67
and see what's left over
for the adult tickets.
Let's scroll a bit
and write that work down.
Scroll, scroll, scroll,
more room, okay.
We know we have
67 student tickets,
and there were 107 tickets
sold in total.
So 107 minus 67.
That's 40.
So there were 40
adult tickets sold.
For our word problem,
we got it.
I think we were dealing with
sporting events in this problem.
Let's scroll back up.
We were.
We sold--
let's write that down,
get this out of the way.
There were 67
student tickets sold,
and there were 40
adult tickets sold.
That gives us our total
of 107 tickets,
and that gives us also
our total of $949.
We got through this system.
I do believe it's time
for you to try one.
Press pause, take your time,
go through this problem,
and see what you get.
When you're ready
to compare
your answers with mine,
press play.

(Describer) Titles:
Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns 15 dollars per hour, and at a clothing store where she earns 12 dollars per hour. Kelsey worked a total of 60 hours last week and earned 825 dollars. How many hours did Kelsey work at each part-time job?

(Describer) Titles:
Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns 15 dollars per hour, and at a clothing store where she earns 12 dollars per hour. Kelsey worked a total of 60 hours last week and earned 825 dollars. How many hours did Kelsey work at each part-time job?

(Describer) Titles:
Kelsey is working two part-time jobs this summer to save money to purchase a car. She works at a computer repair store where she earns 15 dollars per hour, and at a clothing store where she earns 12 dollars per hour. Kelsey worked a total of 60 hours last week and earned 825 dollars. How many hours did Kelsey work at each part-time job?

(female narrator)
Kelsey is working two part-time
jobs this summer
to save money
to purchase a car.
She works at a computer
repair store
where she earns $15 per hour,
and at a clothing store
where she earns $12 per hour.
Kelsey worked 60 hours
last week and earned $825.
How many hours did Kelsey work
at each part-time job?
All right,
let's see how you did.
I do believe
I have the answer here.
I'll show you what I got
and then do the work.
Let's get this out of the way.
So Kelsey worked 35 hours
at the computer repair store,
and 25 hours
at the clothing store.
If you want to see
how I got that answer,
watch me do this.
I do believe--
nope, I didn't.
Let's read it first,
and then go back
and highlight
all of our key information.
"Kelsey is working
two part-time jobs this summer
"to save money
to purchase a car.
"She works
at a computer repair store
"where she earns $15 per hour,
and at a clothing store
"where she earns $12 per hour.
"Kelsey worked 60 hours total
last week and earned $825.
How many hours did Kelsey work
at each part-time job?"
We read through it once,
now let's go back
and let's highlight
all that important information.
Let me switch
to my highlighter.
So Kelsey is working
two part-time jobs this summer
to save money
to purchase a car.
She works at a computer
repair store
where she earns $15 an hour.
So at the computer
repair store, $15 per hour.
And at a clothing store
where she earns $12 per hour.
So clothing store,
$12 per hour.
"Kelsey worked a total
of 60 hours last week
"and she earned $825.
How many hours did Kelsey work
at each part-time job?"
So let's go through this.
Computer repair store,
$15 an hour.
Clothing store,
$12 an hour.
She worked 60 hours,
she earned $825.
We're trying to figure out--
I forgot to highlight that--
how many hours she worked
at each job.
Let's switch
to the pen and see
if you took these steps
to solve this.
I'm going to let X represent
the hours
at the computer store
and let Y represent
the hours
at the clothing store.
You could've reversed that;
either way is fine.
Let's get that answer
out of my way
so I can work down there.
So we're going to let X...
Switch to the pen,
there we go.
X is the computer store--
I'll abbreviate that
for computer--
and Y is the clothing store.
We're doing a lot of scrolling
on this one.
And Y is the clothing store.
So scroll back up.
At the computer repair store
she earns $15 per hour,
and at the clothing store
she earns $12 per hour.
So I'm going to say 15X...
plus 12Y equals 825.
That'll handle the money
that she's earning.
And she worked a total
of 60 hours last week.
So the hours at the computer
store plus the hours
at the clothing store were 60.
That should've
been your system.
Your system should've resembled
something like this.

(Describer) 15x plus 12y equals 825, and x plus y equals 60.

(Describer) 15x plus 12y equals 825, and x plus y equals 60.

(Describer) 15x plus 12y equals 825, and x plus y equals 60.

Let's get some more workspace,
do a little
maneuvering up here.
We can actually
keep that right there.
Step 1: I'm going to keep
using the substitution method
on these.
So I'm going to take
that second equation
and solve it for Y again.
So X plus Y equals 60.
Subtract X from both sides.
So Y equals negative X
plus 60.
So I have that second equation
solved for Y,
which will help me
to be able to substitute
back into that first equation,
okay?
All right, I'm going to get
right underneath this one here.
So that first equation
is 15X plus 12Y equals 825.
So 15 X plus 12Y
equals 825, okay?
And we'd already solved
that second one,
so we're going to substitute
that negative X plus 60
into this equation
for Y.
So 15X plus 12
times negative X plus 60
equals 825.
Let's get some more space
up here.
Let's clean this up.
Let's bring down the 15X.
12 times negative X.
That's negative 12X.
12 times 60.
Just to be safe, let's get
the calculator on that one.
So 12 times 60.
720.
Let's go back
to our work here.
Let's move that
out of the way.
So plus 720 equals 825.
Let's get
some more workspace.
We can combine these two
like terms in the front,
So 15X minus 12X,
that's 3X.
Plus 720 equals 825.
Okay?
We're trying to get
that 3X by itself,
so let's subtract 720
from both sides.
Let's get
a little more workspace.
Getting ahead of myself here.
So that's going to cancel out,
720 minus 720.
There we go.
So let's bring down the 3X.
825 minus 720.
That's 105.
Then our last step:
divide each side by three.
I'll use calculator
for that: 105 divided by 3.
Just to make sure
that's right.
So 105 divided by 3.
35.
Let's go back
to our work.
And we have X equals 35.
Let's go back up to the problem
to see
what meaning that has
in relation
to the word problem.
So if X is 35, and we said
that X represented
the number of hours she worked
at the computer store,
then we know
that she worked 35 hours
at the computer store.
Let's write that down.
Kind of crowded up here
so let's go back to the bottom
and write that down.
So she worked 35 hours...
at the computer store.
And if we go back
to the problem,
I believe it said that she
worked a total of 60 hours.
Right, so she worked
a total of 60 hours.
Let's see how many hours
are left over
for her job
at the clothing store.
Let's scroll back down.
So 60 minus 35.
Let's get
a little more space here.
And that's actually 25.
So she worked 35 hours
at the computer store,
and she worked 25 hours
at the clothing store.
Your work should have been
something similar to that
when you were solving
that real-world problem.
I hope
you're feeling confident
about how to solve
real-world problems
involving systems
of linear equations.
See you back here soon
for some more Algebra 1.
Bye!