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Hi, class.
Today we're going to take a look at problem solving
combined with our systems of equations.
Systems of equations are often a nice method
for solving certain types of story problems,
because many story problems
have more than one thing involved with them.
We're going to take a look at an example
of three typical types of problems
where these systems of equations will be very helpful
in helping you solve the particular situation.
The first one we're taking a look at is this one right here
where it says, "The perimeter of a rectangular garden is 36 feet.
The length is 3 feet more than twice the width.
What are the dimensions of the garden?"
Now, when you're dealing with these,
the first thing you want to notice is
if it describes a situation that you could draw a picture of,
that picture often helps you to solve the situation.
And the first thing I see here is
"Perimeter of a rectangular garden."
I know what a rectangular garden looks like.
So, I'm going to go ahead and make a rectangular garden.
That way, I can label it as I go along.
And I know in a rectangular garden,
I have a length and I have a width.
So I'm going to label those
and then fill some things in as I continue reading.
Now, I notice also that it says, "perimeter" right here.
And there is an equation for perimeter
that is often in the front or back cover of a textbook.
They normally have formulas for geometry.
And you can look that up and then write it down.
And the perimeter formula
is that the perimeter of a rectangular shape
is twice the length plus twice the width. [ P = 2L+ 2w ]
They also give us some valuable information here
that the perimeter is 36 feet.
So that means that P equals 36 feet. [ P = 36 ft ]
That means I can take this 36 feet
and replace the P right here with it
and get an equation of 36 equals 2L plus 2w. [ 36 = 2L+ 2w ]
So, I've already got a ton of information
and I've just made it through that first sentence.
Let's continue on with the next sentence.
It says, "the length is."
In English, "is" translates into an equal sign [ = ] in math.
And I know that I have defined 'L' as length here.
So I'm going to do L equals. [ L = ]
This equals [ = ] right here is representing
that "is" right there in our sentence.
And if I continue reading on, it says, "is 3 feet more."
3 feet more means 3 plus. [ 3 + ]
So the "3 feet more" I can translate into 3 plus [ 3 + ].
And then let's see what it's 3 feet more than.
"It's more than twice the width."
So, the "twice the width" is going to translate into,
since we let w be the width, 2w. [ L = 3 + 2w ]
Now, if I read that sentence once more
just to make sure it matches, the length, which is the L;
'is' is the equals [ = ]; "3 feet more" is 3 plus [ 3 + ];
"twice the width" is the [ 2w ].
So, this equation we've just made right here
fits that second sentence.
Now it says, "What are the dimensions?"
So, we need to look at what we have.
I have an equation right here
which is L equals 3 plus 2w. [ L = 3 + 2w ]
I also have an equation right here
which is 36 equals 2L plus 2w. [ 36 = 2L + 2w ]
It looks like what I've just created is a system of equations.
And since one of the equations has a variable all by itself,
substitution would be a wonderful method for solving this.
So since L equals this 3 plus 2w, [ L = 3 + 2w ]
I'm going to go ahead and take this 3 plus 2w
and replace the L in the other equation with it.
That gives me 36 equals 2 times 3 plus 2w,
replacing the L, plus 2w. [ 36 = 2( 3 + 2w) + 2w ]
Now I just have to solve this and I'll know what my width is
and then I can find out what my length is.
So, distributing my 2, I get 6 plus 4w
and then carry down the rest of the equation.
[ 36 = 6 + 4w + 2w ] Let's combine our like terms.
4w and 2w is 6w. [ 36 = 6 + 6w ]
We now need to move our 6 over,
so I'll subtract 6 from both sides. [ 36 - 6 = 6 - 6 + 6w ]
And I get 30 equals 6w. [ 30 = 6w ]
Divide both sides by 6, [ 30/6 = 6w/6 ]
and our width is 5 feet. [ 5 ft = w ]
So, we know that this is 5 feet.
I can plug that 5 in right here on this other equation
to find my length.
My length will be 3 plus 2 times 5. [ L = 3 + 2(5) ]
2 times 5 is 10. [ 2(5) = 10 ]
Plus 3 is 13. [ L = 3 + 10 ]
So this is 13 feet. [ L = 13 ft ]
And since opposite sides are the same,
this will be 5 feet,
and this will be 13 feet.
And if you add all of those up, you get 36 feet,
which is what the perimeter was.
So, you would want to state in a sentence
that the dimensions of the rectangle are 13 feet for the length
and 5 feet for your width.
Let's take a look at one other type of situation
that's a little bit different than this particular one.
We might not actually have a picture
that represents what we're looking at.
Get rid of our drawings from this,
and then let's take a look at our next one.
Our next one is going to deal with going to the zoo.
And let's say that you're taking a trip to the zoo.
And some children and adults are going to the zoo.
And if 5 adults and 8 children attend the trip to the zoo,
it's going to cost 59 dollars.
However, if we don't have as many people go,
and we have 3 adults and 5 children attend the trip to the zoo,
it's going to cost 36 dollars.
So we want to know: How much is an adult ticket?
and also, How much is a child ticket?
So, since we're looking for adult tickets and children tickets,
we need to define the variables for those.
So, since adult starts with 'a,' why don't we
let a equal the cost of an adult. [ Let a = cost of an adult ]
And since child starts with a 'c,' why don't we
let c equal the cost of a child. [ Let c = cost of a child ]
You can choose whatever variables you want.
I like to use the letters that are what the words start with
unless two of them start with the same letter.
So now, let's go ahead and read through this again
and see what we have.
"If 5 adults." Well, the cost for 5 adults would be 5 times a.
So let's write down 5a.
"And" is normally plus [ + ] when we're translating into math.
"8 children." Well, the cost of 8 children
would be 8 times c because c is the cost of one child.
So that would be 8c.
Then the cost if we have 5 adults and 8 children attend,
the total cost is 59 dollars.
So this 5a plus 8c would equal 59. [ 5a + 8c = 59 ]
Now let's look at the next sentence.
"If 3 adults." That would be a 3a.
"And 5 children attend," that would be 5c,
"...the trip to the zoo would cost 36 dollars,"
so that's going to equal 36. [ 3a + 5c = 36 ]
Notice, we have a nice little system of equations right here.
And both of them,
of the equations, [ 5a + 8c = 59 ][ 3a + 5c = 36 ]
are in standard form. [ Ax + By = C ]
And since they're both in standard form,
the addition method would be a nice method for solving these.
So, I'm going to choose to eliminate my adults.
You can choose to either eliminate adults or children.
If I choose to eliminate the adults,
my coefficients are 3 and 5.
The smallest thing that 3 and 5 go into is 15.
So I'm going to multiply this one by 3 to make that a 15,
and this equation by negative 5 to make that a negative 15
so they cancel.
So this whole equation gets multiplied by 3, [ 3(5a + 8c = 59) ]
and this one gets multiplied by negative 5. [ -5(3a + 5c = 36) ]
So, multiplying the first one by the 3,
we'll write our new equation over here, we get
15a plus 24c equals, [ 15a + 24c =? ]
3 times 59 will be 150 plus 27, which is 177. [15a + 24c = 177 ]
And now for our second equation, [ -5(3a + 5c = 36) ]
negative 5 times 3a is negative 15a. [ -5(3a) = -15a ]
Negative 5 times 5c is negative 25c. [ -5(5c) = -25c ]
And negative 5 times 36 is negative 180. [ -5(36) = -180 ]
Now, we can just draw a line, carry the equal straight down.
177 and negative 180 is negative 3.
The a's cancel [ 15a - 15a =0 ] and we have a negative 1 c,
which you could write as negative 1c or negative c is fine, also.
Now just divide by negative 1, [ -1c/-1 = -3/-1 ]
and the children cost 3 dollars. [ c = $3 ]
Now we can plug that into either of our equations
to find out our adults.
I'm going to go ahead
and plug it into this second equation that we had here.
And that gives me 3a plus 5 times 3 equals 36. [ 3a + 5(3) = 36 ]
5 times 3 is 15. [ 5(3) = 15 ]
Let's move our 15 over
by subtracting 15. [ 3a + 15 - 15 = 36 - 15 ]
Oops, that's an equals sign here.
These cancel. [ 15 - 15 = 0 ]
I get 3a equals 21. [ 3a = 21 ]
Divide both sides by 3 [ 3a/3 = 21/3 ]
and the adults cost 7 dollars. [ a = $7 ]
So we would say that an adult ticket is 7 dollars
and a child ticket is 3 dollars.
So that's another type of a story problem
where these systems will help you out.
Now, let's take a look at one more type of story problem
that systems of equations are very helpful for.
On this last kind of story problem,
I'm also going to show you a table that I use
for organizing my information for these kinds of problems.
For this last one, we're going to be investing money
into two accounts.
Any time you're investing money into two accounts
or you are mixing two mixtures in chemistry
or something like that,
this table I'm going to show you will be very helpful for you.
It's a 3 by 3 grid.
And the first row is really just labels
to tell you what's in the bottom two rows.
So, this first row here, we're going to label,
since if we read this, it says,
"One account is a 4 percent account
and the other is an interest account."
I'm going to label this row here as the 4% account
and the second row here as the 6% account.
The last row is always the total.
Now, let's read our problem and fill in the rest of our grid.
If we take a look at this, it says
you have invested 15,000 dollars into two accounts.
So a total dollar amount is $15,000.
We can put that in one of these total boxes here.
I'm going to put it right here.
So, I have 15,000 dollars in this total box right here.
That means that this row is going to be my money row.
Now, I don't know how much money was in either account,
so since this is my money row,
I'm going to let x be how much goes into my 4% account
and y be how much is in my 6% account.
Now, if we read on, it says,
"at the end of one year, you earn $700 interest."
That means we have a total for interest
which we can put in this other total box.
And it's going to be $700. So this will be my interest row here.
Now let's think about
how you figure out your interest for one year,
if it's just simple interest in your 4% account.
You do that by taking your 4% times how much money's in there.
Well, that would be 4% times x or .04x
would be how much interest you got from that account.
Similarly, the other account would be
6% times the money which is y or .06y.
Everything in our grid is now filled in.
We end up deriving two equations off of this,
one off this row, and one off this row.
And it's always first column plus second column
equals third column.
So we end up with
x plus y equals 15,000 dollars. [ x + y = $15,000.00 ]
And .04x plus .06y equals 700. [ .04x + .06y = $700 ]
Notice that all this did over here was organize our information
which pulled up two nice equations that are both
in standard form. [ x + y = 15000 ] [ .04x + .06y = 700 ]
So addition method will work really nice for these.
Now, on the bottom row, I'd like to get rid of my decimals.
And I have 2 decimal places here and 2 decimal places there.
If I multiply by 100, all the decimals will disappear.
So, I'm going to multiply this by 100. [ 100(.04x + .06y = 700) ]
Now, let's just choose to eliminate our x's.
If I multiply this [ .04x ] by 100, this will be 4x.
So, strategically,
if I multiply this top equation [ x + y = $15,000.00 ]
by negative 4, [ 4(x + y = $15,000.00) ]
the 4x and negative 4x will cancel [ 4x - 4x = 0 ]
and we'll end up with only y's left.
So, let's go ahead and distribute our 100.
We're going to get
4x plus 6y equals 70,000. [ 4x + 6y = 70,000 ]
Now, let's distribute our 4. [ -4(x + y = 15000) ]
We get negative 4x plus negative 4y
equals negative 60,000. [ -4x + -4y = -60,000 ]
Now, we can go ahead and add straight down.
The x's will cancel. [ -4x + 4x = 0 ]
We end up with 2y equals 10,000 dollars. [ 2y = $10,000 ]
Now, just divide both sides by 2, [ 2y/2 = 10000/2 ]
and y is 5,000 dollars. [ y = $5,000 ]
That means we invested $5,000 into the 6% account
and if there was $15,000 total,
that left us with $10,000 in the 4% account.
So that is another type of a problem
where using what we've learned in the chapter
on systems of linear equations
will help you solve a story problem.
You're now ready to go ahead and try some of your story problems
that allow you to do some problem solving
and use your systems of equations.