- [Voiceover] A sketch is drawn that is
six inches by nine inches.
Just looking at our model here,
the sketch is this orange rectangle.
Let's go ahead and label the short length six inches
and the longer length nine inches.
It is put in a frame of uniform width.
So the blue part is the frame with uniform width.
If the area of the sketch is equal to the area of the frame,
what is the width of the frame?
Let's first find the area of the sketch.
Because it's a rectangle we know area is equal to
length times width, so in this case it'd be
nine times six, which equals 54 square inches.
We also know the frame has uniform width,
but because the width is unknown,
let's let the variable X equal the width of the frame.
Which means in our diagram, this would be X inches
and so would this, and the same is true for every corner.
This is X, and this is X inches, and so on.
Next we're told the area of the sketch
is equal to the area of the frame.
We know the sketch has an area of 54 square inches,
which means the area of the frame is also 54 square inches.
So the blue area equals 54 square inches
and the orange area is also equal to 54 square inches.
Now it's time to determine the dimensions of this larger
rectangle, which would be the dimensions
of the outside of the frame.
So we know this length here is six inches.
So if this is X inches and so is this,
this longer length here would be six plus X plus X inches,
or six plus two X inches.
And this longer length here would be nine plus X plus X,
or nine plus two X inches.
And now we'll set up an equation
for the area of the larger rectangle,
which includes the area of the sketch
as well as the area of the frame.
So again, we know the area of rectangle is equal to
length times width, so the area of this larger rectangle
A is equal to the quantity nine plus two X
times the quantity six plus two X.
And we know this product must be the area of the sketch
plus the area of the frame, which would be
54 plus 54 square inches.
So let's substitute 54 plus 54 for A,
and now we have an equation that we can solve for X.
And X will give us the width of the frame.
So 54 plus 54 equals 108, so 108 square inches
is the total area of the sketch and the frame
and this must equal the quantity nine plus two X,
times the quantity six plus two X.
For the next step let's clear the parentheses
from the right side.
When multiplying two binomials we always have four products.
One, two, three, four.
So we have 108 equals, distributing nine
we have 54 plus 18 X.
Distributing two X we have plus 12 X plus four X squared.
Because we have a quadratic equation,
let's set it equal zero and see
if we can solve by factoring.
So let's subtract 108 on both sides,
and simplifying, this difference is zero
so we have zero equals, let's put the square term first
so we have four X squared, we have two X terms,
18 X plus 12 X is 30 X, so plus 30 X,
and finally 54 minus 108 is negative 54, so minus 54.
Let's factor the right side.
The first step in factoring is to factor out
the greatest common factor.
The greatest common factor among these three terms is two.
So if we factor out two we have two times the quantity
two X squared plus 15 X minus 27.
Let's continue on the next slide.
Now we'll factor the trinomial inside the parentheses.
If it does factor, it'll factor into two binomial factors,
and because the square term is two X squared
in this first position, so we'll have a factor of two X
and a factor of X, and now we want to place the factors
and negative 27 in the second positions
so the sum of the inner product and outer product
is equal to positive 15 X.
Let's consider the factors of negative nine
and positive three, as well as the factors of
negative three and positive nine.
If we put minus nine here and plus three here,
notice how the outer product is negative 18 X,
the inner product is positive three X,
and this sum is negative 15 X,
we want a sum of positive 15 X.
And therefore we just need to change the sign
of the negative nine and the positive three.
We'll change the minus nine to plus nine,
and we'll change the plus three to minus three.
Notice how now the outer product is positive 18 X,
the inner product is negative three X,
and now the sum is positive 15 X, which is what we need.
So now we have it factored correctly, we can solve for X.
Because this product is equal to zero,
the factor of two X minus three must equal zero or
the factor of X plus nine must equal zero.
Notice how we don't get a solution for the factor of two
because there's no variable in this factor.
Solving for X here we add three and then divide by two.
So adding three we have two X equals three.
Dividing both sides by two we have X equals three halves.
Solving for X here we subtract nine on both sides,
and we get X equals negative nine.
And remember, X is a length and we can never have
a negative length, and therefore for this problem,
even though X equals negative nine
is a solution to this algebraic equation,
it is not applicable for this application.
So our solution is X equals three halves,
which means three halves, or one and a half,
or 1.5 inches, is the width of the frame.
I hope you found this helpful.