– [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.

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