Question

the length of a rectangle is 5yd less than twice the width. the area of the rectangle is 33yd2. find the dimensions of the rectangle.

Answer 1

Let's say the width is 'w'. Can you write an expression for the length using 'w'? The words are in the problem.

Answer 2

5-w=33?

Answer 3

I don't know then, im confused.

Answer 4

No, forget about the area. Just focus on the length of a side.

Answer 5

'The length of a side is 5 yards less than 2 times the width, w'

Answer 6

5-2w?

Answer 7

Err, almost. The length is 5 yards LESS THAN 2 times the width.

Answer 8

5<2w

Answer 9

...No, you were close the first time. For the length, we have to take away 5 from two times the width.

Answer 10

im so confused. really I don't get this.

Answer 11

If I have 10 dollars, and you have 5 dollars less than I have, how many do you have?

Answer 12

5 dollars lol

Answer 13

Okay. If I have 20 dollars, and you have 5 dollars less than I have, how many do you have?

Answer 14

15

Answer 15

Exactly. And how are you calculating that?

Answer 16

It's just whatever I have - 5, right?

Answer 17

yes

Answer 18

Okay, so if the the length is 5 less than 2 times the width, the length is ...

Answer 19

10? I don't know

Answer 20

we agreed that we would represent the width, whatever it is, even if we don't know the value, by the letter \(w\).

If we want to say the value of twice the width, we write \(2w\)

If we want to say "5 less than twice the width" by the pattern established with the money examples, how would you write that as an expression involving \(w\) and 5?

Answer 21

(you can use a 2 in the expression, of course)

Answer 22

im so confused. im really dumb. 5-2w?

Answer 23

When I said I had 20 dollars and you had 5 less than that, did you figure out
20-5

or

5-20?

Answer 24

2w-5

Answer 25

There you go!

Answer 26

Now we have the width: w
And the length: 2w - 5
What will the be the area of the rectangle?

Answer 27

I agree that it is a bit confusing the first time you encounter that form of expressing "take two times the width and subtract 5"

Answer 28

it says the area is 33?

Answer 29

so the length is 2w-5? and width is w? so im still confused. how do I solve it? how do I find the dimensions of the rectangle?

Answer 30

What's the general form for the area of a rectangle using length and width?

Answer 31

a=lw

Answer 32

Good, can you make an equation using for the area using what we got for width and length?

Answer 33

would be something like 33= (2w-5) x w?

Answer 34

YESYEYSYESYEYS
You got this by the nose.

Answer 35

Now, the hard part is coming up. (OH God).
Can you simplify this equation and set it up like a quadratic equation?

Answer 36

would the answer be 14?

Answer 37

14? No?

Answer 38

oh my gosh, I need to hurry up and finish this.

Answer 39

We have (2w-5)w = 33. We need to solve for w, the width of the rectangle.

Answer 40

Do you know what a quadratic equation is? Ax^2 + Bx + C = 0?

Answer 41

no I don't.

Answer 42

Sigh. This is terrible then. How do you have this problem if you don't know what a quadratic equation is.
Have you seen y = x^2 before?

Answer 43

im sorry, I don't get this at all. this is apart of my homework and I don't understand

Answer 44

Well, I can't help you if you don't even know quadratic equations. That's what this problem is about. And that will be a whole entire new lesson.

Answer 45

Were you born in 1996? How have you not dealt with quadratic equations yet?