Question

The formula for the perimeter of a rectangle is P = 2l + 2w. Find w if P = 68 and l = 10.
A. W = 25
B. W = 15
C. W = 28
D. W = 24

Answer 1

Have you done problems like this before?

Answer 2

Yes, but i have been very sick and missed alot of school and need help

Answer 3

Alright, that's fine. So, they give you an equation with some variables and give you values for some of the variables, right? So have you tried plugging the values in and seeing what it looks like after that?

Answer 4

When i look at this and read what you just said, i get confused, maybe you can draw it out so i can understand better?

Answer 5

Right, so they give you this equation...

P = 2l + 2w

Then they tell you P = 68 and l = 10. Right?

So, then that means:

68 = 2(10) + 2w

Do you understand why?

Answer 6

Hold on let me read this for a second, And thank you for helping me.

Answer 7

No i don't understand why.

Answer 8

Ok, so P, l and w are variables. They aren't *real* numbers that we can multiply, divide, add, and subtract with. The whole point of the problem is to figure out the *real* number that w is supposed to represent.

To help us do that, your problem gave us real numbers for P and l.

Answer 9

So, then we can take the original equation and then plug in the values that they gave us for P and l.

Answer 10

Oh ok.

Answer 11

Do you follow so far?

Answer 12

Yes

Answer 13

Ok, so now we have.

68 = 2(10) + 2w

What do you think we could do to make this easier to understand?

Answer 14

2x10 = 20 so replace the 2(10) with 20?

Answer 15

YES! Very good!

Answer 16

So we have:

68 = 20 + 2w

Now, what next?

Answer 17

Im not sure...

Answer 18

Ok, that's fine. So now we need to try and figure out w. The goal is to get it by itself.

Answer 19

The first thing that's *easiest* to do is to get rid of any numbers not attached to the w. So, let's get rid of the 20 on the right side. Do you know how we could do that?

Answer 20

Move it over?

Answer 21

Hm. Yes, but it's a little bit more than that right? We've got to actually *subtract* it from both sides of the equation.

Answer 22

Ok. Well how are you going to take 20 from 2?

Answer 23

Go negative?

Answer 24

You don't, when you're subtracting a term out this way it doesn't hit the 2.

Answer 25

Oh ok.

Answer 26

Really, what you're doing is this:

68 - 20 = 20 + 2w - 20

Right?

Answer 27

Ok, makes sense

Answer 28

Right, so we would say that the 20 and -20 on the right side are of the same power. Neither have a variable following them, so we combine them and the 20 disappears from that side. Since the 2 has a w attached to it, the 20 can't touch it.

Answer 29

Only multiplication and division can detach and attach variables to numbers.

Answer 30

Ok

Answer 31

So, if we subtract 20 from both sides, what do we get?

Answer 32

48= 20 + 2w?

Answer 33

Very close. The right side of that is wrong, however.

Answer 34

When we subtract 20, we do it to *both* sides. So that 20 shouldn't be there.

Answer 35

Oh ok, so it's 48 + 2w?

Answer 36

Er, actually

48 = 2w

Answer 37

Sorry that's what i meant i pressed shift

Answer 38

It's all good.

Answer 39

So then w're most of the way there right? We've just got to get rid of that 2 and we'll know what w means. Do you have an idea of what we should do next?

Answer 40

48/2?

Answer 41

Yes. But, do remember that if you ever need to do this for an exam, you'll be expected to show the steps in the proper format. So you will need to write something like...

\[\frac{48}{2}=\frac{2w}{2}\]
\[24 = w\]

Answer 42

Good job. :)

Answer 43

Ok Thank you so much.!

Answer 44

np. I've got to go now, but best of luck!