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The perimeter of a rectangle is 38 ft. The length is 5 ft. longer than the width.

Find the dimensions. Write a system of linear equations and solve the resulting system. Let x be the length and y be the width.

Answer the first equation 2x+2y=?

Answer the second equation x=y+?

What is the length ? (in ft.)

What is the width? (in ft.)

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3 Answers

The perimeter is twice the length plus twice the width, or 2x+2y. 

Therefore, 2x+2y = 38 as given.

Also, the length is 5 ft longer than the width so x = y + 5

Knowing these equations are true, you can solve.

Substitute for x: 2(y+5) + 2y = 38

Distribute: 2y + 10 + 2y = 38

Simplify: 4y + 10 = 38

Subtract 10: 4y=28

Divide by 4: y=7

x=y+5 = 12

The length is 12, which is 5 more than the width of 7. The perimeter is 2(12+7) = 2(19) = 38

 

2x+2y=38 x=5+y insert the value of x into the first equation 2(5+y)+2y=38 10+2y+2y=38 4y=28 y=7 Now, insert the value of y into the second equation x=5+y x=5+7 x=12 so the length is 12 and the width is 7

Jamal,

How's it goin? Let's take a look at your problem:

The perimeter of a rectangle is 38 ft. The length is 5 ft. longer than the width.

Ah, a nice tasty word problem. Word problems truly are tasty, but it is an acquired taste, for sure.

Well, we know that a rectangle is a four-sided figure with four right angles, and we know that two of the sides of the rectangle will be of equal length and that the other two sides will be of another equal length. If all four sides share the same length, we would have a square. We are given, though, that this is a rectangle, so let's start out with some definitions.

We'll say that one of the rectangle's longer sides is its "length", and let's define the variable for one of these two sides to be "L".

Let's say that one of the rectangle's shorter sides would be referred to as its "width", and let's define the variable for one of these sides to be "W".

The perimeter is equal to the sum of the lengths of all four of the sides. There are two sides of length, L, and two sides of width, W. Let's give the perimeter the variable "P":

P = 2L + 2W

Easy enough so far. We actually know from our given information that P = 38, and so:

2L + 2W = 38 (Let's name this Equation 1)

Now comes what I find to be the fun part, where we turn words into math. Let's look at the second part of the problem:

"The length is 5 ft. longer than the width."

Well, this is fairly easy to translate, actually, using the variables we defined. Usually, a word like "is" will represent an equal sign:

length = 5ft + width, or:

L = 5 + W (Equation 2)

Great! We have two unknown variables and now have established two equations, so we will be able to solve by substituting. Let's take our equation from above:

2L + 2W = 38

and the equation we just formed:

L = 5 + W

and simply substitute this new equation into the first one! We know that L is equal to W + 5, so everywhere we see L in the first equation, we replace it with (W+5):

2(W+5) + 2W = 38

2W + 10 + 2W = 38

4W + 10 - 10 = 38 - 10

4W = 28

4W / 4 = 28 / 4

W = 7

Nice. We are halfway there. Now we need to solve for L. We can do this very simply by using Equation 2 and plugging in the value of 7 which we just obtained for W:

L = 5 + W

L = 5 + 7

L = 12

Awesome! So W = 7 feet and L = 12 feet. Don't forget to express your units in your final answer! Many teachers will count off if you neglect to do so. In fact, as you move on to college math, most of the time you will express the unit every time you use the variable throughout the problem so you won't lose track.

As a final step, we can check our work. We determined our length, L, by plugging our calculated value for W into Equation 2. Let's plug our value for L into equation 1 and see if we come up with the same value (7 feet) for W. If we do, then we can rest assured that we've done the problem correctly.

2L + 2W = 38

2(12) + 2W = 38

2W + 24 = 38

2W + 24 - 24 = 38 - 24

2W = 14

and, lo and behold:

W= 7

As we expected! I hope this helps and good luck to you in your studies!

-Skyler

Woodbridge Math tutors