Question

The length of a rectangle is 2 more than 3 times the width. If the perimeter is 100, find the length and width of the rectangle.

If l = length and w = width, which of the following systems could be used to solve the problem?

Answer 1

L=17.49635531
W=5.715476066

100 = L X W
L = 2+ 3W
Insert the L equation into the 1st equation, 2+ 3W x W = 100
2+ 3W x W = 100
-2 -2
3W X W = 98
3W^2 = 98
Divide both sides by 3
W^2 = 32.66
Square root both sides
now you have the Width, plug the width into the L X W = 100 equation, then divide 100 by the width, and now you have the length.

Answer 2

The length is 38 units and the width is 12 units.

Explanation:

We have to express the problem in equations.

`l` is gonna be length,
`w` is gonna be width and
`p` is the perimeter.

The length is 2 more than 3 times the width is

`l=2+3w`

The perimeter is 100 refers to

`p=100`

Now, we know that the perimeter of a rectangle is expressed

`p=2(l+w)`

Replacing the first expression in the second equation, we have:

`100=2(2+3w+w)\\100=4+8w\\w=(100-4)/(8)=12`

Now, we replace this value in the first expression to find the length:

`l=2+3(12)=38`

Therefore, the length is 38 units and the width is 12 units.