jrhinehart12
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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?

Respuesta :

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:

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

Step-by-step explanation:

We have to express the problem in equations.

[tex]l[/tex] is gonna be length, [tex]w[/tex] is gonna be width and [tex]p[/tex] is the perimeter.

The length is 2 more than 3 times the width is

[tex]l=2+3w[/tex]

The perimeter is 100 refers to

[tex]p=100[/tex]

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

[tex]p=2(l+w)[/tex]

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

[tex]100=2(2+3w+w)\\100=4+8w\\w=\frac{100-4}{8}=12[/tex]

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

[tex]l=2+3(12)=38[/tex]

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

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