algebra 1 mathmatics

Question

A rectangles length is 3 times longer than its width the perimeter is 32 cm. write a system of equations represnting this situation and find the length of each side

Answer 1

For this problem, I shall establish W as the width of the rectangle and L as the length of the rectangle.

First, you have one equation already.

You know that the rectangle's length is 3 times its width.

Thus,

L = 3W.

Then, the perimeter of a rectangle can be found by adding up all the sides of a rectangle, for which we have,

2L + 2W = 32,

where this equation tells us adding up our 2 lengths and 2 widths gives us a total perimeter of 32 cm.

Therefore, the system of equations we have is:

L = 3W

2L + 2W = 32.

We can solve for W first by plugging in L from our first equation into the second equation:

2L + 2W = 32

2*(3W) + 2W = 32

6W + 2W = 32

8W = 32,

and dividing each side by 8 gives us W = 4.

Then, we can find L by using out first equation, and plugging in what we found for W,

L = 3W = 3(4) = 12.

Thus, we have a rectangle with length 12cm and width 4cm.

Answer 2

Let w be the rectangle's width in centimeters.

Then 3w is the rectangle's length in centimeters.

Therefore the entire perimeter of the rectangle (all four sides) is

w + 3w + w + 3w = 8w (width + length + width + length)

We are given that this perimeter is equal to 32:

8w = 32

Therefore w = 4.

So the width of the rectangle is 4 centimeters and the length is 12 centimeters, for a total perimeter of

4 + 12 + 4 + 12 = 32 centimeters.