If the length of a rectangle is 4 more than 3 times the width, and the perimeter is 32 meters, what are its dimensions?

Please enter your answer as two numbers without the units, separated by a comma.

width --- x

length --- 3x + 4

Perimeter = 2width + 2lengths

take it from there

P = 2L + 2W

P = 2(3W + 4) + 2W

32 = 6W + 8 + 2W

24 = 8W

3 = W

To solve this problem, we can set up equations based on the given information and use algebra to solve for the dimensions of the rectangle.

Let's assume the width of the rectangle is represented by "W" (in meters) and the length is represented by "L" (in meters).

According to the given information, the length of the rectangle is 4 more than 3 times the width. So we can write the equation:

L = 3W + 4

The perimeter of a rectangle is the sum of the lengths of all its sides. In this case, the perimeter is given as 32 meters. Since a rectangle has two pairs of equal length sides (length and width), we can write the equation for the perimeter:

Perimeter = 2L + 2W

Substituting the value of "L" from the first equation, we get:

32 = 2(3W + 4) + 2W

Now, let's simplify and solve for "W":

32 = 6W + 8 + 2W
32 = 8W + 8
8W = 32 - 8
8W = 24

Dividing both sides of the equation by 8, we find:

W = 3

Now, substituting this value back into the first equation, we can solve for "L":

L = 3(3) + 4
L = 9 + 4
L = 13

Therefore, the dimensions of the rectangle are a width of 3 meters and a length of 13 meters.

So, the answer is: 3, 13.