If the length of a rectangle is 8 cm more than 3 times its width and the perimeter is 64 cm what are the dimensions of the rectangle?

This is an algebra problem.

Let's call the length L and the width W

L = 3W + 8 is an equation describing the first part of the problem

2L + 2W = 64 is an equation describing the second part of the problem.

Now, lets substitute the (2L) in the second equation with twice the value of L in the first equation

2(3W + 8) + 2W = 64, and now we'll work out the contents of the brackets.

6W + 16 + 2W = 64, now add the values of W

8W + 16 = 64, now bring all the numeric values to the same side of the equation

8W = 64 - 16

8W = 48, now divide both sides by 8

W = 6

Now we can substitute 6 for W in the first equation

L = 3(6) + 8, now we'll work out the contents of the brackets

L = 18 + 8 now we'll add the numbers

L = 26

So now we have two values and we need to ensure that they add up to 64

2(26) + 2(6) = 52 + 12 = 64

So, the rectangle is 26 x 6