I am stuck on this problem that I have tried to solve so many times! The question is:The length of a rectangle is four more than twice the width. If the perimeter of the rectangle is 32 cm, what are the length and width of the rectangle?

Here is my work:
32=2(2x+4)+2(2w)
32=(4w+8)+(4w)
32= 8w+8
32-8=8w+8-8
24=8w
24/8=8w/8
w=3
l=10
The work seems correct to me but when I check the answer it's wrong! What am I doing wrong?

P = 2L + 2W

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

32 = 6W + 8

24 = 6W

4 = W

In your work, it seems like there is a small error in setting up the equation. Let's go through the problem and solve it step by step.

Let's start by assigning variables for the width and length of the rectangle. We'll use 'w' for the width and 'l' for the length.

From the problem statement, it says that the length of the rectangle is four more than twice the width. So we can write the equation for the length as l = 2w + 4.

Now, let's calculate the perimeter of the rectangle. The perimeter of a rectangle is given by the formula P = 2(l + w). In this case, we have P = 32.

Plugging in the value for l from the first equation, we get 32 = 2(2w + 4) + 2w.

Now, let's simplify the equation:
32 = 4w + 8 + 2w
32 = 6w + 8

Next, let's isolate the variable 'w' by subtracting 8 from both sides:
32 - 8 = 6w + 8 - 8
24 = 6w

Now, divide both sides by 6 to solve for 'w':
24/6 = 6w/6
4 = w

So, the width of the rectangle is 4 cm.

Now, substitute this value back into the first equation to solve for the length:
l = 2w + 4
l = 2(4) + 4
l = 8 + 4
l = 12

Hence, the length of the rectangle is 12 cm.

Therefore, the correct answer is that the width of the rectangle is 4 cm and the length is 12 cm.