I have a question about Algebra polynomials. Question goes about this A rectangular painting is twice as long as it is wide. A frame surrounds the painting. The frame is a constant 2 inches wide and has a area of 196 square inches. Find the dimensions of the painting.

I would have liked to make a visual representation of what i wanted but i don't know how to do that so i am just going to write what i inferred from the question

the painting's length is 2*W (w=width) and the width is just w. the frames width is 2 and the area of the frame is 196. How do i go on with the problem. if its just asking for the dimension of the painting wouldn't that mean that i could just multiply the width of the painting and the length of the painting. I know this is wrong because it gives me other numbers. Can anyone explain me how to solve this.

To solve this problem, we can start by setting up the equation for the combined area of the painting and frame. Let's denote the width of the painting as W and the length as L.

The area of the frame is given as 196 square inches. Since the frame is 2 inches wide and surrounds the entire painting, we can calculate the area of the frame as follows:

Area of the frame = (L + 2)(W + 2) - LW
196 = (L + 2)(W + 2) - LW

Now, let's use the given information that the length of the painting is twice its width:

L = 2W

Substituting this equation into the previous equation, we get:

196 = (2W + 2)(W + 2) - (2W)(W)
196 = (2W + 2)(W + 2) - 2W^2

Expanding the equation further, we have:

196 = 2W^2 + 2W + 4W + 4 - 2W^2
196 = 4W + 4

To isolate the variable W, we can subtract 4 from both sides of the equation:

192 = 4W

Now, divide both sides of the equation by 4:

48 = W

So, the width of the painting is 48 inches.

Since the length of the painting is twice the width, we can find the length by multiplying the width by 2:

L = 2W = 2 * 48 = 96

Therefore, the dimensions of the painting are 48 inches width and 96 inches length.