A rectangular picture is 2 inches longer than it is wide. When a frame 3 inches wide is placed around the picture, the area covered by the picture and frame is 168 square inches greater than the area covered by the picture alone. What are the dimensions of the picture?

Let x = original width, then x+2 = length

(x+3)(x+5) - x(x+2) = 168

Solove for x.

To solve this problem, we can break it down into steps:

Step 1: Let's assume the width of the rectangular picture is "x" inches.
So, according to the given information, the length of the rectangular picture would be "x + 2" inches.

Step 2: Now, the total dimensions of the frame and the picture together would be:
Width = (x + 2) + 2(3) inches
Length = x + 2 + 2(3) inches

Step 3: The area covered by the frame and the picture would be the product of the width and length calculated in the previous step.
Area with frame = (Width)(Length) = [(x + 2) + 6][(x + 2) + 6] square inches

Step 4: The area covered by the picture alone would be the product of the width and length of the rectangular picture.
Area without frame = (Width of the picture)(Length of the picture) = (x)(x + 2) square inches

Step 5: The problem states that the area covered by the picture and frame is 168 square inches greater than the area covered by the picture alone.
So, we can set up an equation:
Area with frame - Area without frame = 168

Step 6: Substitute the expressions for the areas from Step 3 and Step 4 into the equation from Step 5 and simplify:
[(x + 2) + 6][(x + 2) + 6] - (x)(x + 2) = 168

Step 7: Expand and simplify the equation from Step 6:
(x + 8)(x + 8) - x(x + 2) = 168
(x^2 + 16x + 64) - (x^2 + 2x) = 168
x^2 + 16x + 64 - x^2 - 2x = 168
14x + 64 - 168 = 0
14x - 104 = 0
14x = 104
x = 104/14
x = 8

Step 8: Now that we know the value of "x" (width of the picture), we can calculate the length of the picture.
Length = x + 2 = 8 + 2 = 10

Therefore, the dimensions of the rectangular picture are: 8 inches by 10 inches.

Let's denote the width of the picture as "x" inches.

According to the given information, the length of the picture would then be "x + 2" inches because it is 2 inches longer than its width.

To find the dimensions of the picture, we need to determine the area covered by the picture alone and the area covered by the picture and frame.

The area covered by the picture alone is given by the formula:
Area of picture = length × width = x × x = x^2 square inches.

The area covered by the picture and frame is given by the formula:
Area of picture and frame = (length + 2×frame width) × (width + 2×frame width).

In this case, the frame width is 3 inches, so the area covered by the picture and frame is:
Area of picture and frame = (x + 2 + 2×3) × (x + 2×3)
= (x + 8) × (x + 6)
= x^2 + 14x + 48 square inches.

The problem states that the area covered by the picture and frame is 168 square inches greater than the area covered by the picture alone. So, we can set up the following equation:
Area of picture and frame = Area of picture + 168.

Substituting the corresponding formulas, we get:
x^2 + 14x + 48 = x^2 + 168.

By subtracting x^2 from both sides of the equation, we can simplify it to:
14x + 48 = 168.

Next, we subtract 48 from both sides:
14x = 120.

To isolate x, we divide both sides by 14:
x = 120/14.

Simplifying the division, we get:
x ≈ 8.57.

Since the dimensions of the picture must be whole numbers, we can round the width to the nearest whole number:
x ≈ 9.

Therefore, the width of the picture is approximately 9 inches.

Since the length of the picture is 2 inches longer than its width, the length would be:
x + 2 ≈ 9 + 2 = 11.

Therefore, the dimensions of the picture are approximately 9 inches by 11 inches.