Three rectangles have exactly the same area. The dimensions of each rectangle (as length and width) are a and b; a – 3 and b + 2; and a + 3 and b – 1. Find the area of the rectangles.

To find the area of each rectangle, we can use the formula: Area = Length × Width.

Let's start by calculating the area of the first rectangle. The dimensions are a and b, so the area is: Area1 = a × b.

For the second rectangle, the dimensions are a – 3 and b + 2, so the area is: Area2 = (a – 3) × (b + 2).

Finally, for the third rectangle, the dimensions are a + 3 and b – 1, so the area is: Area3 = (a + 3) × (b – 1).

Since all three rectangles have exactly the same area, we can set up the following equations:

Area1 = Area2
a × b = (a – 3) × (b + 2)

Area1 = Area3
a × b = (a + 3) × (b – 1)

Now, let's solve these equations to find the values of a and b.

Expanding the equations, we get:

ab = ab + 2a – 3b – 6
ab = ab + 3a – b – 3

Simplifying the equations, we can cancel out the ab term:

2a – 3b – 6 = 3a – b – 3
2a – 3a = b – 3b + 3 – 6
-a = -2b – 3

Next, let's isolate a or b in one of the equations. Let's solve for a:

a = -2b – 3

Now, substitute this value of a into one of the original equations, for example, the equation:

ab = ab + 3a – b – 3

Substituting a = -2b – 3:

(-2b – 3)b = (-2b – 3)b + 3(-2b – 3) – b – 3

Simplifying:

-2b^2 – 3b = -2b^2 – 3b – 6b – 9 – b – 3

Combining like terms:

-2b^2 – 3b + 2b^2 + 9b + 12 = 0

Rearranging and simplifying:

b^2 + 6b + 12 = 0

Using the quadratic formula or factoring, we find that:

(b + 3)(b + 4) = 0

This gives us two potential values for b: b = -3 or b = -4.

If we plug these values of b back into the equation a = -2b – 3, we can find the corresponding values of a:

For b = -3:
a = -2(-3) – 3
a = 6 – 3
a = 3

For b = -4:
a = -2(-4) – 3
a = 8 – 3
a = 5

Therefore, there are two possible sets of dimensions for the rectangles:

1. Rectangle 1: Length = 3, Width = -3
2. Rectangle 2: Length = 5, Width = -4

To find the area of each rectangle, multiply the length by the width:

1. Area1 = 3 × -3 = -9
2. Area2 = 5 × -4 = -20

However, it is unusual for rectangles to have negative lengths or widths, so we can consider these solutions as extraneous. In practical terms, we cannot have rectangles with negative dimensions.

Therefore, we also conclude that there are no valid dimensions for the rectangles since negative dimensions are not plausible.