Solve this problem.

The length of the smaller rectangle at the right is 1 inch less than twice its width. Both the dimensions of the larger rectangle are 2 inches longer than the smaller rectangle. The area of the shaded region is 86 square inches. What is the area of the smaller rectangle?

Let x = the width of the smaller rectangle.

The length of the smaller rectangle is 2x - 1.

Area is A = lw

So the area of the smaller rectangle is A = (x) (2x - 1) = 2x^2 - x

The larger rectangle's width is two inches more than the width of the smaller rectangle (x+2).

The larger rectangle's length is two inches more than the length of the smaller rectangle:

2x - 1 + 2 = 2x + 1

Area is A = lw

The area of the larger rectangle is A = (x + 2) (2x + 1) = 2x^2 + x + 4x + 2 = 2x^2 + 5x + 2.

The area of the larger rectangle minus the area of the smaller rectangle is 86:

(2x^2 + 5x + 2) - (2x^2 - x) = 86

Rewrite as adding the opposite:

(2x^2 + 5x + 2) + (-2x^2 + x) = 86

Combine like terms:

6x + 2 = 86

6x = 84

x = 14

The area of the smaller rectangle was 2x^2 - x, so

2 (14) ^2 - (14)

2 (196) - 14

392 - 14

378

The area of the smaller rectangle is 378 square inches.