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.