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.

Sadly I cannot answer since you have not included the picture, although I suggest actually discussing this with your group, as the assignment says to do. If you are able to edit this, or create another question with the picture, I’d be happy to answer again!