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.

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?

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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?