Write an absolute value inequality that represents the situation. Then solve the inequality. The difference between the perimeters of the figures is less than or equal to 3.

Answer:[tex]|2x-8|\le 3\\ \\2.5\le x\le 5.5[/tex]Step-by-step explanation:You are given the rectangle with sides of (x + 1) units and 3 units and the square with the side of x units.The perimeter of the rectangle is [tex]P_r=(x+1)+3+(x+1)+3\\ \\=x+1+3+x+1+3\\ \\=2x+8\ units[/tex]The perimeter of the square is [tex]P_S=x+x+x+x\\ \\=4x\ units[/tex]We do not know whose perimeter is 3 units greater (or equal to), so the absolute value of the difference of these perimeters should be less than or equal to 3.Hence,[tex]|4x-(2x+8)|\le 3\\ \\|4x-2x-8|\le 3\\ \\|2x-8|\le 3[/tex]Solve this inequality:[tex]-3\le 2x-8\le 3\\ \\-3+8\le 2x-8+8\le 3+8\\ \\5\le 2x\le 11\\ \\2.5\le x\le 5.5[/tex]