A rectangle has a length that is twice its width. If the width is increased by 3 units and the length is decreased by 2 units, the area becomes 24 square units. What were the original dimensions?
["Title: Solving for the Original Dimensions of a Rectangle: From Proportions to Area Equations", "Creating and analyzing geometric problems helps sharpen problem-solving skills—especially when shapes follow clear mathematical relationships. In this article, we explore a rectangle where the length is twice its width. By adjusting the dimensions and using the resulting area to derive the original values, we reveal the true dimensions of this special rectangle.", "---", "### The Problem Statement", "We are given:\n- A rectangle with length equal to twice its width.\n- When the width is increased by 3 units and the length is decreased by 2 units, the new area is 24 square units.\n- Our goal is to find the original length and width of the rectangle.", "---", "### Step 1: Define Variables Based on Given Relationships", "Let the width of the original rectangle be:\n[\nw \ ext{ units}\n]", "Then, since the length is twice the width:\n[\n\ ext{length} = 2w\n]", "---", "### Step 2: Express the Modified Dimensions", "After modifying the dimensions:\n- New width = ( w + 3 )\n- New length = ( 2w - 2 )", "The area of the modified rectangle is given as 24 square units:\n[\n(w + 3)(2w - 2) = 24\n]", "---", "### Step 3: Expand and Simplify the Equation", "[\n(w + 3)(2w - 2) = 24\n]", "Apply the distributive property (FOIL method):\n[\nw(2w) + w(-2) + 3(2w) + 3(-2) = 24\n]\n[\n2w^2 - 2w + 6w - 6 = 24\n]\n[\n2w^2 + 4w - 6 = 24\n]", "Bring all terms to one side:\n[\n2w^2 + 4w - 30 = 0\n]", "Divide through by 2 to simplify:\n[\nw^2 + 2w - 15 = 0\n]", "---", "### Step 4: Solve the Quadratic Equation", "Factor the quadratic:\n[\nw^2 + 2w - 15 = (w + 5)(w - 3) = 0\n]", "Set each factor equal to zero:\n[\nw + 5 = 0 \quad \Rightarrow \quad w = -5 \quad \ ext{(Not valid; width can’t be negative)}\n]\n[\nw - 3 = 0 \quad \Rightarrow \quad w = 3\n]", "So, the original width is 3 units.", "---", "### Step 5: Find the Original Length", "Since length = (2w):\n[\n\ ext{length} = 2 \ imes 3 = 6 \ ext{ units}\n]", "---", "### Step 6: Verify the Solution", "Original dimensions:\n- Width = 3\n- Length = 6", "Modified dimensions:\n- New width = (3 + 3 = 6)\n- New length = (6 - 2 = 4)", "New area:\n[\n6 \ imes 4 = 24 \ ext{ square units} \quad \ ext{(Matches the given condition)}\n]", "---", "### Conclusion", "Through algebraic modeling and logical expansion, we determined that:\n- The original width of the rectangle was 3 units.\n- The original length was 6 units, forming a rectangle where length is twice the width.", "Understanding how changing dimensions affect area using equations enables precise geometric analysis—perfect for students, teachers, and math enthusiasts alike.", "---", "Keywords: rectangle area, solve for width, solve for length, algebraic geometry, quadratic equation, geometry problems, mathematical modeling.", "---", "Need help visualizing this rectangle? Sketch the original and modified versions to see how dimensions changed and confirm the area relationship. Mathematics becomes clearer with practice!"]
