["Solving the Equation (2(2w + w) = 40) with the Given Relation (l = 2w)", "Understanding how to solve algebraic equations is essential for advancing in math and related disciplines. In this article, we’ll walk through solving the equation (2(2w + w) = 40), using the given relation (l = 2w) as a primary reference point to reinforce context and application. This equation not only demonstrates basic algebraic manipulation but also shows how real-world relationships—like (l = 2w)—can simplify problem-solving in geometry and physics.", "---", "### Step 1: Simplify the Equation Using the Given Relation", "Given:
\n[
\nl = 2w
\n]
\nWe are also provided with:
\n[
\n2(2w + w) = 40
\n]
\nSubstitute (l = 2w) into the expression inside the parentheses:
\n[
\n2w + w = 3w
\n]
\nThus, the equation becomes:
\n[
\n2(3w) = 40
\n]", "---", "### Step 2: Apply Basic Algebraic Properties", "Simplify the left-hand side:
\n[
\n6w = 40
\n]
\nTo isolate (w), divide both sides by 6:
\n[
\nw = \frac{40}{6} = \frac{20}{3}
\n]", "---", "### Step 3: Use the Relation (l = 2w) to Find (l)", "Now, substitute (w = \frac{20}{3}) into (l = 2w):
\n[
\nl = 2 \cdot \frac{20}{3} = \frac{40}{3}
\n]", "---", "### Step 4: Verification", "Double-check by plugging (w = \frac{20}{3}) back into the original equation:
\n[
\n2(2w + w) = 2\left(2 \cdot \frac{20}{3} + \frac{20}{3}\right) = 2\left(\frac{40}{3} + \frac{20}{3}\right) = 2 \cdot \frac{60}{3} = 2 \cdot 20 = 40
\n]
\nThe solution satisfies the equation, confirming accuracy.", "---", "### Why Understanding (l = 2w) Enhances Problem-Solving", "The relation (l = 2w) appears commonly in geometry—especially when dealing with rectangles, where one side is twice the other. Recognizing such patterns allows for faster substitution and reduces error in complex equations. This example illustrates how real-world relationships improve algebraic efficiency and deepen mathematical intuition.", "---", "### Final Answer", "[
\nw = \frac{20}{3} \quad \ ext{and} \quad l = \frac{40}{3}
\n]", "This solution is not only mathematically sound but also practically applicable, particularly in geometric modeling where proportional relationships govern dimensions. Mastering such equations builds a foundation for mastering more advanced algebra and calculus concepts.", "---", "Keywords: solve (2(2w + w) = 40), algebraic equation solving, relation (l = 2w), geometry applications, simplify algebraic expressions, solve linear equations.
\nMeta Description: Solve the equation (2(2w + w) = 40) using (l = 2w) to find (w) and (l). Learn step-by-step algebraic methods with real-world context."]
