L'équation est 2 litres / (5 + x) litres = 20 % = 0,20.
["# Solving L’Équation : 2⁄(5 + x) = 0,20 – A Step-by-Step Guide to Finding x", "If you’re diving into algebra or preparing for a math exam, you may have encountered equations that look tricky at first glance—like 2 ⁄ (5 + x) = 0,20. But don’t worry! This equation simplifies cleanly, and solving it reveals a powerful concept in linear proportion problems. In this article, we’ll break down L’équation est 2 ⁄ (5 + x) = 0,20 step by step, explain how to isolate x, and explore practical applications. Whether you're a student, teacher, or self-learner, mastering this formula helps build strong problem-solving skills.", "---", "## Understanding the Equation", "The equation 2 ⁄ (5 + x) = 0,20 represents a real-world scenario where a fixed volume (2 litres) is divided by a variable quantity (5 + x) litres, and the result equals 20% of the original volume. In fractional form, this matches the percentage transformation:\n2 ⁄ (5 + x) = 20 % = 0,20", "Our goal is to solve for x, the unknown variable representing the denominator — possibly a change in volume, concentration, or scale factor in practical applications.", "---", "## Step-by-Step Solution", "### Step 1: Convert the percentage to decimal\nSince 20% = 0,20, rewrite the equation as:\n2 ⁄ (5 + x) = 0,20", "### Step 2: Eliminate the fraction by cross-multiplication\nMultiply both sides by (5 + x) to isolate the denominator:\n2 = 0,20 × (5 + x)\n[ 2 = 0,20(5 + x) ]", "### Step 3: Distribute the 0,20\n[ 2 = 1 + 0,20x ]", "### Step 4: Subtract 1 from both sides\n[ 2 - 1 = 0,20x ]\n[ 1 = 0,20x ]", "### Step 5: Divide both sides by 0,20\n[ x = \frac{1}{0,20} ]\n[ x = 5 ]", "---", "## Final Answer and Interpretation", "The solution is x = 5. Substituted back into the original equation:\n2 ⁄ (5 + 5) = 2 ⁄ 10 = 0,20, which confirms the equality.", "So, L’équation est vérifiée lorsque x = 5 — confirming that changing the denominator from 5 + x to 10 yields exactly 20% of 2 litres. This solution has broad implications in scaling, dilution, and percentage-based problem solving.", "---", "## Real-World Applications", "Understanding equations like 2 ⁄ (5 + x) = 0,20 equips you to tackle real-life challenges:\n- Scaling Recipes or Formulae: If 20% of 2 litres is required in a reaction, what adjustment in volume (x) achieves this proportion?\n- Concentration Calculations: In chemistry, determining volumes to reach a specific percentage concentration often involves similar algebraic setups.\n- Financial and Proportion Analysis: Comparing ratios in budgeting or resource allocation can rely on solving for unknowns in proportional equations.", "---", "## Why This Equation Matters in Algebra", "This problem illustrates fundamental algebraic strategies: isolating variables, manipulating fractions, and working with percentages—all crucial foundations for more complex equations. By practicing such problems, learners strengthen their ability to model scenarios mathematically.", "---", "## Summary", "- Equation: ( \frac{2}{5 + x} = 0{,}20 )\n- Solution: ( x = 5 )\n- Meaning: When the variable x = 5, the quantity (5 + x) litres produces exactly 20% of 2 litres.\n- Skill Development: Critical for solving proportion problems in science, finance, and daily calculations.", "Mastering this equation transforms a seemingly abstract format into a practical problem-solving tool. Keep practicing—each equation solved builds confidence and clarity in algebra!", "---", "Keywords:\nL’équation est 2 litres / (5 + x) litres = 20 % = 0,20, solve x, algebraic proportion equation, solving percentages algebraically, step-by-step equation solving, mathematics tutorial, real-world math problems."]
