0.30x + 0.10(500 - x) = 0.20 × 500 → 0.3x + 50 - 0.1x = 100 → 0.2x = 50 → x = 250.

["Solving 0.30x + 0.10(500 - x) = 0.20 × 500: A Step-by-Step Guide", "If you’ve ever encountered an equation like 0.30x + 0.10(500 - x) = 0.20 × 500, you’re not alone. This type of linear equation frequently appears in algebra, business modeling, budgeting, and everyday problem-solving. But solving it step-by-step can feel overwhelming—until you break it down. In this article, we’ll uncover how to solve 0.30x + 0.10(500 - x) = 0.20 × 500 = 100, leading elegantly to the solution x = 250.", "---", "### What Is This Equation Trying to Represent?", "This equation is often used in practical scenarios like budgeting, sales projections, or resource allocation. Quite simply, it balances variable contributions toward a target total. The left side combines two contributions:\n- 0.30x: a fixed rate tied to x,\n- 0.10(500 - x): a proportional rate applied to a remaining quantity,\nequal to a known total—500 predicts or a target like 0.20×500=100.", "---", "### Step 1: Simplify Both Sides", "Start by computing the right-hand side:", "[\n0.20 \ imes 500 = 100\n]", "Now rewrite the original equation:", "[\n0.30x + 0.10(500 - x) = 100\n]", "---", "### Step 2: Distribute the 0.10", "Distribute the 0.10 across the parentheses:", "[\n0.30x + 0.10 \ imes 500 - 0.10x = 100\n]", "[\n0.30x + 50 - 0.10x = 100\n]", "Combine like terms:", "[\n(0.30x - 0.10x) + 50 = 100 \quad \Rightarrow \quad 0.20x + 50 = 100\n]", "---", "### Step 3: Isolate the Variable Term", "Subtract 50 from both sides:", "[\n0.20x = 100 - 50 \quad \Rightarrow \quad 0.20x = 50\n]", "---", "### Step 4: Solve for x", "Divide both sides by 0.20:", "[\nx = \frac{50}{0.20} = 250\n]", "---", "### The Solution: x = 250", "This means the value that balances the equation is 250. In real-world terms, if x represents a portion of a total (e.g., sales amount, investment, or usage), it reveals how much of the variable component must be allocated to meet the target.", "---", "### Why This Matters", "Equations like 0.30x + 0.10(500 - x) = 100 demonstrate how weighted contributions sum to a fixed goal—a common theme in financial planning, inventory management, and forecasted budgeting. Understanding each step reduces errors and builds confidence in tackling similar problems.", "---", "### Key Takeaways", "- Always simplify constants and combine like terms first.\n- Balance the equation carefully by isolating the variable.\n- Substitute and compute step-by-step to avoid mistakes.\n- This type of equation models real-life scenarios involving proportional contributions.", "---", "In summary:\nSolving 0.30x + 0.10(500 - x) = 100 yields x = 250, emblematic of how algebra transforms word problems into actionable solutions. Whether analyzing data or planning resources, mastering this method empowers smarter decision-making.", "---", "FAQ: Common Questions About Solving Linear Equations", "- Q: Why do we subtract 50 from both sides?\n A: To isolate the term with x, we eliminate constants on one side, preserving equation balance.", "- Q: Why divide by 0.20 to solve for x?\n A: Division undoes multiplication, reverting (0.20x = 50) to (x = 50 / 0.20).", "- Q: Can this equation model real-world budgets?\n A: Yes! It’s widely used in finance when income or costs depend on variable percentages.", "---", "Mastering these steps ensures you can confidently solve similar equations—whether in homework, finance, or professional analysis. Start practicing with values close to 500, then adjust your numbers—it’s the best way to build fluency!", "---", "Keywords: solving 0.3x + 0.10(500 - x) = 100, linear equation step-by-step, algebra practice, x = 250 solution, real-world equation application, how to solve proportional equations, equation solving guide."]