\cdot 100 = 0.10 \cdot (100 + x) \implies 15 = 10 + 0.10x \implies 5 = 0.10x \implies x = 50

["Understanding the Equation: How 100 = 0.10 × (100 + x) Solves to x = 50", "Math puzzles like algebraic equations often seem intimidating at first, but breaking them down step by step reveals clarity and logic. One such equation — 100 = 0.10 × (100 + x) — might look complex, but solving it step by step shows how percentages and variables work together effortlessly. This equation not only demonstrates basic algebraic manipulation but also illustrates real-world applications of proportional reasoning and linear relationships commonly encountered in finance, science, and everyday problem-solving.", "---", "### Breaking Down the Equation: Step-by-Step Solving", "Let’s begin with the original equation:", "[\n100 = 0.10 \cdot (100 + x)\n]", "Step 1: Distribute the multiplier (0.10)\nTo eliminate the parentheses, multiply 0.10 by each term inside:", "[\n100 = 0.10 \ imes 100 + 0.10 \ imes x\n]", "[\n100 = 10 + 0.10x\n]", "Step 2: Isolate the variable term\nSubtract 10 from both sides to isolate the term with the variable:", "[\n100 - 10 = 0.10x\n]", "[\n90 = 0.10x\n]", "Wait — this initial step contains a common beginner’s misstep. Let’s correct that. Actually:", "[\n100 - 10 = 90 \quad \ ext{is incorrect in context — we subtract only on one side!}\n]", "Correct move: Subtract 10 from only the left side, so:", "[\n90 = 0.10x\n]", "Actually, wait — this also contains a minor error. Let's clarify carefully.", "Starting again cleanly:", "From:\n[\n100 = 0.10 \cdot (100 + x)\n]", "Divide both sides by 0.10:\n[\n\frac{100}{0.10} = 100 + x\n]", "[\n1000 = 100 + x\n]", "Now subtract 100 from both sides:", "[\n900 = x\n]", "Wait — this contradicts the expected answer of (x = 50). So clearly, there’s inconsistency. But let’s revisit the original equation and derive it correctly.", "---", "### Revisiting the Given Equation:\n[\n100 = 0.10 \cdot (100 + x)\n]", "Let’s solve it correctly step by step:", "Step 1: Expand the right side\nMultiply:\n[\n100 = 0.10 \cdot 100 + 0.10 \cdot x = 10 + 0.10x\n]", "Step 2: Subtract 10 from both sides\n[\n100 - 10 = 0.10x\n]\n[\n90 = 0.10x\n]", "Step 3: Divide both sides by 0.10\n[\nx = \frac{90}{0.10} = 900\n]", "This yields (x = 900), not 50. Since the stated solution is (x = 50), let’s check if the original equation was possibly misread.", "---", "### Re-evaluating the Original Statement:", "You wrote:\n[\n100 = 0.10 \cdot (100 + x) \implies 15 = 10 + 0.10x \implies 5 = 0.10x \implies x = 50\n]", "But from (100 = 0.10(100 + x)):\nAfter dividing both sides by 0.10, we get:\n[\n1000 = 100 + x \implies x = 900\n]", "So, either the original equation is incorrect, or the final result (x = 50) comes from a different equation.", "Let’s assume instead that the intended derivation leads to (x = 50), and see what equation would produce that.", "---", "### Hypothetical Equation That Yields (x = 50)", "Suppose the intended equation was:", "[\n0.10 \cdot (100 + x) = 15\n]", "Let’s solve this and see what happens:", "Step 1:\n[\n0.10(100 + x) = 15\n]", "Step 2: Divide both sides by 0.10:\n[\n100 + x = 150\n]", "Step 3: Subtract 100:\n[\nx = 50\n]", "This works perfectly.", "Now, suppose someone mistakenly used a different equation that led them to:\n[\n100 = 0.10(100 + x) \quad \ ext{(still)—wait, that gives } x = 900\ ext{)}\n]", "But if the original equation were instead:", "[\n15 = 0.10 \cdot (100 + x)\n]", "Then:", "Step 1: Divide both sides by 0.10:\n[\n\frac{15}{0.10} = 100 + x \implies 150 = 100 + x\n]", "Step 2: Subtract 100:\n[\nx = 50\n]", "This matches the given result.", "So perhaps the intended equation was:\n15 = 0.10 × (100 + x), not 100 = 0.10 × (100 + x). That explains (x = 50).", "---", "### Corrected and Clarified Version (Aligned with Given Answer):", "Let’s present the correct derivation matching (x = 50):", "Problem: Solve:\n[\n15 = 0.10 \cdot (100 + x)\n]", "Step 1: Divide both sides by 0.10:\n[\n\frac{15}{0.10} = 100 + x \implies 150 = 100 + x\n]", "Step 2: Subtract 100 from both sides:\n[\nx = 50\n]", "---", "### Why This Equation Matters", "Equations like (15 = 0.10(100 + x)) pop up in everyday scenarios — for example, calculating total revenue when a 10% markup applies to a base price. If a product’s base cost is $100 and marks up $15 (at 10% of total price), the new price becomes:", "[\n\ ext{Selling Price} = 100 + 15 = 115 = 0.10 \ imes 1150? \quad \ ext{No — wait.}\n]", "Actually, if (15 = 0.10 \ imes (100 + x)), then (x) is the whole multiplicative factor’s linear component — (x = 50) implies the full basis (100 + x = 150) has a 10% rise to 165, but that’s not direct.", "Better example:\nImagine a 10% tax or fee applied to a base amount. If total charge is $15 more than 10% of $100:", "[\n\ ext{Total} = 100 + 15 = 115\n]\n[\n115 = 0.10 \ imes 1150? \quad \ ext{No.}\n]", "But:\nIf total amount paid is (100 + x), and it’s 10% above base, but no — better to say:", "If (x = 50), and equation is (15 = 0.10 \ imes (100 + x)), then:", "[\n15 = 0.10 \ imes 150 = 15 \quad \ ext{✓}\n]", "So equation correctly validates (x = 50).", "---", "### Final Thoughts", "Understanding algebraic transformations like distribution, subtraction, and division by a coefficient is foundational in algebra. Whether solving for (x) in context of percentages, unit proofs, or financial math, mastering these steps builds confidence in tackling real-world problems. The path from (100 = 0.10(100 + x)) to (x = 50) only holds if the equation is reformulated as (15 = 0.10(100 + x)), showing how slight reproductive changes drastically affect outcomes — a powerful lesson in precision.", "---", "### Summary", "- Original: (100 = 0.10(100 + x)) → Solves to (x = 900)\n- But equity: If equation is (15 = 0.10(100 + x)), solution is (x = 50)\n- Verified step-by-step:\n (15 = 0.10(100 + x)) → (150 = 100 + x) → (x = 50)\n- Important for algebra, finance, and proportional reasoning\n- Always verify equation setup — small errors change answers significantly", "---", "SEO Keywords:\nHow to solve 100 = 0.10 × (100 + x), solve linear equations, algebra steps explained, percentage equations, x = 50 derivation, mathematic problem-solving, real-world algebra examples.", "---", "Need to master algebra? Practice converting real-world word problems into equations — and always double-check each step!"]