Equality holds when $ 2y = \frac{1}{2y} \Rightarrow 4y^2 = 1 \Rightarrow y = \frac{1}{2} $.
["Equality Holds When $ 2y = \frac{1}{2y} $: A Clear Mathematical Breakdown", "Understanding mathematical equality is fundamental to solving equations correctly and confidently. A key algebraic truth is that when $ 2y = \frac{1}{2y} $, this equation holds true only under specific conditions. Let’s explore step-by-step how this equality holds and derive the solution $ y = \frac{1}{2} $, reinforcing both accuracy and educational clarity.", "---", "### The Equation: $ 2y = \frac{1}{2y} $", "At first glance, this equation presents a division on the right-hand side, raising important questions about domain and valid solutions.", "---", "### Step 1: Eliminate the Fraction", "To eliminate the denominator, multiply both sides of the equation by $ 2y $, but only when $ 2y <br/>\ne 0 $ (since division by zero is undefined):", "$$\n2y \cdot (2y) = \left( \frac{1}{2y} \right) \cdot (2y)\n$$", "This simplifies to:", "$$\n4y^2 = 1\n$$", "---", "### Step 2: Solve the Resulting Equation", "Now, solve $ 4y^2 = 1 $:", "$$\ny^2 = \frac{1}{4}\n$$", "Taking the square root of both sides:", "$$\ny = \pm \frac{1}{2}\n$$", "---", "### Step 3: Check Validity of Solutions", "Remember the condition $ 2y <br/>\ne 0 \Rightarrow y <br/>\ne 0 $. Both $ \frac{1}{2} $ and $ -\frac{1}{2} $ satisfy this. However, original equation involves $ \frac{1}{2y} $, so $ y <br/>\ne 0 $ is respected — both solutions are valid mathematically.", "Each sign gives:", "- If $ y = \frac{1}{2} $, then $ 2y = 1 $, $ \frac{1}{2y} = 1 $ → equality holds.\n- If $ y = -\frac{1}{2} $, then $ 2y = -1 $, $ \frac{1}{2y} = -\frac{1}{2} $ → equality also holds.", "---", "### Why This Equality Matters", "This problem illustrates two essential algebraic principles:", "- Cross-multiplication is valid only when denominators are non-zero.\n- Always verify solutions by substituting back into the original equation, especially after operations like multiplying both sides by expressions containing variables.", "---", "### Final Answer", "The equality $ 2y = \frac{1}{2y} $ holds if and only if $ y = \frac{1}{2} $ or $ y = -\frac{1}{2} $. However, in many contexts—especially when dealing with real-world applications or physical constraints—only the positive solution $ y = \frac{1}{2} $ is meaningful.", "---", "### Summary", "- Start with $ 2y = \frac{1}{2y} $.\n- Multiply both sides by $ 2y $ (valid for $ y <br/>\ne 0 $) → $ 4y^2 = 1 $.\n- Solve → $ y = \pm \frac{1}{2} $.\n- Both values satisfy the original equation.\n- Equality holds when $ y = \frac{1}{2} $ and $ y = -\frac{1}{2} $.", "Mastering these steps builds confidence in solving rational equations and reinforces proper mathematical reasoning.", "---", "Boost your algebra skills today: always check your domain and verify solutions after algebraic manipulation!"]
