rac{2x + y}{2x - y} + rac{2x - y}{2x + y} = 3.
["# Solving the Equation ( \frac{2x + y}{2x - y} + \frac{2x - y}{2x + y} = 3 ) – A Complete Step-by-Step Guide", "Understanding algebraic identities and solving rational equations can be a challenge—but mastering them unlocks stronger problem-solving skills, especially in math competitions, engineering, and science fields. In this comprehensive article, we break down how to solve the equation:", "[\n\frac{2x + y}{2x - y} + \frac{2x - y}{2x + y} = 3\n]", "---", "## Mastering the Structure of the Equation", "The left-hand side consists of two rational expressions:", "[\n\frac{2x + y}{2x - y} + \frac{2x - y}{2x + y}\n]", "This form suggests a key algebraic identity:", "[\n\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}\n]", "Let’s apply this insight.", "---", "## Step 1: Substitution to Simplify", "Let\n[\na = 2x + y \quad \ ext{and} \quad b = 2x - y\n]", "Then the equation becomes:", "[\n\frac{a}{b} + \frac{b}{a} = 3\n]", "Apply the identity:", "[\n\frac{a^2 + b^2}{ab} = 3\n]", "Multiply both sides by ( ab ) (noting ( ab <br/>\neq 0 )):", "[\na^2 + b^2 = 3ab\n]", "---", "## Step 2: Expand and Rearrange Algebraically", "First, expand ( a^2 ) and ( b^2 ):", "[\n(2x + y)^2 = 4x^2 + 4xy + y^2\n]\n[\n(2x - y)^2 = 4x^2 - 4xy + y^2\n]", "Add them:", "[\na^2 + b^2 = (4x^2 + 4xy + y^2) + (4x^2 - 4xy + y^2) = 8x^2 + 2y^2\n]", "Now compute ( 3ab ):", "[\n3ab = 3(2x + y)(2x - y) = 3[(2x)^2 - y^2] = 3(4x^2 - y^2) = 12x^2 - 3y^2\n]", "Set the two expressions equal:", "[\n8x^2 + 2y^2 = 12x^2 - 3y^2\n]", "---", "## Step 3: Bring All Terms to One Side", "[\n8x^2 + 2y^2 - 12x^2 + 3y^2 = 0\n]\n[\n-4x^2 + 5y^2 = 0\n]", "Rewriting:", "[\n5y^2 = 4x^2\n]", "---", "## Step 4: Solve for the Relationship Between (x) and (y)", "Divide both sides by (x^2) (assuming (x <br/>\ne 0)):", "[\n5\left(\frac{y}{x}\right)^2 = 4\n]", "[\n\left(\frac{y}{x}\right)^2 = \frac{4}{5}\n]", "Take square roots:", "[\n\frac{y}{x} = \pm \frac{2}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5}\n]", "Thus,", "[\ny = \pm \frac{2\sqrt{5}}{5} x\n]", "---", "## Step 5: Special Case – Exclude (x = 0)", "If (x = 0), then ( a = y ), ( b = -y ), and the original expression becomes:", "[\n\frac{y}{-y} + \frac{-y}{y} = -1 + (-1) = -2 <br/>\ne 3\n]", "So (x = 0) is not allowed.", "Also, check that denominators (2x \pm y <br/>\ne 0):", "- (2x - y <br/>\ne 0 \Rightarrow y <br/>\ne 2x)\n- (2x + y <br/>\ne 0 \Rightarrow y <br/>\ne -2x)", "Substitute ( y = \pm \frac{2\sqrt{5}}{5}x ). These values never equal ( \mp 2x ) unless (x = 0 ), which is already excluded. So the solution set is valid for all (x <br/>\ne 0) with the proportional relationship.", "---", "## Final Answer and Key Insight", "The solution to the equation\n[\n\frac{2x + y}{2x - y} + \frac{2x - y}{2x + y} = 3\n]\nis:", "[\n\boxed{y = \pm \frac{2\sqrt{5}}{5} x \quad \ ext{with} \quad x <br/>\ne 0}\n]", "---", "## Why This Equation Matters", "Equations of this form often appear in optimization, coordinate geometry, and physics when analyzing harmonic means, symmetry in ratios, or reciprocal relationships.", "Understanding their solution reveals not just numerical answers but also elegant symmetry and domain constraints—skills crucial for advanced math and STEM fields.", "---", "### Related Topics to Explore:\n- Solving rational equations\n- Symmetry in algebraic expressions\n- Application of ( \frac{a}{b} + \frac{b}{a} = k ) patterns\n- Geometry of reciprocal functions", "---", "For deeper alignment with math education standards and practical applications, revisit these identities and problem-solving strategies—your toolkit grows stronger every equation you solve!", "---", "Keywords:\nrac{2x + y}{2x - y} + rac{2x - y}{2x + y} = 3, algebra, rational equations, solving equations, x and y, mathematical identities, proportional relationships, equation solving, middle/high school math, contest math."]
