Let \( u = rac{a + b}{a - b} \), then the expression is \( u + rac{1}{u} = 3 \). Multiply both sides by \( u \):
["Let ( u = \frac{a + b}{a - b} ). Solve ( u + \frac{1}{u} = 3 ) — A Step-by-Step Breakdown", "Understanding algebraic identities can transform complex expressions into elegant, solvable forms. In this article, we explore the expression involving a key ratio ( u = \frac{a + b}{a - b} ), revealing how the identity\n[\nu + \frac{1}{u} = 3\n]\nemerges naturally and how to derive it using fundamental algebraic techniques.", "---", "### What Is ( u = \frac{a + b}{a - b} )?", "Let us define a variable ( u ) as:\n[\nu = \frac{a + b}{a - b}\n]\nHere, ( a ) and ( b ) are any real numbers such that ( a <br/>\ne b ) (to avoid division by zero). This substitution simplifies certain symmetric expressions in algebra, geometry, and engineering problems.", "---", "### Starting with the Identity: ( u + \frac{1}{u} = 3 )", "We are given the equation:\n[\nu + \frac{1}{u} = 3\n]\nThis form frequently appears in optimization, function analysis, and reciprocal relationships. Our goal is to verify or use this identity when ( u = \frac{a+b}{a-b} ).", "---", "### Step 1: Multiply Both Sides by ( u )", "To eliminate the fraction, multiply every term on both sides by ( u ), assuming ( u <br/>\ne 0 ) (which holds since ( a + b ) and ( a - b ) cannot both be zero simultaneously unless ( a = b = 0 ), but then ( u ) would be undefined):", "[\nu \cdot u + u \cdot \frac{1}{u} = 3u\n]", "This simplifies to:\n[\nu^2 + 1 = 3u\n]", "Rearranging into standard quadratic form:\n[\nu^2 - 3u + 1 = 0\n]", "---", "### Step 2: Solve the Quadratic Equation (Optional)", "To deepen understanding, solve for ( u ) using the quadratic formula:\n[\nu = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(1)}}{2} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}\n]\nThus, the two real solutions for ( u ) are:\n[\nu = \frac{3 + \sqrt{5}}{2} \quad \ ext{and} \quad u = \frac{3 - \sqrt{5}}{2}\n]\nThese represent exact values consistent with the original equation.", "---", "### Step 3: Reinterpret in Terms of ( a ) and ( b )", "The derivation confirms that when ( u = \frac{a + b}{a - b} ), then\n[\nu + \frac{1}{u} = 3\n]\nis a valid algebraic identity — provided ( u <br/>\ne 0 ) and ( u > 0 ) (since ( \frac{1}{u} ) is defined only when ( u <br/>\ne 0 ), and positivity arises from sign considerations of ( a ) and ( b )).", "---", "### Why This Identity Matters", "Expressions of the form ( u + \frac{1}{u} ) often arise in:", "- Hyperbolic and trigonometric identities, e.g., relating to ( \cosh \ heta + \cosh(-\ heta) = 2\cosh\ heta )\n- Quadratic equation analysis, where symmetric sums appear\n- Optimization problems involving reciprocal terrain\n- Physics and engineering models involving ratios and scalings", "Recognizing patterns like ( u + \frac{1}{u} = 3 ) allows quick validation and simplification.", "---", "### Final Thoughts", "The identity\n[\nu + \frac{1}{u} = 3 \quad \ ext{when} \quad u = \frac{a + b}{a - b}\n]\nis a prime example of how substitution and basic algebra can uncover elegant relationships. By systematically multiplying through by ( u ), we transform a symmetric expression into a solvable quadratic, validating the identity elegantly.", "Whether you're solving equations, modeling systems, or teaching algebra, recognizing such forms saves time and deepens insight.", "---", "Keywords:\nalgebra, equation solving, ( u = \frac{a + b}{a - b} ), identity ( u + \frac{1}{u} = 3 ), quadratic equation, reciprocal expressions, mathematical derivation, elementary algebra, symmetric identities", "---", "Ready to apply this? Try plugging in values for ( a ) and ( b ) — explore how ( u + \frac{1}{u} ) always equals 3 by construction!"]
