2(r+1) = (3 + \sqrt{5})(r - 1)

["Solving the Linear Equation: 2(r + 1) = (3 + √5)(r – 1)", "Understanding how to solve linear equations is a cornerstone of algebra, and today we focus on a specific equation:\n2(r + 1) = (3 + √5)(r – 1)", "Whether you’re a student mastering algebra or someone reinforcing math skills, solving equations with irrational numbers like √5 can feel tricky. But with clear steps, clarity improves, and solutions become straightforward.", "---", "### Step-by-Step Solution", "We begin with the given equation:", "[\n2(r + 1) = (3 + \sqrt{5})(r - 1)\n]", "Step 1: Expand both sides\nFirst, expand the left-hand side:", "[\n2r + 2\n]", "Then expand the right-hand side using distributive property:", "[\n(3 + \sqrt{5})(r - 1) = 3r - 3 + \sqrt{5}r - \sqrt{5}\n]", "So the equation becomes:", "[\n2r + 2 = 3r - 3 + \sqrt{5}r - \sqrt{5}\n]", "---", "Step 2: Bring all terms to one side\nMove all terms to the left-hand side to collect like terms:", "[\n2r + 2 - 3r + 3 - \sqrt{5}r + \sqrt{5} = 0\n]", "Combine like terms:", "- (2r - 3r - \sqrt{5}r = (-1 - \sqrt{5})r)\n- (2 + 3 + \sqrt{5} = 5 + \sqrt{5})", "Resulting equation:", "[\n(-1 - \sqrt{5})r + (5 + \sqrt{5}) = 0\n]", "---", "Step 3: Isolate (r)", "Move constants to the right:", "[\n(-1 - \sqrt{5})r = - (5 + \sqrt{5})\n]", "Multiply both sides by -1:", "[\n(1 + \sqrt{5})r = 5 + \sqrt{5}\n]", "---", "Step 4: Solve for (r)", "Divide both sides by (1 + \sqrt{5}):", "[\nr = \frac{5 + \sqrt{5}}{1 + \sqrt{5}}\n]", "To simplify, rationalize the denominator by multiplying numerator and denominator by the conjugate (1 - \sqrt{5}):", "[\nr = \frac{(5 + \sqrt{5})(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})}\n]", "Compute the denominator:", "[\n(1 + \sqrt{5})(1 - \sqrt{5}) = 1^2 - (\sqrt{5})^2 = 1 - 5 = -4\n]", "Compute the numerator:", "[\n(5 + \sqrt{5})(1 - \sqrt{5}) = 5(1) - 5\sqrt{5} + \sqrt{5}(1) - \sqrt{5} \cdot \sqrt{5} = 5 - 5\sqrt{5} + \sqrt{5} - 5 = (5 - 5) + (-5\sqrt{5} + \sqrt{5}) = -4\sqrt{5}\n]", "So,", "[\nr = \frac{-4\sqrt{5}}{-4} = \sqrt{5}\n]", "---", "### Final Answer", "[\n\boxed{r = \sqrt{5}}\n]", "---", "### Why This Equation Matters", "Equation solving with irrational numbers like √5 appears in many real-world and advanced math contexts—such as coordinate geometry, physics, engineering, and algebra-based modeling. Mastering such problems strengthens algebraic intuition and prepares learners for solving equations with radicals in higher mathematics.", "---", "### Tips for Solving Similar Equations", "- Always expand parentheses carefully\n- Combine like terms methodically\n- Use conjugates when denominators contain square roots\n- Double-check your simplification and sign", "If you're prepping for exams or personally learning algebra, mastering this type of equation builds a solid foundation for future math success.", "---", "Related Keywords for SEO:\n2(r + 1) = (3 + √5)(r – 1), solving linear equations with radicals, irrational numbers in algebra, rationalizing denominators, algebraic equation solving, step-by-step linear equation solution, resolving √5 in equations.", "---", "Conclusion\nSolving 2(r + 1) = (3 + √5)(r – 1) yields r = √5, demonstrating how irrational numbers interact in linear equations. Practice with simple expansions and rationalizing steps will transform abstract symbols into clear, solvable expressions. Keep practicing — algebra becomes intuitive with persistence!"]