Case 1:** \( rac{r+1}{r-1} = rac{3 + \sqrt{5}}{2} \)

["# Case 1: Solving ( \dfrac{r+1}{r-1} = \dfrac{3 + \sqrt{5}}{2} ) — A Step-by-Step Guide", "Solving rational equations like ( \dfrac{r+1}{r-1} = \dfrac{3 + \sqrt{5}}{2} ) is a fundamental algebraic skill with broad applications in science, engineering, and mathematics. This article walks you through Case 1 — solving such an equation step-by-step — offering clear explanations and practical insights to help you tackle similar problems with confidence.", "---", "## Understanding the Equation: ( \dfrac{r+1}{r-1} = \dfrac{3 + \sqrt{5}}{2} )", "At its core, this equation asks: For what value of ( r ) does the rational expression ( \dfrac{r+1}{r-1} ) equal the irrational number ( \dfrac{3 + \sqrt{5}}{2} )?", "Our goal is to isolate ( r ) by eliminating the denominator through algebraic manipulation.", "---", "## Step-by-Step Solution", "### Step 1: Eliminate the Denominator\nStart with the original equation:", "[\n\dfrac{r+1}{r-1} = \dfrac{3 + \sqrt{5}}{2}\n]", "Multiply both sides by ( r - 1 ) (assuming ( r <br/>\ne 1 ), since that would make the denominator zero):", "[\nr + 1 = \dfrac{3 + \sqrt{5}}{2} (r - 1)\n]", "---", "### Step 2: Expand the Right-Hand Side\nDistribute the factor on the right:", "[\nr + 1 = \dfrac{3 + \sqrt{5}}{2} \cdot r - \dfrac{3 + \sqrt{5}}{2}\n]", "---", "### Step 3: Gather All Terms Involving ( r ) on One Side\nSubtract ( \dfrac{3 + \sqrt{5}}{2} r ) from both sides:", "[\nr - \dfrac{3 + \sqrt{5}}{2} r + 1 = -\dfrac{3 + \sqrt{5}}{2}\n]", "Factor ( r ) on the left:", "[\nr \left(1 - \dfrac{3 + \sqrt{5}}{2}\right) + 1 = -\dfrac{3 + \sqrt{5}}{2}\n]", "---", "### Step 4: Simplify the Coefficient of ( r )", "Convert 1 to a fraction with denominator 2:", "[\n1 = \dfrac{2}{2}\n]", "So:", "[\nr \left( \dfrac{2 - (3 + \sqrt{5})}{2} \right) + 1 = -\dfrac{3 + \sqrt{5}}{2}\n]", "Simplify inside the parentheses:", "[\n2 - 3 - \sqrt{5} = -1 - \sqrt{5}\n]", "Thus:", "[\nr \left( \dfrac{ -1 - \sqrt{5} }{2} \right) + 1 = -\dfrac{3 + \sqrt{5}}{2}\n]", "---", "### Step 5: Isolate the Term with ( r )", "Subtract 1 from both sides:", "[\nr \left( \dfrac{ -1 - \sqrt{5} }{2} \right) = -\dfrac{3 + \sqrt{5}}{2} - 1\n]", "Express 1 as ( \dfrac{2}{2} ):", "[\nr \left( \dfrac{ -1 - \sqrt{5} }{2} \right) = -\left( \dfrac{3 + \sqrt{5} + 2}{2} \right) = -\dfrac{5 + \sqrt{5}}{2}\n]", "---", "### Step 6: Solve for ( r )", "Multiply both sides by the reciprocal of the coefficient:", "[\nr = \dfrac{ -\dfrac{5 + \sqrt{5}}{2} }{ \dfrac{ -1 - \sqrt{5} }{2} }\n]", "The denominators 2 cancel:", "[\nr = \dfrac{ -(5 + \sqrt{5}) }{ -(1 + \sqrt{5}) } = \dfrac{5 + \sqrt{5}}{1 + \sqrt{5}}\n]", "---", "### Step 7: Rationalize the Denominator", "Multiply numerator and denominator by the conjugate ( 1 - \sqrt{5} ):", "[\nr = \dfrac{(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 \cdot 1 + 5 \cdot (-\sqrt{5}) + \sqrt{5} \cdot 1 + \sqrt{5} \cdot (-\sqrt{5})\n]\n[\n= 5 - 5\sqrt{5} + \sqrt{5} - 5 = (5 - 5) + (-5\sqrt{5} + \sqrt{5}) = -4\sqrt{5}\n]", "Thus:", "[\nr = \dfrac{ -4\sqrt{5} }{ -4 } = \sqrt{5}\n]", "---", "## Final Answer", "[\n\boxed{r = \sqrt{5}}\n]", "---", "## Verification", "Plug ( r = \sqrt{5} ) back into the original equation:", "[\n\dfrac{\sqrt{5} + 1}{\sqrt{5} - 1}, \quad \ ext{and} \quad \dfrac{3 + \sqrt{5}}{2}\n]", "Rationalize the denominator by multiplying numerator and denominator by ( \sqrt{5} + 1 ):", "[\n\dfrac{(\sqrt{5}+1)^2 }{(\sqrt{5})^2 - 1^2} = \dfrac{(5 + 2\sqrt{5} + 1)}{5 - 1} = \dfrac{6 + 2\sqrt{5}}{4} = \dfrac{3 + \sqrt{5}}{2}\n]", "This matches the right-hand side, confirming the solution is correct.", "---", "## Why This Case Matters", "- Algebraic Mastery: Demonstrates techniques such as cross-multiplication, coefficient isolation, and rationalization — essential for more complex equations.\n- Irrational Solutions: Shows how real-world constants like ( \sqrt{5} ) appear in equations and how to solve for them cleanly.\n- Applications: Useful in physics, finance, and geometry where rational equations model rates, proportions, or geometric ratios.", "---", "## Summary", "Case 1 in solving ( \dfrac{r+1}{r-1} = \dfrac{3 + \sqrt{5}}{2} ) involves careful algebraic manipulation—clear steps that eliminate the denominator, isolate ( r ), simplify, and rationalize. The solution confirms ( r = \sqrt{5} ), a key irrational value rooted in number theory. This example strengthens foundational algebra skills vital for advanced problem-solving across STEM fields.", "Whether you're a student mastering algebra or a professional applying equations daily, mastering such cases ensures you handle complex expressions with clarity and precision."]