Solution: We are given $ a - b = 6 $ and $ a^2 + b^2 = 130 $.

["Solving the System: How to Find $ a $ and $ b $ Given $ a - b = 6 $ and $ a^2 + b^2 = 130 $", "When faced with the equations\n$$\na - b = 6 \quad \ ext{and} \quad a^2 + b^2 = 130,\n$$\nmany might wonder how to efficiently solve for $ a $ and $ b $. This problem is a classic algebra challenge often used in math competitions, interviews, and advanced problem-solving contexts. This article breaks down the step-by-step solution and explains the underlying mathematical principles — all while optimizing for SEO with clear structure and keyword focus.", "---", "### Understanding the Problem", "We are given:\n1. $ a - b = 6 $\n2. $ a^2 + b^2 = 130 $", "Our goal is to find the real numbers $ a $ and $ b $ satisfying both equations. This type of problem is ideal for testing algebraic manipulation, substitution, and quadratic identities.", "---", "### Step-by-Step Solution", "Step 1: Use substitution from the first equation\nFrom $ a - b = 6 $, express $ a $ in terms of $ b $:\n$$\na = b + 6\n$$", "Step 2: Substitute into the second equation\nReplace $ a $ with $ b + 6 $ in $ a^2 + b^2 = 130 $:\n$$\n(b + 6)^2 + b^2 = 130\n$$", "Expand $ (b + 6)^2 $:\n$$\nb^2 + 12b + 36 + b^2 = 130\n$$", "Combine like terms:\n$$\n2b^2 + 12b + 36 = 130\n$$", "Step 3: Simplify the quadratic equation\nSubtract 130 from both sides:\n$$\n2b^2 + 12b - 94 = 0\n$$", "Divide through by 2 to simplify:\n$$\nb^2 + 6b - 47 = 0\n$$", "Step 4: Solve the quadratic using the quadratic formula\nUse the formula $ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} $, with $ A = 1 $, $ B = 6 $, $ C = -47 $:\n$$\nb = \frac{-6 \pm \sqrt{6^2 - 4(1)(-47)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 188}}{2} = \frac{-6 \pm \sqrt{224}}{2}\n$$", "Simplify $ \sqrt{224} $:\n$$\n\sqrt{224} = \sqrt{16 \ imes 14} = 4\sqrt{14}\n$$", "So,\n$$\nb = \frac{-6 \pm 4\sqrt{14}}{2} = -3 \pm 2\sqrt{14}\n$$", "Step 5: Find corresponding $ a $ values\nRecall $ a = b + 6 $:", "- If $ b = -3 + 2\sqrt{14} $, then $ a = 3 + 2\sqrt{14} $\n- If $ b = -3 - 2\sqrt{14} $, then $ a = 3 - 2\sqrt{14} $", "---", "### Final Answer\nThe solutions are:\n$$\n(a, b) = (3 + 2\sqrt{14},\ -3 + 2\sqrt{14}) \quad \ ext{or} \quad (3 - 2\sqrt{14},\ -3 - 2\sqrt{14})\n$$", "---", "### Why This Problem Matters in Algebra and Problem Solving", "Problems involving sum of squares and linear differences are fundamental in algebraic modeling. They appear in geometry (distance from origin), statistics (variance identities), and optimization. Mastering these techniques improves your ability to handle systems of equations efficiently.", "---", "### SEO Keywords for Search Optimization\n- Solve $ a - b = 6 and $ a^2 + b^2 = 130\n- Algebra problem solution $ a $ and $ b $ given linear and quadratic equations\n- How to solve equations with difference and sum of squares\n- Step-by-step system of equations with square terms\n- Algebra tutorial: Find $ a $ and $ b $ from $ a - b = 6 $ and $ a^2 + b^2 = 130 $", "---", "### Conclusion", "By applying substitution and the quadratic formula, we efficiently solve for $ a $ and $ b $ in just a few clear steps. This method not only yields precise values but also enhances your algebraic toolkit. Whether for homework, interviews, or self-study, mastering this system prepares you for more complex mathematical challenges ahead.", "---", "*Keywords: solve $ a - b = 6 $, $ a^2 + b^2 = 130 $, algebra solution, quadratic equations, substitution method, system of equations."]