Solution: We are given $ x + y = 12 $ and $ x^2 + y^2 = 80 $.

["Solving the System: How to Use $ x + y = 12 $ and $ x^2 + y^2 = 80 $ to Find $ x $ and $ y $", "When given two equations — $ x + y = 12 $ and $ x^2 + y^2 = 80 $ — many learners feel puzzled. However, with the right algebraic approach, you can efficiently solve for the values of $ x $ and $ y $. This article walks you through the step-by-step solution and explains how to use these equations in algebra, math problems, and even real-world applications.", "---", "### Understanding the Given Equations", "We are given:", "1. $ x + y = 12 $\n2. $ x^2 + y^2 = 80 $", "These equations are useful in various mathematical contexts, including geometry, optimization, and systems modeling. Our goal is to find possible values of $ x $ and $ y $ that satisfy both equations.", "---", "### Step 1: Use Algebraic Identity to Simplify", "To link the sum of variables and the sum of their squares, recall the identity:", "[\n(x + y)^2 = x^2 + y^2 + 2xy\n]", "From equation (1), $ x + y = 12 $. So,", "[\n(x + y)^2 = 12^2 = 144\n]", "Substitute into the identity:", "[\n144 = x^2 + y^2 + 2xy\n]", "Now use equation (2), $ x^2 + y^2 = 80 $:", "[\n144 = 80 + 2xy\n]", "---", "### Step 2: Solve for $ xy $", "Subtract 80 from both sides:", "[\n64 = 2xy\n]", "Divide by 2:", "[\nxy = 32\n]", "Now we know:", "- $ x + y = 12 $\n- $ xy = 32 $", "These are the sum and product of two numbers — classic values for forming a quadratic equation.", "---", "### Step 3: Form a Quadratic Equation", "If $ x $ and $ y $ are roots of a quadratic equation, the equation is:", "[\nt^2 - (x + y)t + xy = 0\n]", "Substitute known values:", "[\nt^2 - 12t + 32 = 0\n]", "---", "### Step 4: Solve the Quadratic Equation", "Use the quadratic formula:", "[\nt = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(32)}}{2(1)} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2}\n]", "[\nt = \frac{12 \pm 4}{2}\n]", "So,", "[\nt = \frac{12 + 4}{2} = 8 \quad \ ext{or} \quad t = \frac{12 - 4}{2} = 4\n]", "---", "### Step 5: Final Solution", "Thus, the values of $ x $ and $ y $ are:", "[\nx = 8, \quad y = 4 \quad \ ext{or} \quad x = 4, \quad y = 8\n]", "Both satisfy the original equations:", "- $ 8 + 4 = 12 $ ✅\n- $ 8^2 + 4^2 = 64 + 16 = 80 $ ✅\n- $ 4 + 8 = 12 $ ✅\n- $ 4^2 + 8^2 = 16 + 64 = 80 $ ✅", "---", "### Why This Solution Matters (Applications)", "Equations of this form appear in:", "- Geometry: Finding side lengths or distances in right triangles.\n- Economics: Analyzing revenue and cost combinations.\n- Data Science: Working with correlation and variance in datasets.", "Understanding how to manipulate sum and sum-of-squares relationships helps simplify problems and find optimal solutions.", "---", "### Key Takeaways", "- Use $ (x + y)^2 $ and the identity to eliminate $ x^2 + y^2 $.\n- From $ x + y $ and $ xy $, construct a quadratic equation.\n- Solve using the quadratic formula.\n- Both variables are symmetric — order matters only in interpretation.", "---", "### Try It Yourself", "Next time you encounter $ x + y = a $ and $ x^2 + y^2 = b $, remember:", "1. Square the sum: $ (x + y)^2 = a^2 $\n2. Plug into identity: $ a^2 = b + 2xy $\n3. Solve for $ xy $\n4. Form $ t^2 - at + xy = 0 $\n5. Solve using quadratic formula", "This method works every time!", "---", "Summary: Given $ x + y = 12 $ and $ x^2 + y^2 = 80 $, the solution is $ x = 8, y = 4 $ (or vice versa). Use algebraic identities and quadratic equations to solve such systems efficiently. These skills form the foundation for tackling more advanced math problems.", "---", "Keywords: solve $ x + y = 12 $ and $ x^2 + y^2 = 80 $, algebraic system solution, quadratic equations, sum and product of roots, math problem-solving, algebra tutorials, $ x $ and $ y $ values."]