a + b = 12 \quad \text{and} \quad a^2 + b^2 = 80

["Solving the Equation: + b = 12 and a² + b² = 80 – A Step-by-Step Guide", "Mathematics presents us with intriguing challenges, and combining simple linear equations with nonlinear expressions often reveals elegant solutions. One such problem is:\n$$\na + b = 12 \ ag{1}\n$$\n$$\na^2 + b^2 = 80 \ ag{2}\n$$\nOur goal is to find the values of $a$ and $b$ that satisfy both equations—and uncover deeper insights into their relationship.", "### Step 1: Express One Variable in Terms of the Other\nFrom equation (1), solve for $b$:\n$$\nb = 12 - a\n$$", "### Step 2: Substitute into the Second Equation\nReplace $b$ in equation (2):\n$$\na^2 + (12 - a)^2 = 80\n$$\nExpand the squared term:\n$$\na^2 + (144 - 24a + a^2) = 80\n$$\nCombine like terms:\n$$\n2a^2 - 24a + 144 = 80\n$$\nSubtract 80 from both sides:\n$$\n2a^2 - 24a + 64 = 0\n$$\nDivide throughout by 2:\n$$\na^2 - 12a + 32 = 0\n$$", "### Step 3: Solve the Quadratic Equation\nUse the quadratic formula:\n$$\na = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(32)}}{2(1)} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2} = \frac{12 \pm 4}{2}\n$$\nThis gives two solutions:\n$$\na = \frac{12 + 4}{2} = 8 \quad \ ext{and} \quad a = \frac{12 - 4}{2} = 4\n$$", "### Step 4: Find Corresponding Values of $b$\nIf $a = 8$, then $b = 12 - 8 = 4$.\nIf $a = 4$, then $b = 12 - 4 = 8$.", "Thus, the solutions are $(a, b) = (8, 4)$ or $(4, 8)$.", "### Step 5: Verify the Solution\nCheck $a^2 + b^2 = 80$:\n- $8^2 + 4^2 = 64 + 16 = 80$ ✅\n- $4^2 + 8^2 = 16 + 64 = 80$ ✅", "Both pairs satisfy all original equations.", "### Why This Problem Matters\nUnderstanding how to solve systems involving linear and quadratic terms builds foundational skills for algebra, physics, and engineering. This specific case demonstrates how substituting variables and simplifying equations leads directly to solutions—key competencies in mathematical problem-solving.", "### Final Insight\nThe equation $a + b = 12$ defines a line, while $a^2 + b^2 = 80$ represents a circle of radius $\sqrt{80}$ centered at the origin. Their intersections—$(4,8)$ and $(8,4)$—highlight where these two geometric shapes meet, providing both numerical and visual clarity.", "Key takeaway: By combining substitution and algebra, complex relationships can be tested and solved efficiently—essential knowledge for students and professionals alike.", "---", "Related SEO Keywords:\nSolve quadratic equations, linear and quadratic systems, algebra problem solving, a + b = 12, a² + b² = 80, mathematical identities, substitution method, systems of equations, geometry and algebra intersection, quadratic solutions."]