We now solve the system $ a + b = 35 $, $ a^2 + b^2 = 625 $. Use identity:
["How to Solve the System $ a + b = 35 $ and $ a^2 + b^2 = 625 $: A Clear Algebraic Approach", "When faced with the system of equations:\n$$\na + b = 35 \quad \ ext{(1)}\n$$\n$$\na^2 + b^2 = 625 \quad \ ext{(2)}\n$$\nmany may hesitate, but using a powerful algebraic identity makes the solution straightforward and elegant.", "### Using the Identity: $ (a + b)^2 = a^2 + 2ab + b^2 $", "From equation (1), we know:\n$$\na + b = 35 \Rightarrow (a + b)^2 = 35^2 = 1225\n$$", "From equation (2):\n$$\na^2 + b^2 = 625\n$$", "Substitute into the identity:\n$$\n(a + b)^2 = a^2 + b^2 + 2ab\n$$\n$$\n1225 = 625 + 2ab\n$$", "Now solve for $ ab $:\n$$\n1225 - 625 = 2ab \Rightarrow 600 = 2ab \Rightarrow ab = 300\n$$", "### Summary of what we have:\n- $ a + b = 35 $\n- $ ab = 300 $", "This means $ a $ and $ b $ are the roots of the quadratic equation:\n$$\nx^2 - (a + b)x + ab = 0 \Rightarrow x^2 - 35x + 300 = 0\n$$", "### Solve the quadratic:\nUse the quadratic formula:\n$$\nx = \frac{35 \pm \sqrt{(-35)^2 - 4 \cdot 1 \cdot 300}}{2}\n= \frac{35 \pm \sqrt{1225 - 1200}}{2}\n= \frac{35 \pm \sqrt{25}}{2}\n= \frac{35 \pm 5}{2}\n$$", "So,\n$$\nx = \frac{40}{2} = 20 \quad \ ext{or} \quad x = \frac{30}{2} = 15\n$$", "Thus, the solutions are $ a = 20, b = 15 $ or $ a = 15, b = 20 $.", "### Final Answer:\nThe system $ a + b = 35 $, $ a^2 + b^2 = 625 $ has solutions $ a = 15 $, $ b = 20 $ (or vice versa), derived efficiently using the identity $ (a + b)^2 = a^2 + b^2 + 2ab $.", "This method avoids complex substitution and reveals the elegant symmetry in symmetric equations — a powerful tool in algebra!"]
