Solution: Group and complete the square: $ 4(x^2 + 2x) - 9(y^2 - 2y) = 11 $. Complete squares: $ 4[(x+1)^2 - 1] - 9[(y-1)^2 - 1] = 11 $. Expand: $ 4(x+1)^2 - 4 - 9(y-1)^2 + 9 = 11 \Rightarrow 4(x+1)^2 - 9(y-1)^2 = 6 $. Rewrite: $ \frac{(x+1)^2}{\frac{3}{2}} - \frac{(y-1)^2}{\frac{2}{3}} = 1 $. Center at $ (-1, 1) $. \boxed{(-1, 1)}1. **A chemistry experiment requires 50 mL of a 10% salt solution. If you only have a 20% salt solution and distilled water, how much of each should be mixed to obtain

["Solving Salt Solution Mixing Problems: A Step-by-Step Chemistry Approach", "When preparing precise saline solutions for experiments, accurately mixing concentrated stock solutions with water is essential. Consider a scenario in chemistry: preparing 50 mL of a 10% salt solution using only 20% (stock) salt solution and distilled water. How much of each should be combined?", "### Step 1: Define Variables\nLet ( x ) = volume in mL of 20% salt solution used.\nThen, ( 50 - x ) = volume of distilled water added (since total volume must be 50 mL and water has 0% salt).", "### Step 2: Set Up the Mass Balance Equation\nThe total amount of salt in the final solution must equal 10% of 50 mL:\n[\n\ ext{Salt from stock} + \ ext{Salt from water} = \ ext{Salt in final solution}\n]", "Since distilled water contributes no salt:\n[\n0.20x + 0 = 0.10 \ imes 50\n]\n[\n0.20x = 5\n]", "### Step 3: Solve for ( x )\n[\nx = \frac{5}{0.20} = 25\n]\nThus, 25 mL of the 20% solution is used.", "### Step 4: Determine Volume of Water\n[\n50 - 25 = 25 \ ext{ mL of distilled water}\n]", "### Final Answer\nTo prepare 50 mL of a 10% salt solution, mix 25 mL of the 20% salt solution with 25 mL of distilled water.", "---", "This method follows the principle of conservation of mass in mixing problems, ensuring accuracy crucial in chemical experiments. Mixing by volume rather than mass accounts for salt concentration differences effectively.", "\boxed{25\ ext{ mL of }20%\ ext{ solution and }25\ ext{ mL water}}"]