diff --git a/lib/solver_service.dart b/lib/solver_service.dart
index c8a9403..acd7c57 100644
--- a/lib/solver_service.dart
+++ b/lib/solver_service.dart
@@ -56,13 +56,12 @@ class SolverService {
     final steps = <CalculationStep>[];
     steps.add(
       CalculationStep(
-        title: "第一步：表达式求值",
-        explanation: "这是一个标准的数学表达式，我们将直接计算其结果。",
+        title: '第一步：表达式求值',
+        explanation: '这是一个标准的数学表达式，我们将直接计算其结果。',
         formula: input,
       ),
     );
 
-    // **FIXED**: Correct usage of the math_expressions library
     Parser p = Parser();
     Expression exp = p.parse(input);
     ContextModel cm = ContextModel();
@@ -75,134 +74,118 @@ class SolverService {
   CalculationResult _solveLinearEquation(String input) {
     final steps = <CalculationStep>[];
     steps.add(
-      CalculationStep(title: "原方程", explanation: "这是一元一次方程。", formula: input),
+      CalculationStep(
+        title: '原方程',
+        explanation: '这是一元一次方程。',
+        formula: '\$\$$input\$\$',
+      ),
     );
 
-    // 解析方程: ax+b=cx+d
     final parts = _parseLinearEquation(input);
     final a = parts.a, b = parts.b, c = parts.c, d = parts.d;
 
-    // 移项合并
     final newA = a - c;
     final newD = d - b;
 
     steps.add(
       CalculationStep(
-        title: "第一步：移项",
-        explanation: "将所有含 x 的项移到等式左边，常数项移到右边。",
+        title: '第一步：移项',
+        explanation: '将所有含 x 的项移到等式左边，常数项移到右边。',
         formula:
-            "${a}x ${c >= 0 ? '-' : '+'} ${c.abs()}x = $d ${b >= 0 ? '-' : '+'} ${b.abs()}",
+            '\$\$${a}x ${c >= 0 ? '-' : '+'} ${c.abs()}x = $d ${b >= 0 ? '-' : '+'} ${b.abs()}\$\$',
       ),
     );
 
     steps.add(
       CalculationStep(
-        title: "第二步：合并同类项",
-        explanation: "合并等式两边的项。",
-        formula: "${newA}x = $newD",
+        title: '第二步：合并同类项',
+        explanation: '合并等式两边的项。',
+        formula: '\$\$${newA}x = $newD\$\$',
       ),
     );
 
-    // 求解x
     if (newA == 0) {
-      if (newD == 0) {
-        return CalculationResult(steps: steps, finalAnswer: "有无穷多解");
-      } else {
-        return CalculationResult(steps: steps, finalAnswer: "无解");
-      }
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: newD == 0 ? '有无穷多解' : '无解',
+      );
     }
 
     final x = newD / newA;
     steps.add(
       CalculationStep(
-        title: "第三步：求解 x",
-        explanation: "两边同时除以 x 的系数 ($newA)。",
-        formula: "x = $newD / $newA",
+        title: '第三步：求解 x',
+        explanation: '两边同时除以 x 的系数 ($newA)。',
+        formula: '\$\$x = \frac{$newD}{$newA}\$\$',
       ),
     );
 
-    return CalculationResult(steps: steps, finalAnswer: "x = $x");
+    return CalculationResult(steps: steps, finalAnswer: '\$\$x = $x\$\$');
   }
 
   /// 3. 求解一元二次方程 (升级版)
   CalculationResult _solveQuadraticEquation(String input) {
     final steps = <CalculationStep>[];
 
-    // 整理成 ax^2+bx+c=0 的形式并提取系数
     final eqParts = input.split('=');
     if (eqParts.length != 2) throw Exception("方程格式错误，应包含一个 '='。");
 
     final leftCoeffs = _parsePolynomial(eqParts[0]);
     final rightCoeffs = _parsePolynomial(eqParts[1]);
 
-    // 移项合并
     final a = (leftCoeffs[2] ?? 0) - (rightCoeffs[2] ?? 0);
     final b = (leftCoeffs[1] ?? 0) - (rightCoeffs[1] ?? 0);
     final c = (leftCoeffs[0] ?? 0) - (rightCoeffs[0] ?? 0);
 
     if (a == 0) {
-      // 如果 a=0, 这实际上是一个一元一次方程
-      return _solveLinearEquation("${b}x+${c}=0");
+      return _solveLinearEquation('${b}x+$c=0');
     }
 
-    // **FIXED**: Corrected typo 'ax' to 'a'
     steps.add(
       CalculationStep(
-        title: "第一步：整理方程",
-        explanation: "将方程整理成标准形式 ax^2+bx+c=0。",
+        title: '第一步：整理方程',
+        explanation: r'将方程整理成标准形式 $ax^2+bx+c=0$。',
         formula:
-            "${a}x^2 ${b >= 0 ? '+' : ''} ${b}x ${c >= 0 ? '+' : ''} $c = 0",
+            '\$\$${a}x^2 ${b >= 0 ? '+' : ''} ${b}x ${c >= 0 ? '+' : ''} $c = 0\$\$',
       ),
     );
 
-    // 尝试因式分解 (十字相乘法)
     if (a == a.round() && b == b.round() && c == c.round()) {
       final factors = _tryFactorization(a.toInt(), b.toInt(), c.toInt());
       if (factors != null) {
         steps.add(
           CalculationStep(
-            title: "第二步：因式分解法 (十字相乘)",
-            explanation: "我们发现可以将方程分解为两个一次因式的乘积。",
+            title: '第二步：因式分解法 (十字相乘)',
+            explanation: '我们发现可以将方程分解为两个一次因式的乘积。',
             formula: factors.formula,
           ),
         );
-
         steps.add(
           CalculationStep(
-            title: "第三步：求解",
-            explanation: "分别令每个因式等于 0，解出 x。",
-            formula: "解得 ${factors.solution}",
+            title: '第三步：求解',
+            explanation: '分别令每个因式等于 0，解出 x。',
+            formula: '解得 ${factors.solution}',
           ),
         );
-
         return CalculationResult(steps: steps, finalAnswer: factors.solution);
-      } else {
-        steps.add(
-          CalculationStep(
-            title: "第二步：选择解法",
-            explanation: "尝试因式分解失败，我们选择使用求根公式法。",
-            formula: "\$\\Delta = b^2 - 4ac\$",
-          ),
-        );
       }
-    } else {
-      steps.add(
-        CalculationStep(
-          title: "第二步：选择解法",
-          explanation: "系数包含小数，我们使用求根公式法。",
-          formula: "\$\\Delta = b^2 - 4ac\$",
-        ),
-      );
     }
 
-    // 使用求根公式法
+    steps.add(
+      CalculationStep(
+        title: '第二步：选择解法',
+        explanation: '无法进行因式分解，我们选择使用求根公式法。',
+        formula: r'\$\$\Delta = b^2 - 4ac\$\$',
+      ),
+    );
+
     final delta = b * b - 4 * a * c;
     steps.add(
       CalculationStep(
-        title: "第三步：计算判别式 (Delta)",
-        explanation: "\$\\Delta = ($b)^2 - 4 \\cdot ($a) \\cdot ($c) = $delta",
-        // **FIXED**: Removed unnecessary braces for linter warning
-        formula: "\$\\Delta = $delta",
+        title: '第三步：计算判别式 (Delta)',
+        explanation:
+            '\$\$\Delta = b^2 - 4ac = ($b)^2 - 4 \cdot ($a) \cdot ($c) = $delta\$\$',
+        formula: '\$\$\Delta = $delta\$\$',
       ),
     );
 
@@ -211,40 +194,40 @@ class SolverService {
       final x2 = (-b - sqrt(delta)) / (2 * a);
       steps.add(
         CalculationStep(
-          title: "第四步：应用求根公式",
+          title: '第四步：应用求根公式',
           explanation:
-              "因为 \$\\Delta > 0\$，方程有两个不相等的实数根。公式: \$x = \\frac{-b \\pm \\sqrt{\\Delta}}{2a}\$",
+              r'因为 $\Delta > 0$，方程有两个不相等的实数根。公式: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$。',
           formula:
-              "x₁ = ${x1.toStringAsFixed(4)}, x₂ = ${x2.toStringAsFixed(4)}",
+              '\$\$x_1 = ${x1.toStringAsFixed(4)}, \quad x_2 = ${x2.toStringAsFixed(4)}\$\$',
         ),
       );
       return CalculationResult(
         steps: steps,
         finalAnswer:
-            "x₁ = ${x1.toStringAsFixed(4)}, x₂ = ${x2.toStringAsFixed(4)}",
+            '\$\$x_1 = ${x1.toStringAsFixed(4)}, \quad x_2 = ${x2.toStringAsFixed(4)}\$\$',
       );
     } else if (delta == 0) {
       final x = -b / (2 * a);
       steps.add(
         CalculationStep(
-          title: "第四步：应用求根公式",
-          explanation: "因为 \$\Delta = 0\$，方程有两个相等的实数根。",
-          formula: "x₁ = x₂ = ${x.toStringAsFixed(4)}",
+          title: '第四步：应用求根公式',
+          explanation: r'因为 $\Delta = 0$，方程有两个相等的实数根。',
+          formula: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
         ),
       );
       return CalculationResult(
         steps: steps,
-        finalAnswer: "x₁ = x₂ = ${x.toStringAsFixed(4)}",
+        finalAnswer: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
       );
     } else {
       steps.add(
         CalculationStep(
-          title: "第四步：判断解",
-          explanation: "因为 \$\Delta < 0\$，该方程在实数范围内无解。",
-          formula: "无实数解",
+          title: '第四步：判断解',
+          explanation: r'因为 $\Delta < 0$，该方程在实数范围内无解。',
+          formula: '无实数解',
         ),
       );
-      return CalculationResult(steps: steps, finalAnswer: "无实数解");
+      return CalculationResult(steps: steps, finalAnswer: '无实数解');
     }
   }
 
@@ -262,32 +245,44 @@ class SolverService {
 
     steps.add(
       CalculationStep(
-        title: "原始方程组",
-        explanation: "这是一个二元一次方程组，我们将使用加减消元法求解。",
+        title: '原始方程组',
+        explanation: '这是一个二元一次方程组，我们将使用加减消元法求解。',
         formula:
-            "${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1  ---(1)\n${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2  ---(2)",
+            '''
+
+egin{cases}
+${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 & (1) \\
+${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 & (2)
+\\end{cases}
+
+''',
       ),
     );
 
     final det = a1 * b2 - a2 * b1;
     if (det == 0) {
-      if (a1 * c2 - a2 * c1 == 0) {
-        return CalculationResult(steps: steps, finalAnswer: "有无穷多解");
-      } else {
-        return CalculationResult(steps: steps, finalAnswer: "无解");
-      }
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: a1 * c2 - a2 * c1 == 0 ? '有无穷多解' : '无解',
+      );
     }
 
     final newA1 = a1 * b2, newC1 = c1 * b2;
     final newA2 = a2 * b1, newC2 = c2 * b1;
 
-    // **FIXED**: Wrapped calculations in braces {} for string interpolation
     steps.add(
       CalculationStep(
-        title: "第一步：消元",
-        explanation: "为了消去变量 y，将方程(1)两边乘以 $b2\$，方程(2)两边乘以 $b1\$。",
+        title: '第一步：消元',
+        explanation: '为了消去变量 y，将方程(1)两边乘以 $b2，方程(2)两边乘以 $b1。',
         formula:
-            "${newA1}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC1  ---(3)\n${newA2}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC2  ---(4)",
+            '''
+
+egin{cases}
+${newA1}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC1 & (3) \\
+${newA2}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC2 & (4)
+\\end{cases}
+
+''',
       ),
     );
 
@@ -296,79 +291,96 @@ class SolverService {
 
     steps.add(
       CalculationStep(
-        title: "第二步：相减",
-        explanation: "将方程(3)减去方程(4)，得到一个只含 x 的方程。",
+        title: '第二步：相减',
+        explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
         formula:
-            "($newA1 - $newA2)x = $newC1 - $newC2 \n=> ${xCoeff}x = $constCoeff",
+            '\$\$($newA1 - $newA2)x = $newC1 - $newC2 \Rightarrow ${xCoeff}x = $constCoeff\$\$',
       ),
     );
 
     final x = constCoeff / xCoeff;
     steps.add(
       CalculationStep(
-        title: "第三步：解出 x",
-        explanation: "求解上述方程得到 x 的值。",
-        formula: "x = $x",
+        title: '第三步：解出 x',
+        explanation: '求解上述方程得到 x 的值。',
+        formula: '\$\$x = $x\$\$',
       ),
     );
 
-    // **FIXED**: Added a check for b1 to avoid division by zero if we substitute into an equation without y.
     if (b1.abs() < 1e-9) {
-      // If b1 is very close to 0, use the second equation
       final yCoeff = b2;
       final yConst = c2 - a2 * x;
       final y = yConst / yCoeff;
       steps.add(
         CalculationStep(
-          title: "第四步：回代求解 y",
-          explanation: "将 x = $x 代入原方程(2)中。",
-          // **FIXED**: Corrected string interpolation for calculations and substitutions
+          title: '第四步：回代求解 y',
+          explanation: '将 x = $x 代入原方程(2)中。',
           formula:
-              "${a2}(${x}) + ${b2}y = $c2 \n=> ${a2 * x} + ${b2}y = $c2 \n=> ${b2}y = $c2 - ${a2 * x} \n=> ${b2}y = ${c2 - a2 * x}",
+              '''
+
+\\begin{aligned}
+$a2($x) + ${b2}y &= $c2 \\
+${a2 * x} + ${b2}y &= $c2 \\
+${b2}y &= $c2 - ${a2 * x} \\
+${b2}y &= ${c2 - a2 * x}
+\\end{aligned}
+
+''',
         ),
       );
       steps.add(
         CalculationStep(
-          title: "第五步：解出 y",
-          explanation: "求解得到 y 的值。",
-          formula: "y = $y",
+          title: '第五步：解出 y',
+          explanation: '求解得到 y 的值。',
+          formula: '\$\$y = $y\$\$',
         ),
       );
-      return CalculationResult(steps: steps, finalAnswer: "x = $x, y = $y");
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: '\$\$x = $x, \quad y = $y\$\$',
+      );
     } else {
-      // Default case, substitute into the first equation
       final yCoeff = b1;
       final yConst = c1 - a1 * x;
       final y = yConst / yCoeff;
       steps.add(
         CalculationStep(
-          title: "第四步：回代求解 y",
-          explanation: "将 x = $x 代入原方程(1)中。",
+          title: '第四步：回代求解 y',
+          explanation: '将 x = $x 代入原方程(1)中。',
           formula:
-              "${a1}(${x}) + ${b1}y = $c1 \n=> ${a1 * x} + ${b1}y = $c1 \n=> ${b1}y = $c1 - ${a1 * x} \n=> ${b1}y = ${c1 - a1 * x}",
+              '''
+
+\\begin{aligned}
+$a1($x) + ${b1}y &= $c1 \\
+${a1 * x} + ${b1}y &= $c1 \\
+${b1}y &= $c1 - ${a1 * x} \\
+${b1}y &= ${c1 - a1 * x}
+\\end{aligned}
+
+''',
         ),
       );
       steps.add(
         CalculationStep(
-          title: "第五步：解出 y",
-          explanation: "求解得到 y 的值。",
-          formula: "y = $y",
+          title: '第五步：解出 y',
+          explanation: '求解得到 y 的值。',
+          formula: '\$\$y = $y\$\$',
         ),
       );
-      return CalculationResult(steps: steps, finalAnswer: "x = $x, y = $y");
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: '\$\$x = $x, \\quad y = $y\$\$',
+      );
     }
   }
 
   /// ---- 辅助函数 ----
 
-  /// 展开表达式，例如 (x-1)(x+2) -> x^2+x-2
   String _expandExpressions(String input) {
     String result = input;
-    // 循环直到没有更多的表达式可以展开
     while (true) {
       String oldResult = result;
 
-      // 模式1: k*(ax+b)^2, 例如 4(x-1)^2 or (x-1)^2
       final powerMatch = RegExp(
         r'(-?\d*\.?\d*)?\(([^)]+)\)\^2',
       ).firstMatch(result);
@@ -376,11 +388,7 @@ class SolverService {
         final kStr = powerMatch.group(1);
         double k = 1.0;
         if (kStr != null && kStr.isNotEmpty) {
-          if (kStr == '-') {
-            k = -1.0;
-          } else {
-            k = double.parse(kStr);
-          }
+          k = kStr == '-' ? -1.0 : double.parse(kStr);
         }
 
         final factor = powerMatch.group(2)!;
@@ -388,18 +396,16 @@ class SolverService {
         final a = coeffs[1] ?? 0;
         final b = coeffs[0] ?? 0;
 
-        // (ax+b)^2 = a^2*x^2 + 2ab*x + b^2
         final newA = k * a * a;
         final newB = k * 2 * a * b;
         final newC = k * b * b;
 
         final expanded =
-            "${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}${newC}";
-        result = result.replaceFirst(powerMatch.group(0)!, "($expanded)");
-        continue; // 重新开始循环以处理嵌套表达式
+            '${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
+        result = result.replaceFirst(powerMatch.group(0)!, '($expanded)');
+        continue;
       }
 
-      // 模式2: (ax+b)(cx+d), 例如 (x-3)(x+2)
       final factorMulMatch = RegExp(
         r'\(([^)]+)\)\(([^)]+)\)',
       ).firstMatch(result);
@@ -414,21 +420,17 @@ class SolverService {
         final c = coeffs2[1] ?? 0;
         final d = coeffs2[0] ?? 0;
 
-        // (ax+b)(cx+d) = ac*x^2 + (ad+bc)*x + bd
         final newA = a * c;
         final newB = a * d + b * c;
         final newC = b * d;
 
         final expanded =
-            "${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}${newC}";
-        result = result.replaceFirst(factorMulMatch.group(0)!, "($expanded)");
-        continue; // 重新开始循环
+            '${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
+        result = result.replaceFirst(factorMulMatch.group(0)!, '($expanded)');
+        continue;
       }
 
-      // 如果在这次迭代中没有改变，则跳出循环
-      if (result == oldResult) {
-        break;
-      }
+      if (result == oldResult) break;
     }
     return result;
   }
@@ -440,18 +442,18 @@ class SolverService {
     final leftCoeffs = _parsePolynomial(parts[0]);
     final rightCoeffs = _parsePolynomial(parts[1]);
 
-    final a = leftCoeffs[1] ?? 0.0;
-    final b = leftCoeffs[0] ?? 0.0;
-    final c = rightCoeffs[1] ?? 0.0;
-    final d = rightCoeffs[0] ?? 0.0;
-
-    return LinearEquationParts(a, b, c, d);
+    return LinearEquationParts(
+      (leftCoeffs[1] ?? 0.0),
+      (leftCoeffs[0] ?? 0.0),
+      (rightCoeffs[1] ?? 0.0),
+      (rightCoeffs[0] ?? 0.0),
+    );
   }
 
   Map<int, double> _parsePolynomial(String side) {
     final coeffs = <int, double>{};
     final pattern = RegExp(
-      r'([+-]?(?:\d*\.?\d*))?x(?:\^(\d+))?|([+-]?\d*\.?\d+)',
+      r'([+-]?(?:\d*\.?\d*)?)x(?:\^(\d+))?|([+-]?\d*\.?\d+)',
     );
     var s = side.startsWith('+') || side.startsWith('-') ? side : '+$side';
 
@@ -463,11 +465,7 @@ class SolverService {
         String coeffStr = match.group(1) ?? '+';
         double coeff = 1.0;
         if (coeffStr.isNotEmpty && coeffStr != '+') {
-          if (coeffStr == '-') {
-            coeff = -1.0;
-          } else {
-            coeff = double.parse(coeffStr);
-          }
+          coeff = coeffStr == '-' ? -1.0 : double.parse(coeffStr);
         } else if (coeffStr == '-') {
           coeff = -1.0;
         }
@@ -486,7 +484,7 @@ class SolverService {
     final xMatch = RegExp(r'([+-]?\d*\.?\d*)x').firstMatch(parts[0]);
     if (xMatch != null) {
       final coeff = xMatch.group(1);
-      if (coeff == null || coeff == '+') {
+      if (coeff == null || coeff.isEmpty || coeff == '+') {
         a = 1.0;
       } else if (coeff == '-') {
         a = -1.0;
@@ -497,7 +495,7 @@ class SolverService {
     final yMatch = RegExp(r'([+-]?\d*\.?\d*)y').firstMatch(parts[0]);
     if (yMatch != null) {
       final coeff = yMatch.group(1);
-      if (coeff == null || coeff == '+') {
+      if (coeff == null || coeff.isEmpty || coeff == '+') {
         b = 1.0;
       } else if (coeff == '-') {
         b = -1.0;
@@ -523,10 +521,7 @@ class SolverService {
     return null;
   }
 
-  // **FIXED**: Simplified check method
-  bool check(int m, int n, int b) {
-    return m + n == b;
-  }
+  bool check(int m, int n, int b) => m + n == b;
 
   ({String formula, String solution}) formatFactor(int m, int n, int a) {
     int common = gcd(n.abs(), a.abs());
@@ -538,25 +533,23 @@ class SolverService {
     final a2 = a ~/ den;
     final c2 = m ~/ a2;
 
-    // **FIXED**: Correctly handle coefficients of 1
     final f1Part1 = a1 == 1 ? 'x' : '${a1}x';
-    final f1 = c1 == 0 ? f1Part1 : "$f1Part1 ${c1 >= 0 ? '+' : ''} $c1";
+    final f1 = c1 == 0 ? f1Part1 : '$f1Part1 ${c1 >= 0 ? '+' : ''} $c1';
 
     final f2Part1 = a2 == 1 ? 'x' : '${a2}x';
-    final f2 = c2 == 0 ? f2Part1 : "$f2Part1 ${c2 >= 0 ? '+' : ''} $c2";
+    final f2 = c2 == 0 ? f2Part1 : '$f2Part1 ${c2 >= 0 ? '+' : ''} $c2';
 
-    // **FIXED**: Renamed variables to lowerCamelCase
     final int x1Num = -c1, x1Den = a1;
     final int x2Num = -c2, x2Den = a2;
 
-    final sol1 = x1Den == 1 ? "$x1Num" : "$x1Num/$x1Den";
-    final sol2 = x2Den == 1 ? "$x2Num" : "$x2Num/$x2Den";
+    final sol1 = x1Den == 1 ? '$x1Num' : '\\frac{$x1Num}{$x1Den}';
+    final sol2 = x2Den == 1 ? '$x2Num' : '\\frac{$x2Num}{$x2Den}';
 
     final solution = x1Num * x2Den == x2Num * x1Den
-        ? "x₁ = x₂ = $sol1"
-        : "x₁ = $sol1, x₂ = $sol2";
+        ? 'x_1 = x_2 = $sol1'
+        : 'x_1 = $sol1, \\quad x_2 = $sol2';
 
-    return (formula: "(${f1})(${f2}) = 0", solution: solution);
+    return (formula: '\$\$($f1)($f2) = 0\$\$', solution: '\$\$$solution\$\$');
   }
 
   int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);