📚 Chapter Review 3: Equations and Inequalities | 第三章复习：方程与不等式
Welcome to your comprehensive chapter review on equations and inequalities, a cornerstone of the Edexcel A Level Mathematics syllabus. In this article, we will revisit the essential concepts, solving techniques, and strategic approaches that you need to master for success in both Pure Mathematics and applied contexts. From linear equations to quadratic inequalities, simultaneous systems to discriminant analysis, every topic is broken down into clear explanations, paired examples, and bilingual commentary to reinforce your understanding.
欢迎来到方程与不等式的全面章节复习，这是爱德思 A Level 数学大纲的基石。在本文中，我们将重温基本概念、求解技巧以及你必须掌握的策略性方法，以在纯数学和应用背景中取得成功。从线性方程到二次不等式、联立方程组到判别式分析，每一个主题都被拆解为清晰的解释、配对的例子和双语注释，以强化你的理解。
1. Solving Linear Equations | 解线性方程
ne side of the equation using inverse operations, ensuring that any operation performed on one side is also applied to the other. For example, to solve 2x + 3 = 7, subtract 3 from both sides to get 2x = 4, then divide by 2 to find x = 2.
线性方程是最简单的代数方程形式，其中变量的最高次数为1。目标是使用逆运算将未知数隔离在方程的一侧，确保在一侧执行的任何运算同样应用于另一侧。例如，解方程 2x + 3 = 7，两边减去3得到 2x = 4，然后除以2得到 x = 2。
t common denominator to eliminate denominators before proceeding. For instance, (x/2) + (x/3) = 5 becomes 3x + 2x = 30 after multiplying by 6, giving 5x = 30, so x = 6.
始终通过将解代入原方程来检验答案。在涉及分数的方程中，先乘以最小公分母消去分母再继续求解。例如，(x/2) + (x/3) = 5 乘以6后变为 3x + 2x = 30，得到 5x = 30，因此 x = 6。
2. Quadratic Equations: Factorisation | 二次方程：因式分解法
tion, which relies on rewriting the expression as a product of two linear factors. The solutions, or roots, are then found by setting each factor equal to zero. For example, x² - 5x + 6 = 0 factorises to (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
当方程形式为 ax² + bx + c = 0 时，称为二次方程。最有效的求解方法之一便是因式分解法，它依靠将表达式重写为两个一次因式的乘积。然后通过令每个因式等于零来求得解或根。例如，x² - 5x + 6 = 0 分解为 (x - 2)(x - 3) = 0，得到 x = 2 或 x = 3。
To factorise, look for two numbers that multiply to give ac (the product of the coefficient of x² and the constant term) and add to give b. If the coefficient a ≠ 1, use splitting the middle term or other techniques. Always check your factorisation by expanding the factors to ensure they match the original expression.
因式分解时，寻找两个数相乘等于 ac（x² 系数与常数项的乘积）且相加等于 b。若系数 a ≠ 1，使用分裂中项或其他技巧。务必通过展开因式来检验因式分解是否正确，确保与原表达式一致。
3. Quadratic Formula and Completing the Square | 二次公式与配方法
any quadratic in the standard form ax² + bx + c = 0, the solutions are given by x = [ -b ± √b² - 4ac ] / 2a. This formula is derived from completing the square and works universally, even when roots are irrational or complex.
并非所有二次方程都能整齐地分解，这时便需要用到二次公式。对于标准形式 ax² + bx + c = 0 的任何二次方程，其解由 x = [ -b ± √b² - 4ac ] / 2a 给出。该公式由配方法推导而来，并且通用，即使根是无理数或复数也适用。
Completing the square is both an alternative solving method and a way to find the vertex of a parabola. The process involves rewriting the expression a(x + p)² + q form. For example, x² + 6x + 1 becomes (x + 3)² - 8. This format is fundamental for understanding the graph's turning point and is often used in deriving the quadratic formula itself.
配方法既是一种替代求解方法，也是寻找抛物线顶点的方式。其过程涉及将表达式重写为 a(x + p)² + q 的形式。例如，x² + 6x + 1 变为 (x + 3)² - 8。该形式对于理解图像的转折点至关重要，并且常用于推导二次公式本身。
4. Discriminant and Nature of Roots | 判别式与根的性质
The expression under the square root in the quadratic formula, b² - 4ac, is called the discriminant (often denoted by Δ). It reveals critical information about the nature of the roots without solving the equation. If the discriminant is positive, the quadratic has two distinct real roots. If it is zero, there is exactly one real root (a repeated root), and if it is negative, there are no real roots—only two complex conjugate roots.
二次公式中平方根下的表达式 b² - 4ac 称为判别式（通常用 Δ 表示）。它无需解方程即可揭示有关根的性质的关键信息。若判别式为正，二次方程有两个不同的实根。若为零，则恰好有一个实根（重根）；若为负，则没有实根，只有两个共轭复根。
e equation x² + kx + 9 = 0 has equal roots, set the discriminant to zero: k² - 4(1)(9) = 0 ⇒ k² = 36 ⇒ appear in examination questions combining algebraic manipulation and graph interpretation.
判别式也可用于确定系数的条件。例如，要求方程 x² + kx + 9 = 0 有等根时的 k 值，可令判别式为零：k² - 4(1)(9) = 0 ⇒ k² = 36 ⇒ k = ±6。此类问题常出现在结合代数操作与图像解释的考试题目中。
5. Solving Quadratic Inequalities | 解二次不等式
ratic expression on intervals determined by those roots. After factorising (x - 5)(x + 1) < 0, the critical values are x = -1 and x = 5. A sign diagram or sketch of the parabola shows the expression is negative between the roots, so the solution is -1 < x < 5.
诸如 x² - 4x - 5 < 0 的二次不等式，可以通过首先求出相应方程的根，然后分析由这些根划分的区间上二次表达式的符号来求解。因式分解后得 (x - 5)(x + 1) < 0，临界值为 x = -1 和 x = 5。使用符号表或抛物线草图可知表达式在两个根之间为负，因此解为 -1 < x < 5。
ful when multiplying or dividing an inequality by a negative number, as this reverses the inequality sign. For rational inequalities involving quadratics, bring all terms to one side and use a similar interval analysis.
请记住，若不等式是严格的（< 或 >），临界值不包含在内；若是非严格的（≤ 或 ≥），则包含。此外，当不等式两边乘以或除以一个负数时，务必注意反转不等号。对于涉及二次式的有理不等式，将所有项移到一边并使用类似的区间分析。
6. Simultaneous Equations: Linear and Quadratic | 联立方程：一次与二次
range the linear equation for one variable, substitute into the quadratic, and solve the resulting single-variable equation. For example, given y = 2x + 1 and x² + y² = 25, substitute to obtain x² + (2x + 1)² = 25, simplify to 5x² + 4x - 24 = 0, and solve using factorisation or the quadratic formula.
许多 A Level 题目涉及求解一个方程为一次、另一个方程为二次的联立方程组。标准方法是代入法：将线性方程改写为一个变量的表达式，代入二次方程中，并求解所得的单变量方程。例如，给定 y = 2x + 1 和 x² + y² = 25，代入得到 x² + (2x + 1)² = 25，化简为 5x² + 4x - 24 = 0，然后通过因式分解或二次公式求解。
Always find the corresponding values of the other variable for each solution. Discard any extraneous solutions if they arise from restrictions in the original context (e.g., geometric constraints). Graphical interpretation is also valuable: the solutions represent the intersection points of a straight line and a curve (circle, parabola, etc.).
始终为每个解求出另一变量的相应值。如果因原始情境中的限制（例如几何约束）产生增根，应予舍去。图形解释也很重要：解代表一条直线与一条曲线（圆、抛物线等）的交点。
7. Simultaneous Equations: Two Linear | 联立方程：两个一次方程
od, you multiply equations by constants so that the coefficients of one variable become equal or opposite, then add or subtract to eliminate that variable. For instance, 3x + 2y = 12 and 4x - 2y = 2 can be added directly to get 7x = 14, so x = 2, then back-substitute to find y = 3.
同时求解两个一次方程是一项基本技能，通常通过消元法或代入法处理。在消元法中，将方程乘以常数，使其中一个变量的系数相等或互为相反数，然后相加或相减以消去该变量。例如，3x + 2y = 12 和 4x - 2y = 2 可以直接相加得到 7x = 14，因此 x = 2，然后回代求出 y = 3。
Substitution is useful when one equation is already solved for one variable. Always verify your solution by plugging both values into the original equations. These techniques extend to systems with three or more variables, but for A Level, two-variable systems are the typical requirement.
当其中一个方程已经解出某个变量时，代入法很有用。始终通过将两个值代入原方程来验证解。这些技巧可扩展到三个或更多变量的方程组，但对于 A Level，通常要求的是两个变量的方程组。
8. Linear Inequalities and Regions | 一次不等式与区域
Linear inequalities extend the concepts of linear equations by introducing inequality symbols. Solving them similarly involves isolating the variable, but careful attention must be paid to the direction of the inequality when multiplying or dividing by a negative number. The solution is often expressed as a range on the number line or in set notation.
一次不等式通过引入不等号扩展了一次方程的概念。求解方式类似，涉及隔离变量，但当乘以或除以负数时，必须仔细关注不等号的方向。解通常表示为数轴上的区间或集合符号。
When dealing with two or more linear inequalities simultaneously, the solution is the intersection of the regions. Graphically, each inequality defines a half-plane, and the feasible region is the area that satisfies all constraints. This topic provides a bridge to linear programming, which appears in Decision Mathematics, but the core skill of shading regions and testing points is vital for Pure Mathematics questions.
当同时处理两个或更多一次不等式时，解为各区域的交集。在图形上，每个不等式定义一个半平面，可行区域为满足所有约束的区域。这一主题为决策数学中的线性规划搭建了桥梁，但绘制区域和检验点的核心技能对于纯数学题目至关重要。
9. Solving Equations Involving Fractions (Rational Equations) | 解涉及分数的方程（有理方程）
Rational equations contain algebraic fractions with variables in denominators. The safest method is to find the lowest common denominator of all fractions, multiply every term by it, and simplify to a polynomial equation. However, it is essential to check for extraneous solutions that would make any original denominator zero, as these are invalid.
有理方程包含分母中有变量的代数分式。最安全的方法是找到所有分式的最小公分母，乘以每一项，并化简为多项式方程。然而，必须检验是否有使原分母为零的增根，这些都是无效的。
0 and x+2 = -3 ≠ 0, so the solution is valid. Always state the domain restrictions upfront to avoid mistakes.
示例：解方程 2/(x-1) = 1/(x+2)。两边乘以 (x-1)(x+2) 得 2(x+2) = x-1，即 2x+4 = x-1，推出 x = -5。检验：分母 x-1 = -6 ≠ 0，x+2 = -3 ≠ 0，因此解有效。务必提前声明定义域限制以避免错误。
10. Modelling with Equations and Inequalities | 方程与不等式的建模
Real-world problems often translate into mathematical models involving equations or inequalities. The process requires defining variables, constructing an equation or inequality representing the constraints or relationships, solving it, and then interpreting the solution in the original context. Common contexts include geometry (areas, perimeters), physics (distance-speed-time), and economics (cost, revenue).
现实问题往往转化为涉及方程或不等式的数学模型。该过程需要定义变量，构建表示约束或关系的方程或不等式，求解，然后在原始情境中解释解。常见情境包括几何（面积、周长）、物理（距离-速度-时间）和经济学（成本、收入）。
For instance, if a rectangle' 0 → (w + 8)(w - 6) < 0, giving -8 < w < 6. Since width must be positive, 0 < w < 6. Modelling requires careful checks of the solution's feasibility.
例如，如果一个矩形的长比宽多 2 m，且其面积必须小于 48 m²，则可建立不等式 w(w + 2) < 48，然后求解二次不等式 w² + 2w - 48 < 0 → (w + 8)(w - 6) < 0，得到 -8 < w < 6。由于宽度必须为正，所以 0 < w < 6。建模要求仔细检查解的可行性。
11. Common Pitfalls and Exam Tips | 常见陷阱与应试技巧
One frequent mistake is forgetting to reverse the inequality sign when multiplying or dividing by a negative. Another is mishandling the discriminant, such as confusing the conditions for real and equal roots. Always write the quadratic in standard form before identifying a, b, c. When using the quadratic formula, especially with negative coefficients, bracketing helps prevent sign errors.
一个常见错误是乘或除以负数时忘记反转不等号。另一个是对判别式的处理不当，例如混淆实根与等根的条件。在确定 a、b、c 之前，务必将二次式写为标准形式。使用二次公式时，尤其是遇到负系数时，使用括号有助于避免符号错误。
In simultaneous equations, substitution errors occur if brackets are not expanded carefully. For inequalities, always draw a sign diagram or sketch graph to confirm the solution intervals. In exam settings, show clear steps: rearrangements, factorisation steps, discriminant calculations, and final solution sets using correct notation (set notation or interval notation as required).
在联立方程中，若括号展开不慎，就会出现代入错误。对于不等式，务必绘制符号表或草图以确认解区间。在考试中，要展示清晰的步骤：移项、因式分解步骤、判别式计算，以及使用正确符号（要求的集合符号或区间符号）表示最终解集。
12. Linking to Graphs and Further Studies | 与图表及后续学习的联系
(x). The inequality f(x) > 0 corresponds to regions where the graph lies above the x-axis. This visual approach strengthens your ability to understand transformations, asymptotes, and calculus applications such as finding intervals of increase or decrease.
方程和不等式与图形表示密切相关。f(x) = 0 的解是图像 y = f(x) 的 x 截距。不等式 f(x) > 0 对应于图像位于 x 轴上方的区域。这种视觉方法能增强您理解变换、渐近线以及微积分应用（如寻找递增或递减区间）的能力。
Mastery of this chapter paves the way for more advanced topics like logarithms, exponentials, and trigonometric equations, where similar solving principles apply. The logical thinking and algebraic manipulation skills you refine here will be invaluable across the entire A Level Mathematics curriculum and beyond.
掌握本章内容将为学习对数、指数和三角方程等更高阶主题铺平道路，这些主题应用了类似的求解原理。您在此过程中磨练的逻辑思维和代数操作技能，将在整个 A Level 数学课程及以后的学习中发挥难以估量的作用。
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