📚 Edexcel Pure Maths Mixed Exercise 3: Equations and Inequalities | Edexcel 纯数学综合练习 3：方程与不等式
Mixed Exercise 3 is the consolidation task at the end of Chapter 3 in the Edexcel AS/A Level Pure Mathematics course. It brings together linear and quadratic equations, discriminant conditions, simultaneous equations, inequalities, and contextual modelling. Students who complete it carefully gain a stronger command of algebraic manipulation and examination technique.
综合练习 3 是 Edexcel AS/A Level 纯数学课程第 3 章末尾的巩固练习，汇集了线性方程与二次方程、判别式条件、联立方程、不等式以及实际建模。认真完成该练习的学生能够更好地掌握代数运算和考试技巧。
1. Specification Map and Skills Tested | 考纲范围与考查技能
This mixed exercise aligns with Edexcel Pure Mathematics Year 1, Chapter 3. The main assessment objectives are solving quadratic equations by factorising, completing the square, and using the quadratic formula; interpreting the discriminant; solving linear and quadratic inequalities; solving one linear and one quadratic simultaneous equation; and applying these skills to word problems.
本综合练习对应 Edexcel 纯数学第一册第 3 章。主要考查目标是：通过因式分解、配方法和求根公式解二次方程；解释判别式；解线性与二次不等式；求解一个线性方程和一个二次方程组成的联立方程组；以及将这些技能应用到文字题中。
The questions often mix two or more concepts in one item, so you must recognise which method is needed before starting. For example, a modelling question may require you to form a quadratic inequality and then use the discriminant to check whether a solution is possible.
题目常在一问中混合两个或更多概念，因此开始前必须先判断需要哪种方法。例如，一道建模题可能要求你列出二次不等式，再用判别式检查解是否存在。
2. Solving Linear Equations Confidently | 熟练掌握线性方程
Linear equations in Mixed Exercise 3 usually involve fractions, brackets, or parameters. Clear algebraic steps prevent sign mistakes. For example, to solve 3(x − 2) + 4 = 2x + 7, expand first, collect like terms, then isolate x.
综合练习 3 中的线性方程通常含有分数、括号或参数。清晰的代数步骤可以防止符号错误。例如，解 3(x − 2) + 4 = 2x + 7 时，先去括号，再合并同类项，最后分离出 x。
If the equation includes a denominator, multiply every term by the lowest common denominator. Always check whether a denominator could be zero, because that value must be excluded from the solution set.
如果方程含有分母，应先将每一项乘以最小公分母。注意检查分母是否可能为零，因为使分母为零的值必须从解集中排除。
3(x − 2) + 4 = 2x + 7 → 3x − 6 + 4 = 2x + 7 → 3x − 2 = 2x + 7 → x = 9
3. Quadratic Equations: Three Reliable Methods | 二次方程：三种可靠方法
For a quadratic equation ax² + bx + c = 0, there are three standard methods. Factorising is fastest when the expression factorises neatly over integers. Completing the square is useful when the coefficient of x² is 1 and when the equation must be written in vertex form. The quadratic formula always works, even when factorising is difficult or impossible.
对于二次方程 ax² + bx + c = 0，有三种标准方法。当二次式能够整式分解时，因式分解最快。当 x² 的系数为 1 且需要写成顶点式时，配方法很有用。求根公式总是适用，即使因式分解困难或不可能。
x = (−b ± √(b² − 4ac)) ÷ 2a
For example, solve x² + 6x + 5 = 0 by factorising: (x + 1)(x + 5) = 0, so x = −1 or x = −5. The same answer can be obtained by completing the square: x² + 6x + 9 = 4, giving (x + 3)² = 4, so x + 3 = ±2.
例如，用因式分解解 x² + 6x + 5 = 0：(x + 1)(x + 5) = 0，所以 x = −1 或 x = −5。用配方法也可得到相同答案：x² + 6x + 9 = 4，即 (x + 3)² = 4，所以 x + 3 = ±2。
4. Using the Discriminant Correctly | 正确使用判别式
The discriminant Δ = b² − 4ac comes from the quadratic formula. It tells you the nature of the roots without solving the equation. When Δ > 0, there are two distinct real roots; when Δ = 0, there is one repeated real root; when Δ < 0, there are no real roots.
判别式 Δ = b² − 4ac 来自求根公式。它可以在不求根的情况下判断根的性质：当 Δ > 0 时，有两个不相等的实根；当 Δ = 0 时，有一个重根；当 Δ < 0 时，没有实根。
| Discriminant condition | Nature of roots |
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root |
| Δ < 0 | No real roots |
A typical exam question asks for the value of k that gives a repeated root. For x² + kx + 4 = 0, set Δ = k² − 16 equal to 0, giving k = ±4. State both values unless the question restricts k.
典型的考试题会要求求使方程有重根的 k 值。例如对于 x² + kx + 4 = 0，令 Δ = k² − 16 = 0，得 k = ±4。除非题目对 k 有限制，否则应写出两个值。
5. Linear Inequalities and Sign Reversal | 线性不等式与符号反向
Solving linear inequalities is similar to solving linear equations, but there is one critical difference: when you multiply or divide both sides by a negative number, you must reverse the inequality sign. Forgetting this step is one of the most common errors in Mixed Exercise 3.
解线性不等式与解线性方程类似，但有一个关键区别：当两边同时乘或除以一个负数时，必须反向不等式符号。忘记这一步是综合练习 3 中最常见的错误之一。
For example, solve −2x + 3 ≤ 9. Subtract 3: −2x ≤ 6. Divide by −2 and reverse: x ≥ −3. In interval notation, the solution is x ∈ [−3, ∞).
例如，解 −2x + 3 ≤ 9。减 3 得 −2x ≤ 6。除以 −2 并反向得 x ≥ −3。用区间表示，解为 x ∈ [−3, ∞)。
−2x + 3 ≤ 9 → −2x ≤ 6 → x ≥ −3
6. Quadratic Inequalities and Critical Values | 二次不等式与临界值
To solve a quadratic inequality such as x² − 5x + 6 ≤ 0, first solve the corresponding equation to find the critical values. Here, x² − 5x + 6 = 0 gives (x − 2)(x − 3) = 0, so the critical values are x = 2 and x = 3. Because the coefficient of x² is positive, the graph opens upward and the expression is negative or zero between the roots.
要解二次不等式 x² − 5x + 6 ≤ 0，先解对应方程求临界值。此处 x² − 5x + 6 = 0 得 (x − 2)(x − 3) = 0，所以临界值为 x = 2 和 x = 3。由于 x² 的系数为正，图像开口向上，表达式在两个根之间小于或等于零。
Therefore, the solution is 2 ≤ x ≤ 3, or x ∈ [2, 3]. For strict inequalities, use open intervals: x² − 5x + 6 < 0 gives x ∈ (2, 3). Always check the inequality symbol carefully at the final step.
因此，解为 2 ≤ x ≤ 3，即 x ∈ [2, 3]。对于严格不等式，应使用开区间：x² − 5x + 6 < 0 的解为 x ∈ (2, 3)。最后一步务必仔细检查不等式符号。
7. Simultaneous Equations: Linear and Quadratic | 线性与二次联立方程
When one equation is linear and the other is quadratic, the safest method is substitution. Rearrange the linear equation to make x or y the subject, substitute into the quadratic equation, and solve the resulting quadratic in one variable. Then substitute back to find the paired values.
当一个方程为线性、另一个为二次时，最稳妥的方法是代入法。先将线性方程变形为 x 或 y 为主项，代入二次方程，解出得到的单变量二次方程。然后代回求对应的配对值。
Example: solve y = 2x + 1 and x² + y² = 13. Substitute y = 2x + 1 into the circle equation: x² + (2x + 1)² = 13. Expand to get 5x² + 4x − 12 = 0. Factorising gives (5x − 6)(x + 2) = 0, so x = 6/5 or x = −2. The paired solutions are (6/5, 17/5) and (−2, −3).
例题：解 y = 2x + 1 与 x² + y² = 13。将 y = 2x + 1 代入圆的方程：x² + (2x + 1)² = 13。展开得 5x² + 4x − 12 = 0。因式分解为 (5x − 6)(x + 2) = 0，得 x = 6/5 或 x = −2。对应解为 (6/5, 17/5) 与 (−2, −3)。
x² + (2x + 1)² = 13 → 5x² + 4x − 12 = 0 → (5x − 6)(x + 2) = 0
8. Contextual Modelling and Validation | 实际建模与验证
Word problems in Mixed Exercise 3 require you to define a variable, write an equation or inequality from the situation, solve it, and then interpret the answer in the original context. Always reject solutions that do not make sense, such as negative lengths or times.
综合练习 3 中的文字题要求你定义变量，根据题意写出方程或不等式，求解，然后结合原始情境解释答案。必须舍去不合理的解，例如负的长度或时间。
For example, a rectangular garden has length x metres and width (x − 3) metres. If the area must be at least 40 m², write x(x − 3) ≥ 40. This becomes x² − 3x − 40 ≥ 0, with roots x = 8 and x = −5. Since the length and width must be positive, x ≥ 8 is the only valid solution.
例如，一个矩形花园的长为 x 米，宽为 (x − 3) 米。若面积必须至少为 40 平方米，写出 x(x − 3) ≥ 40。化为 x² − 3x − 40 ≥ 0，其根为 x = 8 和 x = −5。由于长和宽必须为正，唯一有效解是 x ≥ 8。
Modelling questions often also ask for a maximum or minimum value. You can use completing the square or the vertex formula x = −b ÷ 2a to locate the turning point before interpreting it in context.
建模题还常要求求最大值或最小值。可以用配方法或顶点公式 x = −b ÷ 2a 找到最值点，再结合实际意义解释。
9. Common Errors and How to Avoid Them | 常见错误与避免方法
The same mistakes appear repeatedly in mixed exercises. Recognising them early reduces lost marks. The table below lists some frequently seen errors, their consequences, and a quick prevention strategy.
同样的错误在综合练习中反复出现。尽早识别这些错误可以减少失分。下表列出了一些常见错误、后果和快速预防策略。
| Common error | Consequence | Prevention |
| Forgetting to reverse the sign when dividing by a negative number | Incorrect inequality solution | Underline the negative divisor before dividing |
| Wrong pairing in simultaneous equations | 更多咨询请联系16621398022（同微信）
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