AQA A-Level数学 线性不等式考点全解析 | AQA A-Level Maths: Linear Inequalities Explained

AQA A-Level数学 线性不等式考点全解析

开篇：Example 3.4.2 与 AQA 第三章的定位

教材中的 Example 3.4.2 出自 AQA A-Level 数学（7357）Year 1 纯数部分第三章"方程与不等式"（Equations and Inequalities）第 4 节的第 2 个例题。第三章的学习主线是：先解决联立方程（3.1 至 3.3），再转向不等式（3.4 线性不等式、3.5 二次不等式），最后落到图像与区域（3.6 至 3.7）。其中 3.4 节线性不等式是全书最早出现的"变号规则"场景，也是后面解二次不等式、画可行域的基础。

Example 3.4.2 comes from Section 3.4 of Chapter 3, Equations and Inequalities, in the pure mathematics part of the AQA A-Level Maths Year 1 sequence. The chapter flows from simultaneous equations (Sections 3.1 to 3.3), through linear inequalities (3.4) and quadratic inequalities (3.5), to graphs and regions (3.6 and 3.7). Section 3.4 is the first place where the sign-flipping rule appears, and it underpins everything that follows, including sketching feasible regions.

本文不照抄课本例题原题，而是围绕 3.4 节要求掌握的技能展开：线性不等式的标准解法、变号规则、双重不等式、解集的数轴与区间表示，再补充与联立方程、图像区域的衔接以及 AQA 试卷的答题规范。每个知识点配一个带完整步骤的例题，学完即可直接做书后 3.4 节练习和历年真题。

Rather than reproducing the textbook question, this guide covers the skills Section 3.4 expects you to master: the standard solving procedure, the sign-flip rule, double inequalities, and solution sets on number lines and in interval notation, plus the links to simultaneous equations, graphs and AQA exam technique. Every topic comes with a fully worked example so you can move straight to the end-of-section exercises and past-paper questions.

一、不等式符号与数轴表示：<、≤、>、≥ 的规范

线性不等式的解集通常用三种方式表达：数轴、区间和集合记号。数轴上，大于或小于用空心圆点（open circle）表示端点不含，大于等于或小于等于用实心圆点（filled circle）表示端点包含，方向用箭头标出。区间写法中，x > 4 记作 (4, ∞)，x ≤ -3 记作 (-∞, -3]，圆括号表示开区间、方括号表示闭区间。集合记号则写成 {x : x > 4} 的形式，AQA 考纲要求三种写法都掌握。

A linear inequality solution set can be written three ways: on a number line, in interval notation, and in set notation. On a number line, strict inequalities use an open circle to show the endpoint is excluded, while inclusive inequalities use a filled circle; an arrow shows the direction. In interval notation, x > 4 becomes (4, ∞) and x ≤ -3 becomes (-∞, -3], with round brackets for open intervals and square brackets for closed ones. Set notation writes the condition as {x : x > 4}. The AQA specification expects fluency in all three forms.

Worked Example 例题

在数轴上表示 x ≤ 2 并用区间与集合记号写出解集。解：数轴上 2 处画实心圆点，箭头指向左侧；区间写法 (-∞, 2]；集合写法 {x : x ≤ 2}。

Show x ≤ 2 on a number line and write the solution set in interval and set notation. Solution: draw a filled circle at 2 with an arrow pointing left; interval form is (-∞, 2]; set form is {x : x ≤ 2}.

二、线性不等式求解三步法：以 3x - 5 > 7 为例

解线性不等式与解一元一次方程几乎完全相同：移项、合并同类项、两边同乘同除，唯一的区别是乘除负数时要变号。以 3x - 5 > 7 为例：第一步，两边同时加 5 得 3x > 12；第二步，两边同时除以 3 得 x > 4；第三步，把解集表示出来，区间形式为 (4, ∞)。任何一步都保持不等号方向不变，因为每一步两边加或乘的都是正数。

Solving a linear inequality is almost identical to solving a linear equation: rearrange, collect like terms, then multiply or divide both sides, with the single extra rule that multiplying or dividing by a negative flips the inequality sign. Take 3x - 5 > 7: first add 5 to both sides to get 3x > 12, then divide both sides by 3 to get x > 4, and finally write the set as (4, ∞). The direction of the inequality never changes here because every step adds or multiplies by a positive quantity.

Worked Example 例题

解不等式 2x + 3 < 15。解：两边减 3 得 2x < 12；两边除以 2 得 x < 6。解集为 (-∞, 6)，数轴上 6 处画空心圆点。

Solve 2x + 3 < 15. Solution: subtract 3 from both sides to get 2x < 12; divide by 2 to get x < 6. The solution set is (-∞, 6), drawn with an open circle at 6.

三、变号规则：乘除负数时不等号方向必须翻转

这是 3.4 节最核心也最容易丢分的规则：不等式两边同时乘以或除以一个负数，不等号方向反转。例如 -2x + 3 ≥ 11：先两边减 3 得 -2x ≥ 8，再两边除以 -2，此时必须把 ≥ 变成 ≤，得到 x ≤ -4。直观理解：数轴上 -2 和 -4 的大小关系在乘以 -1 后颠倒，例如 2 < 5 两边乘 -1 得 -2 > -5，方向确实反了。

This is the most important rule in Section 3.4 and the easiest place to lose marks: when you multiply or divide both sides of an inequality by a negative number, the inequality sign reverses. For -2x + 3 ≥ 11, subtract 3 from both sides to get -2x ≥ 8, then divide by -2 and flip ≥ to ≤, giving x ≤ -4. The intuition: the order of -2 and -4 on the number line is reversed after multiplication by -1, just as 2 < 5 becomes -2 > -5 when both sides are multiplied by -1.

Worked Example 例题

解不等式 5 - 2x < 1。解：两边减 5 得 -2x < -4；两边除以 -2 并变号得 x > 2。解集为 (2, ∞)。

Solve 5 - 2x < 1. Solution: subtract 5 from both sides to get -2x < -4; divide by -2 and reverse the sign to get x > 2. The solution set is (2, ∞).

四、双重不等式求解：-3 ≤ 2x + 1 < 7 完整流程

双重不等式（double inequality）把两个不等式合并成一个式子，AQA 常考。标准解法是对三边同时做同样的运算：-3 ≤ 2x + 1 < 7，先三边同时减 1 得 -4 ≤ 2x < 6，再三边同时除以 2 得 -2 ≤ x < 3。注意每个操作都要同时作用于左、中、右三部分，漏掉任何一边都会出错。另一种做法是把原式拆成 -3 ≤ 2x + 1 且 2x + 1 < 7 两个不等式分别解，再取交集，结果相同。

A double inequality combines two inequalities into one statement and appears regularly on AQA papers. The standard method applies the same operation to all three parts at once: for -3 ≤ 2x + 1 < 7, subtract 1 from every part to get -4 ≤ 2x < 6, then divide every part by 2 to get -2 ≤ x < 3. Every operation must act on the left, middle and right simultaneously; missing any part is a common error. Alternatively, split the statement into -3 ≤ 2x + 1 and 2x + 1 < 7, solve each, and take the intersection, which gives the same result.

Worked Example 例题

解 1 < 3x + 4 ≤ 10。解：三边减 4 得 -3 < 3x ≤ 6；三边除以 3 得 -1 < x ≤ 2。数轴上 -1 画空心、2 画实心。

Solve 1 < 3x + 4 ≤ 10. Solution: subtract 4 from all three parts to get -3 < 3x ≤ 6; divide by 3 to get -1 < x ≤ 2. On the number line, use an open circle at -1 and a filled circle at 2.

五、与联立方程的衔接：3.1 至 3.3 的方程组基础

不等式不是孤立考点，它与第三章前面的联立方程共用一套代数技能。线性联立方程组用消元法（elimination）或代入法（substitution）求解：例如 2x + y = 5 与 x - y = 1 相加消去 y 得 3x = 6，即 x = 2，代回得 y = 1。当一个方程含二次项（如圆 x² + y² = 13 与直线 y = x + 1）时，用代入法并借助判别式 b² - 4ac 判断交点个数：判别式大于 0 有两个交点，等于 0 相切，小于 0 无交点。这个判别式思路直接通向 3.5 节的二次不等式。

Inequalities share their algebra skills with the simultaneous equations covered earlier in the chapter. Linear pairs are solved by elimination or substitution: adding 2x + y = 5 and x - y = 1 cancels y to give 3x = 6, so x = 2, and substituting back gives y = 1. When one equation is quadratic, such as the circle x² + y² = 13 with the line y = x + 1, substitute the linear expression into the quadratic and use the discriminant b² - 4ac to count intersections: positive means two points, zero means one tangent point, negative means none. This discriminant reasoning leads directly into quadratic inequalities in Section 3.5.

Worked Example 例题

判断直线 y = x + 1 与圆 x² + y² = 13 的交点个数。解：代入得 x² + (x+1)² = 13，展开 2x² + 2x - 12 = 0，判别式 = 4 + 96 = 100 > 0，所以有两个交点。

Find how many times the line y = x + 1 meets the circle x² + y² = 13. Solution: substituting gives x² + (x+1)² = 13, which expands to 2x² + 2x - 12 = 0; the discriminant is 4 + 96 = 100, which is positive, so there are two intersection points.

六、图像与区域：y > 2x + 1 的阴影表示

含两个变量的线性不等式对应平面上的半平面区域。画 y > 2x + 1 时，先画边界线 y = 2x + 1，严格不等式用虚线（dashed line）表示边界不在解集内，非严格不等式用实线（solid line）；然后判断阴影方向，最稳妥的方法是取一个测试点（通常取原点 (0,0)）：代入后不等式成立则原点所在侧为解区。判定某点是否在区域内同样用代入法：点 (2,5) 代入 y > 2x + 1 得 5 > 5 不成立，且它在边界线上，而严格不等式不含边界，所以该点不在区域内。

A linear inequality in two variables corresponds to a half-plane in the coordinate plane. To sketch y > 2x + 1, first draw the boundary line y = 2x + 1, using a dashed line for strict inequalities because the boundary is excluded, or a solid line for inclusive ones; then decide which side to shade by testing a point, usually the origin. A point is in the region if its coordinates satisfy the inequality: substituting (2,5) into y > 2x + 1 gives 5 > 5, which is false, and the point lies on the boundary anyway, so it is not in the region.

Worked Example 例题

判定点 (0,0) 与 (1,4) 是否在区域 y ≤ 3 - x 内。解：原点代入得 0 ≤ 3 成立，所以在区域内；(1,4) 代入得 4 ≤ 2 不成立，所以不在区域内。

Decide whether (0,0) and (1,4) lie in the region y ≤ 3 - x. Solution: the origin gives 0 ≤ 3, which is true, so it is inside; (1,4) gives 4 ≤ 2, which is false, so it is outside.

七、常见错误盘点：五个最容易丢分的地方

第一，乘除负数忘记变号，这是 3.4 节失分的第一大原因，解题时可以先在草稿上写出"除以 -2，变号"的提示。第二，移项符号错误，例如把 -3 从右边移到左边忘记变号，建议把移项当作"两边同加同减"来写。第三，解双重不等式时只对两边做运算而漏掉中间，或者拆开后忘记取交集。第四，数轴端点画反：严格不等式用空心、非严格用实心，方向箭头指向解集一侧。第五，区间端点写错，x < 4 对应 (4, ∞) 而不是 [4, ∞)，开闭括号必须与空心实心一致。

First, forgetting to reverse the sign when multiplying or dividing by a negative is the biggest source of lost marks in Section 3.4; write a note such as divide by -2, flip the sign before you do it. Second, sign errors when rearranging, for example moving -3 to the other side without changing its sign; writing every step as add or subtract on both sides avoids this. Third, operating on only two of the three parts of a double inequality, or splitting it and forgetting to take the intersection. Fourth, drawing the wrong endpoint: open circles for strict inequalities, filled circles for inclusive ones, with the arrow pointing into the solution set. Fifth, writing the wrong interval endpoint: x < 4 is (4, ∞), not [4, ∞), and the bracket type must match the circle type.

八、AQA 考试策略：题型分布与答题规范

在 AQA A-Level 数学（7357）中，不等式考点主要出现在 Paper 1（纯数，无计算器）和 Paper 2（纯数，可带计算器），单题通常 2 到 4 分，多以"解不等式并写出解集"或"判断区域"的形式出现；国际版 OxfordAQA 9660 则在 AS 的 MA01 与 A2 的 MA03 两卷中考查。答题时务必做到三点：完整写出移项与变号步骤，AQA 评分对方法分（method mark）很慷慨；解集同时用数轴或区间给出，避免因表达不完整扣分；选择题中可用代入法快速检验选项，例如把每个选项的端点代回原不等式验证。

On the AQA A-Level Maths specification (7357), inequalities appear mainly on Paper 1, pure mathematics without a calculator, and Paper 2, pure mathematics with a calculator, usually worth 2 to 4 marks as solve the inequality and write the solution set or shade a region; on the international OxfordAQA 9660 route they are assessed in MA01 at AS and MA03 at A2. In your answer, always show the full rearrangement and sign-flip steps, because AQA marks method generously; present the solution set on a number line or in interval form so the answer is unambiguous; and in multiple-choice questions test each option by substituting endpoints back into the original inequality.

九、四周复习计划：从例题到真题

第一周：重做课本 3.1 至 3.3 节的联立方程习题，确保消元法、代入法与判别式熟练，这是不等式的地基。第二周：专攻 3.4 与 3.5 节，每天 8 道线性不等式、4 道双重不等式，把变号规则练成肌肉记忆，并开始接触二次不等式。第三周：完成 3.6 与 3.7 节的图像区域练习，每道题都画数轴或坐标系并标注开闭端点，再用近五年 AQA 真题中的不等式小题检验。第四周：限时模考整套 Paper 1，统计不等式题用时与失分，对照评分细则逐条订正，考前三天只复习错题本与区间写法规范。

Week 1: redo the simultaneous equation exercises in Sections 3.1 to 3.3 until elimination, substitution and the discriminant feel automatic, since they are the foundation. Week 2: focus on Sections 3.4 and 3.5, doing eight linear and four double inequalities a day so the sign-flip rule becomes automatic, then start quadratic inequalities. Week 3: finish the graph and region questions in Sections 3.6 and 3.7, drawing a number line or coordinate plane for every question with correct open and filled endpoints, then test yourself with inequality questions from the last five years of AQA papers. Week 4: sit a full timed Paper 1, log your time and lost marks on inequality questions, mark yourself against the official mark scheme, and spend the last three days on your error log and interval notation rules.

十、关键双语术语对照

不等式 inequality | 线性不等式 linear inequality | 二次不等式 quadratic inequality | 解集 solution set | 数轴 number line | 实心圆点 filled circle | 空心圆点 open circle | 区间 interval | 区间端点 endpoint | 开区间 open interval | 闭区间 closed interval | 变号 reverse the sign | 双重不等式 double inequality | 联立方程 simultaneous equations | 消元法 elimination | 代入法 substitution | 判别式 discriminant | 区域 region | 边界线 boundary line | 虚线 dashed line | 实线 solid line | 阴影 shading | 交集 intersection

These terms appear repeatedly in AQA past papers. Learning them in pairs lets you read questions faster, answer in the examiner's language, and avoid confusing interval notation with set notation when the question asks for a specific form.

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