Linear Inequalities: Solving and Number Line Representation | 线性不等式：求解与数轴表示

📚 Linear Inequalities: Solving and Number Line Representation | 线性不等式：求解与数轴表示

Linear inequalities are a fundamental topic in A-Level mathematics. They extend the concept of equations by introducing relational symbols such as <, >, ≤, and ≥. Unlike solving equations, where the solution is a specific value or set of values, solving an inequality yields a range of possible values. This article will guide you through the essential methods for solving linear inequalities and representing their solutions accurately on a number line.

线性不等式是A-Level数学中的一个基础课题。它通过引入 <、>、≤ 和 ≥ 等关系符号，扩展了方程的概念。与求解方程得到特定值或一组成员值不同，解不等式得到的是一组可能的取值范围。本文将引导你掌握求解线性不等式的基本方法，以及如何在数轴上准确表示解集。

1. Understanding Inequality Symbols | 理解不等式符号

The four basic inequality symbols form the foundation of this topic. The symbol ‘>’ means “greater than,” ‘<' means "less than," '≥' means "greater than or equal to," and '≤' means "less than or equal to." It is essential to distinguish between strict inequalities ( > and < ) and non-strict inequalities ( ≥ and ≤ ), as this difference affects both the solution set and its representation on the number line.

四个基本不等式符号构成了本课题的基础。符号 ‘>’ 表示“大于”，’<' 表示“小于”，'≥' 表示“大于或等于”，'≤' 表示“小于或等于”。务必区分严格不等式（> 和 <）与非严格不等式（≥ 和 ≤），因为这一区别既影响解集，也影响其在数轴上的表示。

For example, the expression x > 3 means that x can take any value strictly greater than 3, but not 3 itself. In contrast, x ≥ 3 includes 3 as a valid solution. Visualising these differences early will help you avoid common mistakes later.

例如，表达式 x > 3 表示 x 可以取任何严格大于 3 的值，但不能取 3 本身。相比之下，x ≥ 3 则包含 3 作为有效解。及早直观地理解这些差异，有助于你之后避免常见错误。

2. Key Properties of Inequalities | 不等式的基本性质

Solving linear inequalities relies on a set of properties that are similar to those used for equations, with one critical exception. You may add or subtract the same quantity from both sides of an inequality without changing its direction. You may also multiply or divide both sides by a positive number without changing the direction of the inequality sign.

求解线性不等式依赖于一系列与方程类似的性质，但有一个关键例外。你可以在不等式两边同时加上或减去同一个量，而不改变不等号的方向。你也可以在不等式两边同时乘以或除以一个正数，而保持不等号方向不变。

However, if you multiply or divide both sides of an inequality by a negative number, the inequality sign must be reversed. For example, if -2x < 6, then dividing both sides by -2 gives x > -3. This reversal is the most common source of error among students, so it must be memorised and applied with care.

然而，如果在不等式两边同时乘以或除以一个负数，则必须将不等号方向反转。例如，若 -2x < 6，两边同时除以 -2 得到 x > -3。这一反转是学生最常见错误来源，必须牢记并谨慎应用。

If a < b and c > 0, then ac < bc
If a < b and c < 0, then ac > bc

The reciprocal property is also worth noting. If a and b have the same sign and a < b, then 1/a > 1/b. This property is less frequently tested but may appear in more challenging exam questions.

倒数性质也值得注意。若 a 和 b 同号且 a < b，则 1/a > 1/b。这一性质虽然考查频率较低，但可能在更具挑战性的考试题目中出现。

3. Solving Single-Variable Linear Inequalities | 求解单变量线性不等式

Solving a single-variable linear inequality follows a procedure very similar to solving a linear equation. First, expand any brackets and combine like terms on each side of the inequality. Then, use addition or subtraction to collect all variable terms on one side and constant terms on the other. Finally, multiply or divide to isolate the variable, remembering to reverse the inequality sign if you multiply or divide by a negative number.

求解单变量线性不等式的步骤与解线性方程非常相似。首先，展开所有括号并在不等式两边合并同类项。然后，通过加法或减法将所有含变量的项集中到一边，常数项集中到另一边。最后，通过乘法或除法将变量系数化为 1，记住如果乘以或除以负数，需要反转不等号方向。

Consider the following worked example. Solve the inequality 3x – 5 > x + 7. First, subtract x from both sides to obtain 2x – 5 > 7. Next, add 5 to both sides to get 2x > 12. Finally, divide both sides by 2 to obtain x > 6. The solution set is all real numbers greater than 6.

请看以下示例。求解不等式 3x – 5 > x + 7。首先，两边同时减去 x，得到 2x – 5 > 7。接着，两边同时加上 5，得到 2x > 12。最后，两边同时除以 2，得到 x > 6。因此解集是所有大于 6 的实数。

Let us examine a second example involving a negative coefficient. Solve the inequality 5 – 2x ≤ 11. First, subtract 5 from both sides to obtain -2x ≤ 6. Now, divide both sides by -2. Since we are dividing by a negative number, the inequality sign must be reversed, giving x ≥ -3. This example illustrates the critical step that distinguishes inequality solving from equation solving.

再看一个涉及负系数的例子。求解不等式 5 – 2x ≤ 11。首先，两边同时减去 5，得到 -2x ≤ 6。然后，两边同时除以 -2。因为我们除以的是负数，所以必须反转不等号方向，得到 x ≥ -3。这个例子说明了不等式求解与方程求解的关键区别。

4. Representing Solutions on a Number Line | 在数轴上表示解集

A number line provides a clear visual representation of the solution set of an inequality. For a single variable x, we draw a horizontal line, mark the critical value(s), and shade the region that satisfies the inequality. The type of circle used at the boundary point distinguishes strict from non-strict inequalities.

数轴为不等式的解集提供了清晰的视觉表示。对于单变量 x，我们画一条水平线，标注关键值（临界点），并涂阴影标出满足不等式的区域。边界点处使用的圆点类型区分了严格不等式与非严格不等式。

For x > a or x < a, draw an open circle (hollow circle) at a to show that a is not included in the solution set.
对于 x > a 或 x < a，在 a 处画一个空心圆圈，表示 a 不包含在解集中。
For x ≥ a or x ≤ a, draw a closed circle (filled circle) at a to show that a is included in the solution set.
对于 x ≥ a 或 x ≤ a，在 a 处画一个实心圆点，表示 a 包含在解集中。

For example, to represent x > 2, place an open circle at 2 and draw an arrow pointing to the right. This indicates all values greater than 2 are solutions. To represent x ≤ -1, place a closed circle at -1 and draw an arrow pointing to the left, indicating all values less than or equal to -1 are solutions.

例如，要表示 x > 2，在 2 处画空心圆圈，并向右侧画箭头。这表示所有大于 2 的值都是解。要表示 x ≤ -1，在 -1 处画实心圆点，并向左侧画箭头，表示所有小于或等于 -1 的值都是解。

When the inequality is compound, such as 1 ≤ x < 4, the number line representation uses both types of circles: a closed circle at 1 and an open circle at 4, with the segment between them shaded. This visual representation is essential in exams, as it quickly communicates the solution set to the examiner.

当不等式为复合形式时，如 1 ≤ x < 4，数轴表示需要同时使用两种圆点：在 1 处画实心圆点，在 4 处画空心圆圈，并涂阴影连接两者之间的线段。在考试中数轴表示至关重要，因为它能迅速向阅卷人传达解集。

5. Interval Notation | 区间符号

Interval notation is a concise alternative to number line representation. It uses square brackets [ ] to denote inclusive endpoints and round brackets ( ) to denote exclusive endpoints. For example, the solution x ≥ 3 can be written as [3, ∞), while x > 3 is written as (3, ∞). The symbol ∞ represents infinity, and the minus sign before it, -∞, represents negative infinity; infinity is always paired with a round bracket, as infinity is never a finite endpoint that can be included.

区间符号是数轴表示的一种简洁替代方案。它使用方括号 [ ] 表示包含端点，圆括号 ( ) 表示排除端点。例如，解集 x ≥ 3 可以写成 [3, ∞)，而 x > 3 则写成 (3, ∞)。符号 ∞ 表示无穷大，加负号为 -∞ 表示负无穷大；无穷大始终搭配圆括号，因为无穷大不是可以包含在内的有限端点。

For a bounded interval such as -2 < x ≤ 5, the interval notation is (-2, 5]. The round bracket at -2 shows that -2 is not included, while the square bracket at 5 shows that 5 is included. This notation is widely used in higher-level mathematics and Cambridge/Edexcel exam papers, so proficiency in converting between inequalities, interval notation, and number lines is essential.

对于有界区间，如 -2 < x ≤ 5，其区间符号为 (-2, 5]。-2 处的圆括号表示不包含 -2，而 5 处的方括号表示包含 5。这一符号在高等数学和Edexcel考试卷中被广泛使用，因此熟练掌握不等式、区间符号与数轴之间的相互转换至关重要。

Set notation: {x : x > a} | 集合符号：{x : x > a}

In Edexcel A-Level, you may also encounter set notation. The solution set of an inequality is often expressed as {x : condition}, such as {x : x > 3} or {x : -1 ≤ x < 4}. Understanding all three forms of representation — inequality, interval, and set notation — ensures you can interpret questions correctly and present answers in the required format.

在 Edexcel A-Level 考试中，你可能会遇到集合符号。不等式的解集通常表示为 {x : 条件}，例如 {x : x > 3} 或 {x : -1 ≤ x < 4}。理解所有三种表示形式——不等式、区间和集合符号——可确保你正确理解题意，并按所要求的格式呈现答案。

6. Solving Double Inequalities | 求解双重不等式

A double inequality is a single statement that combines two inequalities, such as 2 < 3x + 1 < 10 or -1 ≤ 5 - 2x < 9. Solving a double inequality involves performing the same operation on all three parts of the inequality simultaneously. The goal is to isolate the variable x in the middle section.

双重不等式是一个同时包含两个不等式的单一表达，例如 2 < 3x + 1 < 10 或 -1 ≤ 5 - 2x < 9。求解双重不等式时，需要对不等式的三个部分同时执行相同的运算。目标是让变量 x 单独位于中间部分。

Consider the example 2 < 3x + 1 < 10. First, subtract 1 from all three parts to obtain 1 < 3x < 9. Then, divide all three parts by 3 to get 1/3 < x < 3. The solution is the open interval (1/3, 3). Note that the operations are applied to the left part, the middle part, and the right part simultaneously to maintain the validity of the inequality.

以例 2 < 3x + 1 < 10。首先，三部分同时减去 1，得到 1 < 3x < 9。然后，三部分同时除以 3，得到 1/3 < x < 3。解为开区间 (1/3, 3)。注意，运算必须同时施加于左部、中部和右部，以保证不等式仍然成立。

Now consider a more demanding example involving a negative multiplier: -1 ≤ 5 – 2x < 9. Subtract 5 from all three parts to get -6 ≤ -2x < 4. Now divide by -2. Since we are dividing by a negative number, we must reverse both inequality signs, which gives 3 ≥ x > -2. This can be rewritten in the conventional ascending form as -2 < x ≤ 3.

现在来看一个涉及负数乘数的更具挑战性的例子：-1 ≤ 5 – 2x < 9。三部分同时减去 5，得到 -6 ≤ -2x < 4。然后同时除以 -2。由于我们除以的是一个负数，必须反转两个不等号方向，得到 3 ≥ x > -2。可将其改写为约定的递增形式：-2 < x ≤ 3。

7. Systems of Linear Inequalities | 线性不等式组

A system of linear inequalities consists of two or more inequalities that must be satisfied simultaneously. Solving such a system means finding the intersection of the individual solution sets — that is, the set of values that satisfy every inequality in the system. This concept is closely related to linear programming, a topic often examined in A-Level decision mathematics.

线性不等式组由两个或更多必须同时满足的不等式构成。求解不等式组意味着找出各个解集的交集，即同时满足系统中每一个不等式的值的集合。这一概念与线性规划密切相关，后者是A-Level数学中常考的专题。

For example, consider the system: x > 1 and x ≤ 5. The first inequality gives the interval (1, ∞), and the second gives (-∞, 5]. Their intersection is (1, 5]. On a number line, we would shade the region between 1 and 5, using an open circle at 1 and a closed circle at 5.

例如，考虑不等式组：x > 1 且 x ≤ 5。第一个不等式给出区间 (1, ∞)，第二个给出 (-∞, 5]。它们的交集为 (1, 5]。在数轴上，我们将 1 到 5 之间的区域涂阴影，在 1 处用空心圆圈，在 5 处用实心圆点。

When the inequalities are more complex, such as 2x + 1 > 5 and 3x – 4 ≤ 8, solve each inequality separately first. The first gives x > 2, and the second gives x ≤ 4. The solution to the system is therefore 2 < x ≤ 4. It is advisable to draw a number line for each individual inequality and then determine the common shaded region.

当不等式更复杂时，例如 2x + 1 > 5 和 3x – 4 ≤ 8，应首先分别求解每个不等式。第一个给出 x > 2，第二个给出 x ≤ 4。因此不等式组的解为 2 < x ≤ 4。建议为每个不等式分别画数轴，然后找出共同涂阴影的区域。

8. Word Problems and Modelling | 应用题与建模

Linear inequalities are frequently used to model real-world situations involving constraints or limits. In such problems, you must translate the given conditions into mathematical inequalities, solve them, and interpret the solution in the original context. Common scenarios include budgeting, speed limits, age restrictions, and production capacities.

线性不等式常用于对涉及约束或限制的现实情境进行建模。在这类问题中，你需要将给定条件转化为数学不等式，求解，并根据原始情境解读结果。常见场景包括预算限制、速度限制、年龄限制和生产能力等。

Consider the following example. A student has £50 to spend on books and stationery. Each book costs £12 and each item of stationery costs £3. The student wants to buy at least 2 books. Let b represent the number of books and s the number of stationery items. The inequalities are: 12b + 3s ≤ 50 and b ≥ 2. If the student buys 3 books, then 36 + 3s ≤ 50, which gives 3s ≤ 14, so s ≤ 14/3. Since s must be an integer, the maximum number of stationery items is 4.

请看以下示例。一个学生有 50 英镑用于购买书籍和文具。每本书 12 英镑，每件文具 3 英镑。学生想至少买 2 本书。设 b 代表书的数量，s 代表文具数量。不等式为：12b + 3s ≤ 50 且 b ≥ 2。如果学生买 3 本书，则 36 + 3s ≤ 50，即 3s ≤ 14，所以 s ≤ 14/3。由于 s 必须是整数，最多可买 4 件文具。

In examination contexts, word problems require you to define variables clearly, state the inequalities explicitly, and justify your final answer. Marks are typically awarded for each of these steps, so showing full working is crucial. Remember that real-world quantities often need to be integers, non-negative, or subject to other practical constraints.

在考试情境中，应用题要求你清晰地定义变量，明确写出不等式，并解释最终答案。每个步骤通常都会分配分数，因此展示完整过程至关重要。请记住，现实中的数量往往需要是整数、非负数，或受其他实际约束的限制。

9. Common Mistakes and Exam Pitfalls | 常见错误与考试陷阱

Several systematic errors recur when students solve linear inequalities. The most serious is forgetting to reverse the inequality sign when multiplying or dividing by a negative number. Another common error is reversing the sign when rearranging a double inequality. For instance, when solving -2 < 3 - x, some students incorrectly add x to both sides but fail to correctly track the direction of the inequality.

学生在求解线性不等式时会反复出现若干系统性错误。最严重的是在乘除负数时忘记反转不等号方向。另一个常见错误是在重组双重不等式时反转了符号方向。例如，求解 -2 < 3 - x 时，有些学生在两边同时加上 x，但没有正确跟踪不等号的朝向。

Mistake 1: Dividing by a negative number without reversing the inequality sign — always check the sign of the multiplier or divisor.
错误一：除以负数时不反转不等号方向——务必检查乘数或除数的正负号。
Mistake 2: Misplacing open and closed circles on the number line — remember that < and > use open circles, while ≤ and ≥ use closed circles.
错误二：在数轴上弄错空心圆与实心圆的位置——记住 < 和 > 用空心圆，≤ 和 ≥ 用实心圆。
Mistake 3: Incorrectly writing interval notation — infinity must always use round brackets, and endpoints must reflect inclusivity.
错误三：区间符号书写错误——无穷大必须始终使用圆括号，端点必须准确反映是否包含。
Mistake 4: Multiplying or dividing each term of a double inequality inconsistently — apply the same operation to all three parts.
错误四：对双重不等式的各部分不一致地执行乘法或除法——必须对三个部分同时施加相同运算。

Many students also lose marks by not writing the final answer in the requested format. The exam question may specify ‘write your answer in interval notation’ or ‘show your answer on a number line.’ Always read such instructions carefully and present your solution accordingly. Even if your algebraic manipulation is correct, an incorrectly formatted final answer may not receive full credit.

许多学生还会因为未按要求格式写出最终答案而失分。考试题目可能指定“用区间符号写出答案”或“在数轴上表示答案”。务必仔细阅读此类指令，并据此呈现你的解答。即使代数变形完全正确，格式不对的最终答案也可能无法获得满分。

10. Worked Exam-Style Examples | 考试风格例题精解

Let us work through a set of examination-style questions step by step to consolidate the methods introduced in this article. Each example is designed to mirror the style of Edexcel A-Level questions on linear inequalities.

让我们逐步解析一组考试风格例题，以巩固本文介绍的方法。每道例题都旨在贴近 Edexcel A-Level 线性不等式题目的风格。

Example A | 例题 A：Solve the inequality 4(x – 2) + 3 > 2x + 1 and represent the solution on a number line.

例题 A：求解不等式 4(x – 2) + 3 > 2x + 1，并在数轴上表示解集。

First, expand the bracket: 4x – 8 + 3 > 2x + 1, which simplifies to 4x – 5 > 2x + 1. Subtract 2x from both sides: 2x – 5 > 1. Add 5 to both sides: 2x > 6. Divide by 2: x > 3. The solution set is all values greater than 3. On the number line, place an open circle at 3 and shade to the right.

首先，展开括号：4x – 8 + 3 > 2x + 1，即 4x – 5 > 2x + 1。两边同时减去 2x：2x – 5 > 1。两边同时加上 5：2x > 6。除以 2：x > 3。解集为所有大于 3 的值。在数轴上，在 3 处画空心圆圈并向右侧涂阴影。

Solution: x > 3 | 解：x > 3
Interval notation: (3, ∞) | 区间符号：(3, ∞)

Example B | 例题 B：Solve the double inequality -5 ≤ 2x – 3 < 7.

例题 B：求解双重不等式 -5 ≤ 2x – 3 < 7。

Add 3 to all three parts: -2 ≤ 2x < 10. Divide all three parts by 2: -1 ≤ x < 5. The solution set is [-1, 5). On the number line, place a closed circle at -1 and an open circle at 5, shading the interval between them.

三部分同时加上 3：-2 ≤ 2x < 10。三部分同时除以 2：-1 ≤ x < 5。解集为 [-1, 5)。在数轴上，在 -1 处画实心圆点，在 5 处画空心圆圈，并涂阴影连接两者间的区间。

Example C | 例题 C：Find the set of values of x for which both 3x – 7 > 2 and 2x + 1 ≤ 11 hold simultaneously.

例题 C：求同时满足 3x – 7 > 2 和 2x + 1 ≤ 11 的 x 的取值范围。

Solve the first inequality: 3x – 7 > 2 gives 3x > 9, so x > 3. Solve the second inequality: 2x + 1 ≤ 11 gives 2x ≤ 10, so x ≤ 5. The intersection of these two solution sets is 3 < x ≤ 5, written in interval notation as (3, 5]. Both endpoints are correctly marked: open at 3, closed at 5.

求解第一个不等式：3x – 7 > 2 得 3x > 9，所以 x > 3。求解第二个不等式：2x + 1 ≤ 11 得 2x ≤ 10，所以 x ≤ 5。这两个解集的交集为 3 < x ≤ 5，用区间符号写作 (3, 5]。两个端点均正确标记：3 处空心，5 处实心。

11. Summary and Strategic Advice | 总结与策略建议

Linear inequalities are a core skill that appears throughout the A-Level mathematics curriculum, often as part of larger problems in algebra, coordinate geometry, and optimisation. Mastery of this topic requires fluency in algebraic manipulation, careful attention to the direction of the inequality sign, and the ability to present solutions in multiple equivalent forms: inequality form, interval notation, and number line diagrams.

线性不等式是贯穿A-Level数学课程的核心技能，常常作为代数、坐标几何和最优化等更大问题的一部分出现。掌握本课题需要熟练的代数变形能力、对不等式方向的细致关注，以及能以多种等价形式呈现解的能力：不等式形式、区间符号和数轴图形。

Before the examination, practise the following: solving inequalities with brackets and fractions, solving double inequalities with negative coefficients, and translating word problems into inequality systems. For each solution, draw the corresponding number line and write the interval notation. This practice builds the habits that prevent careless errors in the exam hall.

备考期间，请针对以下内容进行练习：求解含有括号和分数的不等式，求解带有负系数的双重不等式，以及将文字题转化为不等式组。对每个解，画出相应的数轴并写出区间符号。这样的练习能帮助养成良好习惯，防止考试时出现粗心错误。

Final tip: Always substitute a value from your solution set back into the original inequality as a check. This quick verification not only confirms your answer but also helps you identify any sign-reversal mistakes before submission.

终极提示：始终从你的解集中选取一个值代回原不等式进行检验。这种快速验证不仅能确认答案的正确性，还能帮助你在提交之前发现任何符号反转的错误。

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Linear Inequalities: Solving and Number Line Representation | 线性不等式：求解与数轴表示