📚 GCSE AQA Maths Inequalities Revision | GCSE AQA 数学：不等式 考点精讲
Inequalities are an essential part of the GCSE AQA Mathematics syllabus. They allow us to describe a range of values rather than a single number, and they appear in both algebra and problem-solving contexts. Mastering inequalities involves understanding symbols, solving them like equations (but with a crucial twist), and representing the solution sets on a number line. This revision guide will walk you through every key skill you’ll need for the exam, from basic symbols to tackling double inequalities and word problems, with some Higher Tier extensions.
不等式是 GCSE AQA 数学大纲中的重要部分。它们让我们能够描述一个数值范围而不仅仅是单个数字，并且出现在代数和问题求解的情境中。掌握不等式需要理解符号、像解方程一样求解不等式（但有一个关键转折），以及在数轴上表示解集。这份复习指南将带你逐一攻克考试所需的每一项关键技能，从基本符号到处理双不等式和文字题，并包含一些高等级延伸内容。
1. Inequality Symbols and Meanings | 不等式符号与含义
Inequalities compare two quantities and show that one is smaller or larger, or possibly equal. The four fundamental symbols are:
不等式用于比较两个量，表明其中一个较小或较大，或者可能相等。四个基本符号是：
< means “less than”. For example, 3 < 8.
> means “greater than”. For example, 7 > 2.
≤ (or ≤) means “less than or equal to”. For instance, x ≤ 5 means x can be 5 or any number smaller.
≥ (or ≥) means “greater than or equal to”. For instance, y ≥ -2 means y can be -2 or any number larger.
< 表示 “小于”。例如，3 < 8。
> 表示 “大于”。例如，7 > 2。
≤ (或 ≤) 表示 “小于或等于”。例如，x ≤ 5 意味着 x 可以是 5 或任何更小的数。
≥ (或 ≥) 表示 “大于或等于”。例如，y ≥ -2 意味着 y 可以是 -2 或任何更大的数。
The symbol ≠ is also sometimes used to mean “not equal to”, as in x ≠ 4, but the focus of linear inequalities is on range comparisons using the four main signs.
有时也会使用符号 ≠ 表示 “不等于”，比如 x ≠ 4，但线性不等式的重点是使用四个主要符号进行范围比较。
2. Representing Inequalities on a Number Line | 不等式的数轴表示
Any inequality can be visualised on a number line. This is an exam skill that tests your understanding of open and closed circles, as well as the direction of shading.
任何不等式都可以在数轴上直观表示。这是一项考查你对空心圆和实心圆以及阴影方向理解的考试技能。
For a strict inequality (using < or >), draw an open circle at the boundary number. The open circle shows that the number itself is not included. Then shade (draw an arrow) in the direction of the allowed values.
For an inclusive inequality (using ≤ or ≥), draw a closed (filled) circle at the boundary number, meaning the number is included, and then shade the appropriate side.
对于严格不等式（使用 < 或 >），在边界数字处画一个空心圆。空心圆表示该数字本身不包含在内。然后沿所允许的值的方向涂上阴影（画一个箭头）。
对于包含等号的不等式（使用 ≤ 或 ≥），在边界数字处画一个实心（填充）圆，表示该数字包含在内，然后对相应的一侧涂上阴影。
Example: Represent x > 2 on a number line. Place an open circle at 2 and draw an arrow pointing to the right. If it were x ≤ -1, place a closed circle at -1 and draw an arrow to the left.
例如：在数轴上表示 x > 2。在 2 处放置一个空心圆，并向右画一个箭头。如果是 x ≤ -1，在 -1 处放置一个实心圆，并向左画一个箭头。
In exams, you may label a few reference numbers to make the line clear, such as marking … -2, -1, 0, 1, 2, 3, 4 …
在考试中，你可以标注几个参考数字以使数轴清晰，例如标记 … -2, -1, 0, 1, 2, 3, 4 …。
3. Solving Simple Linear Inequalities | 解简单线性不等式
Solving a linear inequality is very similar to solving a linear equation. You can add, subtract, multiply or divide both sides by the same positive number without changing the inequality direction.
解线性不等式与解线性方程非常相似。你可以对两边同时加、减、乘或除以同一个正数，而不改变不等号的方向。
Step-by-step example: Solve x – 4 > 1. Add 4 to both sides: x > 5. The solution is all numbers greater than 5.
逐步求解示例：解 x – 4 > 1。两边加 4：x > 5。解是所有大于 5 的数。
Another example: Solve 3y ≤ 12. Divide both sides by 3: y ≤ 4. The solution set includes 4 and all numbers less than 4.
另一个示例：解 3y ≤ 12。两边除以 3：y ≤ 4。解集包含 4 以及所有小于 4 的数。
Always remember to check your solution by picking a value that satisfies the inequality and substituting it back into the original.
务必通过选取一个满足不等式的值并代回原式来检验你的解。
4. The Flip Rule: Multiplying or Dividing by a Negative Number | 翻转规则：乘以或除以负数
This is the single most important rule in inequalities: when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign.
这是不等式中最重要的一条规则：当你对不等式的两边乘以或除以一个负数时，你必须反转不等号的方向。
For example, start with the true statement -2 < 4. Multiply both sides by -3: (-2)×(-3) = 6, and 4×(-3) = -12. The new inequality must be 6 > -12, so the < sign has flipped to >.
例如，从真命题 -2 < 4 开始。两边乘以 -3：(-2)×(-3) = 6，而 4×(-3) = -12。新的不等式必须为 6 > -12，因此 < 号已翻转成 >。
Apply this in algebra: Solve -2x > 8. Divide both sides by -2, and reverse the sign: x < -4.
在代数中运用此规则：解 -2x > 8。两边除以 -2，并反转不等号：x < -4。
A common mistake is to flip the sign only sometimes or to forget it altogether. Always highlight the negative coefficient step in your working.
一个常见错误是有时翻转了符号，有时又忘了，或者干脆彻底忘记。在你的解题步骤中要始终突出标记负系数的处理步骤。
5. Solving Two-Step Inequalities | 解两步不等式
Two-step inequalities involve more than one operation. You will typically add or subtract first, and then multiply or divide. Stay alert for a negative coefficient in the final step.
两步不等式包含超过一种运算。通常你会先进行加减，然后再进行乘除。在最后一步要警惕负系数的出现。
Example: Solve 4x + 6 ≤ 2. Subtract 6 from both sides: 4x ≤ -4. Then divide both sides by 4 (positive, so no flip): x ≤ -1.
示例：解 4x + 6 ≤ 2。两边减去 6：4x ≤ -4。然后两边除以 4（正数，因此不翻转）：x ≤ -1。
Example with a negative coefficient: Solve 12 – 3x < 9. Subtract 12 from both sides: -3x < -3. Now divide by -3 and flip the sign: x > 1.
负系数示例：解 12 – 3x < 9。两边减去 12：-3x < -3。现在除以 -3 并翻转不等号：x > 1。
Represent the solution x > 1 on a number line with an open circle at 1 and shading to the right.
在数轴上表示解 x > 1，在 1 处画空心圆，并向右涂上阴影。
6. Double Inequalities | 双不等式
A double inequality combines two inequalities into a single statement, such as 1 < 2x + 3 ≤ 7. This means 2x + 3 is greater than 1 AND less than or equal to 7.
双不等式将两个不等式合并成一个单独的表达式，例如 1 < 2x + 3 ≤ 7。这意味着 2x + 3 大于 1 并且小于或等于 7。
To solve, you can perform the same operation on all three parts simultaneously (left, middle, right). Subtract 3 from all parts: 1 – 3 < 2x + 3 – 3 ≤ 7 – 3, which simplifies to -2 < 2x ≤ 4. Then divide everything by 2: -1 < x ≤ 2.
求解时，你可以同时对三部分（左、中、右）进行相同的运算。所有部分减 3：1 – 3 < 2x + 3 – 3 ≤ 7 – 3，化简为 -2 < 2x ≤ 4。然后所有部分除以 2：-1 < x ≤ 2。
If the middle term had a negative coefficient, you would divide all parts by a negative number, which would require reversing both inequality symbols. For instance, 2 < -3x ≤ 9 divided by -3 becomes -2/3 > x ≥ -3, which is often rewritten as -3 ≤ x < -2/3.
如果中间项有负系数，则所有部分都要除以一个负数，这要求两个不等号都反转。例如，2 < -3x ≤ 9 除以 -3 变为 -2/3 > x ≥ -3，这通常被重写为 -3 ≤ x < -2/3。
Alternatively, you can split a double inequality into two separate inequalities and solve them independently, then find the overlap of the solutions. This method can reduce sign-flip errors.
或者，你可以将一个双不等式拆分成两个单独的不等式并分别求解，然后找出解的交集。这种方法可以减少符号翻转的错误。
7. Inequalities with Brackets | 带括号的不等式
When an inequality contains brackets, expand them first, just as you would in an equation. Then collect like terms and isolate the variable.
当不等式中含有括号时，首先要展开括号，就像解方程时那样。然后合并同类项并隔离变量。
Example: Solve 3(x – 2) > x + 4. Expand the left side: 3x – 6 > x + 4. Subtract x from both sides: 2x – 6 > 4. Add 6 to both sides: 2x > 10. Divide by 2: x > 5.
示例：解 3(x – 2) > x + 4。展开左边：3x – 6 > x + 4。两边减去 x：2x – 6 > 4。两边加 6：2x > 10。除以 2：x > 5。
If there are fractions inside the brackets, you might first multiply both sides by a suitable number to clear denominators, but be careful with signs.
如果括号内有分数，你可以首先在两边乘以一个适当的数以去掉分母，但要注意符号。
Remember to treat the inequality symbol like an equals sign throughout the expansion and simplification steps, except if you ever multiply or divide by a negative, where the flip applies.
记住，在整个展开和化简的步骤中，将不等号像等号一样对待，除非你进行了乘以或除以负数的操作，这时需要进行翻转。
8. Graphing Solutions on a Number Line (Exam Technique) | 数轴上作图表示解集（考试技巧）
Many exam questions ask you to solve an inequality and then represent the solution on a number line. Make sure your number line is neat, clearly labelled, and uses the correct circle style.
许多考题要求你解一个不等式，然后在数轴上表示解集。确保你的数轴整洁、标注清晰，并使用正确的圆圈样式。
For instance, solve 5 – 2x ≥ 1. Subtract 5: -2x ≥ -4. Divide by -2 and flip: x ≤ 2. On the number line, place a closed (filled) circle at 2 and draw a bold arrow pointing to the left.
例如，解 5 – 2x ≥ 1。减 5：-2x ≥ -4。除以 -2 并翻转：x ≤ 2。在数轴上，在 2 处放置一个实心圆，并向左画一个粗箭头。
Sometimes you will need to show integer solutions on a number line by marking individual points. In such cases, use small solid dots on the appropriate whole numbers.
有时你需要通过标记单个点来在数轴上显示整数解。在这种情况下，在相应的整数上使用小实心圆点。
Always double-check the direction of the arrow. If the inequality is x < -3, the arrow goes left; if x > -3, the arrow goes right.
务必仔细检查箭头的方向。如果不等式是 x < -3，箭头向左；如果是 x > -3，箭头向右。
9. Solving Systems of Linear Inequalities | 解线性不等式组
Sometimes you must find values that satisfy more than one inequality at the same time. The solution set is the intersection (overlap) of the individual solution sets.
有时你必须找出同时满足多个不等式的值。解集是各个单独解集的交集（重叠部分）。
Example: List all integer values of n such that n > -2 and n ≤ 3. The first inequality gives n = -1, 0, 1, 2, 3, 4, 5, … The second gives n = …, -1, 0, 1, 2, 3. The integers that satisfy both are -1, 0, 1, 2, 3. Notice that -2 is not included because n > -2 is strict.
示例：列出所有满足 n > -2 且 n ≤ 3 的整数值 n。第一个不等式给出 n = -1, 0, 1, 2, 3, 4, 5, …，第二个不等式给出 n = …, -1, 0, 1, 2, 3。同时满足两者的整数是 -1, 0, 1, 2, 3。注意 -2 没有包含在内，因为 n > -2 是严格的。
Always write the final answer in a clear form, such as “x is an integer and -1 < x < 4” or list the possible values: 0, 1, 2, 3.
始终以清晰的形式写出最终答案，例如 “x 为整数且 -1 < x < 4″，或者列出可能的值：0, 1, 2, 3。
You may be asked to represent the overlapping region on a number line. This involves drawing the two individual representations and highlighting the common part, for example by a thicker line segment.
你可能会被要求在数轴上表示重叠区域。这需要画出两个单独的表示，并突出公共部分，例如用一条更粗的线段。
10. Word Problems Involving Inequalities | 涉及不等式的文字题
Inequalities commonly appear in real-world contexts. Key phrases to recognise include “at least” (≥), “at most” (≤), “fewer than” (<), and “more than” (>).
不等式通常出现在现实情境中。需要识别的关键短语包括 “至少” (≥)、”最多” (≤)、”少于” (<) 以及 “多于” (>)。
Example: A school trip requires that a coach carries no more than 50 students. Currently 38 students have signed up. Write an inequality for the number of additional students, a, and solve it. The inequality is 38 + a ≤ 50. Solving gives a ≤ 12. So at most 12 more students can join.
示例：学校旅行要求一辆巴士运载不超过 50 名学生。目前已有 38 名学生报名。写出关于额外学生人数 a 的不等式，并求解。该不等式是 38 + a ≤ 50。求解得 a ≤ 12。因此最多还能加入 12 名学生。
Another typical problem: A rectangular garden has a width of 3 metres. The perimeter must be greater than 20 metres. Write and solve an inequality for the length, L. Perimeter = 2×(L + 3) > 20. Simplify: 2L + 6 > 20, so
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