📚 IGCSE AQA Maths: Inequalities Revision Guide | IGCSE AQA 数学：不等式 考点精讲
Inequalities are a key topic in the IGCSE AQA Mathematics syllabus. They appear in various forms – from solving simple linear inequalities to shading regions on graphs and tackling quadratic inequalities. Mastering inequalities will not only boost your algebra skills but also help you with real-life problem-solving and graphical interpretation. This guide walks you through every essential concept, notation and method you need to ace your exam.
不等式是 IGCSE AQA 数学大纲中的重点主题。它们以多种形式出现——从解简单的一次不等式到在图形上描绘区域，再到处理二次不等式。掌握不等式不仅会提升你的代数能力，还能帮助你在实际问题解决和图形解读中得心应手。本指南将带你逐一梳理每一个关键概念、符号和方法，助你在考试中取得高分。
1. What are Inequalities? | 不等式是什么？
An inequality compares two expressions and shows that one is greater than, less than or equal to the other. Instead of the equals sign, we use symbols like > (greater than), < (less than), ≥ (greater than or equal to) and ≤ (less than or equal to). A strict inequality uses > or <, while a non-strict inequality includes the possibility of equality.
不等式用于比较两个表达式，表明其中一个大于、小于或等于另一个。我们使用符号 >（大于）、<（小于）、≥（大于或等于）以及 ≤（小于或等于）来代替等号。严格不等式使用 > 或 <，而非严格不等式包含相等的可能性。
For example, 2x + 3 > 7 means “two times x plus three is greater than seven”. Inequalities can describe a whole range of values rather than a single solution. This makes them powerful for modelling constraints in everyday situations such as budgets, speed limits and measurements.
例如，2x + 3 > 7 表示“2 乘以 x 加 3 大于 7”。不等式可以描述一整段数值范围，而不仅仅是一个解。这使得它们在模拟日常生活中的约束条件（如预算、限速和测量值）时十分有用。
2. Representing Inequalities on a Number Line | 在数轴上表示不等式
Inequalities can be visualised using a number line. A solid circle ● is used for ≤ or ≥ to show that the endpoint is included. An open circle ○ is used for < or > to show the endpoint is not included. An arrow indicates that all numbers in that direction are part of the solution set.
不等式可以用数轴直观地表示。实心圆 ● 用于 ≤ 或 ≥，表示端点包含在内。空心圆 ○ 用于 < 或 >，表示端点不包含在内。箭头表示该方向上的所有数字都属于解集。
For the inequality x ≤ 4, you would draw a solid circle at 4 and shade the line to the left. For x > −2, you would draw an open circle at −2 and shade to the right. Combined inequalities such as −3 < x ≤ 1 are represented by an open circle at −3, a solid circle at 1, and a line connecting them.
对于不等式 x ≤ 4，你应在 4 处画一个实心圆，并向左描线。对于 x > −2，你应在 −2 处画一个空心圆，并向右描线。复合不等式如 −3 < x ≤ 1，则用 −3 处的空心圆、1 处的实心圆，并用线段将它们连接起来表示。
3. Solving Linear Inequalities | 解一元一次不等式
Solving linear inequalities follows the same steps as solving linear equations: use inverse operations to isolate the variable on one side. However, there is one golden rule – when you multiply or divide both sides by a negative number, you must reverse the inequality sign.
解一元一次不等式的步骤与解一次方程相同：使用逆运算将变量移到一侧。但有一条黄金法则——当你将两边同时乘以或除以一个负数时，必须反转不等号方向。
Example: Solve 5 − 2x ≥ 11. First subtract 5 from both sides: −2x ≥ 6. Divide both sides by −2, remembering to reverse the sign: x ≤ −3. The solution is all x less than or equal to −3. Always check by testing a value, e.g. x = −4: 5 − 2(−4) = 13, which is ≥ 11.
例子：解 5 − 2x ≥ 11。首先两边减去 5：−2x ≥ 6。然后两边除以 −2，注意反转符号：x ≤ −3。解集为所有小于或等于 −3 的 x。验证时可代入一个数值，例如 x = −4：5 − 2(−4) = 13，确实 ≥ 11。
4. Solving Double Inequalities | 解双重不等式
A double inequality like −4 < 3x + 2 ≤ 8 combines two inequalities in one statement. To solve it, keep the three parts balanced. Perform the same operation on all three parts simultaneously. In this example, subtract 2 from each part: −6 < 3x ≤ 6. Then divide all parts by 3: −2 < x ≤ 2. The solution is all numbers greater than −2 and less than or equal to 2.
双重不等式如 −4 < 3x + 2 ≤ 8 将两个不等式合并在一个表达式中。求解时，要在三个部分之间保持平衡，同时对三部分执行相同的运算。在这个例子中，每部分减去 2：−6 < 3x ≤ 6。然后每部分除以 3：−2 < x ≤ 2。解集为所有大于 −2 且小于或等于 2 的数。
You can represent this on a number line with an open circle at −2, a closed circle at 2 and a line between them. Double inequalities are often used for error bounds and measurement tolerances.
你可以在数轴上用 −2 处的空心圆、2 处的实心圆以及两圆之间的线段来表示。双重不等式常被用于误差界限和测量公差。
5. Inequalities with Unknowns on Both Sides | 两边含未知数的不等式
When the unknown appears on both sides of the inequality, collect the variable terms on one side and the constants on the other. The same rule about reversing the sign when multiplying or dividing by a negative applies.
当未知数出现在不等式的两边时，把变量项移到一边，常数项移到另一边。乘以或除以负数时反转符号的规则同样适用。
Solve 4x − 7 < 2x + 5. Subtract 2x from both sides: 2x − 7 < 5. Add 7: 2x < 12. Divide by 2: x < 6. The solution does not involve a negative coefficient in the final step, so the sign stays the same.
解 4x − 7 < 2x + 5。两边减去 2x：2x − 7 < 5。加 7：2x < 12。除以 2：x < 6。最后一步没有涉及负系数，因此符号方向不变。
For 5 − 3x ≥ x + 9, add 3x to both sides to have a positive coefficient: 5 ≥ 4x + 9. Then subtract 9: −4 ≥ 4x. Divide by 4: −1 ≥ x, which is the same as x ≤ −1. The sign was not reversed because we divided by a positive 4.
对于 5 − 3x ≥ x + 9，可在两边加 3x 以获得正系数：5 ≥ 4x + 9。然后减 9：−4 ≥ 4x。除以 4：−1 ≥ x，等价于 x ≤ −1。因为我们除的是正数 4，符号没有反转。
6. Integer Solutions to Inequalities | 不等式的整数解
An exam question might ask for the integer values that satisfy an inequality or a set of inequalities. After solving algebraically, list all integers within the solution range. For instance, if −2 < x ≤ 3, the integer solutions are −1, 0, 1, 2, 3. Note that −2 is not included because of the strict inequality.
考试中可能会要求找出满足某个不等式或不等式组的整数值。在代数求解之后，列出解集范围内的所有整数。例如，如果 −2 < x ≤ 3，整数解为 −1、0、1、2、3。注意 −2 由于是严格不等式而不包含在内。
For two inequalities like x ≥ −1 and x < 4, combine them to −1 ≤ x < 4. The integer solutions are −1, 0, 1, 2, 3. Always check the endpoints carefully: if the inequality is x ≤ 2.5, the integers that satisfy it are …, 0, 1, 2.
对于像 x ≥ −1 和 x < 4 这样的两个不等式，可将其合并为 −1 ≤ x < 4。整数解为 −1、0、1、2、3。务必仔细检查端点：若不等式为 x ≤ 2.5，满足它的整数是……0、1、2。
7. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities involve an x² term and are solved by finding the roots of the corresponding quadratic equation, then determining which intervals satisfy the inequality. First, rearrange to have 0 on one side. Then factorise or use the quadratic formula to find the critical values.
二次不等式包含 x² 项，求解方法是先求出对应二次方程的根，再确定哪些区间满足不等式。首先将式子移项使一边为 0，然后因式分解或使用求根公式找出临界值。
For x² − 5x + 6 < 0, factorise to (x − 2)(x − 3) < 0. The critical values are x = 2 and x = 3. Sketch a parabola (positive coefficient of x², so a ∪ shape) that crosses the x-axis at 2 and 3. The expression is negative between the roots. Hence the solution is 2 < x < 3.
对于 x² − 5x + 6 < 0，因式分解得 (x − 2)(x − 3) < 0。临界值为 x = 2 和 x = 3。画出 x² 系数为正的抛物线（呈 ∪ 形），在 x = 2 和 3 处与 x 轴相交。表达式在两根之间取负值，因此解为 2 < x < 3。
If the inequality were x² − 5x + 6 ≥ 0, the solution would be x ≤ 2 or x ≥ 3, because the parabola is non-negative outside the interval between the roots. Always be careful with strict vs non-strict inequalities and draw a quick sketch to confirm.
若不等式为 x² − 5x + 6 ≥ 0，解则为 x ≤ 2 或 x ≥ 3，因为抛物线在两根区间之外取非负值。务必注意严格与非严格不等式的区别，并快速画草图加以确认。
8. Graphical Inequalities | 图形不等式
You can display inequalities on a coordinate grid. A linear inequality like y > 2x + 1 divides the plane into two regions. First, draw the boundary line y = 2x + 1. Use a dashed line for > or < and a solid line for ≥ or ≤. Then shade the region that satisfies the inequality – test a point such as (0,0) to decide which side to shade.
你可以在坐标网格上展示不等式。例如 y > 2x + 1 这样的线性不等式会将平面分成两个区域。首先画出边界线 y = 2x + 1。对于 > 或 < 使用虚线，对于 ≥ 或 ≤ 使用实线。然后为满足不等式的区域涂上阴影——可选取一个点如 (0,0) 进行测试，以确定涂哪一侧。
For y > 2x + 1, the line is dashed because points on the line are not included. Testing (0,0): 0 > 2(0) + 1 → 0 > 1 is false, so shade the opposite side. Graphical inequalities often combine to define a region that satisfies several conditions simultaneously.
对于 y > 2x + 1，由于线上的点不包含在内，因此使用虚线。测试 (0,0)：0 > 2(0) + 1 → 0 > 1 为假，因此将另一侧涂上阴影。图形不等式常常组合在一起，定义一个同时满足多个条件的区域。
9. Shading Regions Defined by Inequalities | 描绘由不等式定义的区域
A common exam question gives two or more inequalities and asks you to shade the region where they are all true. For example, y ≤ x + 2, y > −1 and x < 4. Draw each boundary line using the correct style (solid for ≤, dashed for > and <). Shade the unwanted regions or explicitly highlight the intersection region as instructed.
常见的考题会给出两个或更多不等式，要求你描出它们同时成立的区域。例如，y ≤ x + 2, y > −1 以及 x < 4。使用合适的线型（对 ≤ 用实线，对 > 和 < 用虚线）画出各边界线。按题目要求涂掉不符合的区域，或明确标出交集区域。
To avoid confusion, many students prefer shading the region that does not satisfy each inequality, leaving the required region unshaded. However, always follow what the question asks. If it says “shade the region that satisfies all the inequalities”, then shade only the final overlap.
为避免混淆，许多学生喜欢先涂掉不符合每个不等式的区域，让所求区域保持空白。但务必遵循题目要求。如果题目说“为满足所有不等式的区域涂上阴影”，那么只需为最终的交集区域涂上阴影。
You may also need to write down the inequalities that define a given shaded region. Look for boundaries: vertical or horizontal lines (x = constant, y = constant) and sloping lines. Decide solid or dashed, and use a test point to determine the correct inequality sign.
你可能还需要写出定义一个给定阴影区域的不等式。观察边界：垂直线或水平线（x = 常数, y = 常数）以及斜线。判断是实线还是虚线，并利用测试点确定正确的不等号方向。
10. Real-life Applications of Inequalities | 不等式在现实中的应用
Inequalities model real constraints: a mobile phone plan costs £12 plus £0.05 per minute and you want to spend less than £25. The inequality is 12 + 0.05m < 25, leading to m < 260 minutes. Inequalities are also used in error intervals. If a length is 15 cm to the nearest cm, the true length L satisfies 14.5 ≤ L < 15.5.
不等式可以模拟现实约束条件：某手机套餐月租£12，每分钟通话费£0.05，你希望花费低于£25。不等式为 12 + 0.05m < 25，解得 m < 260 分钟。不等式也可用于误差区间：若一段长度四舍五入到最近厘米为 15 cm，则真实长度 L 满足 14.5 ≤ L < 15.5。
In optimisation problems, you set up inequalities representing limits on resources, time or money, then find the feasible region on a graph. The maximum or minimum values often occur at the vertices of this region.
在优化问题中，你需要建立资源、时间或资金限制的不等式，然后在图形上找出可行域。最大值或最小值常出现在该区域的顶点处。
Understanding inequalities gives you the language to describe ranges, limits and conditions accurately – a skill that reaches far beyond your IGCSE exam.
理解不等式能让你精确地描述范围、界限和条件——这项技能的应用远远超出了 IGCSE 考试本身。
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