📚 IGCSE Mathematics: Linear Equations and Inequalities | IGCSE 数学：线性方程与不等式
Linear equations and inequalities are central to the IGCSE Mathematics syllabus. They appear in algebra questions, word problems, graphs, and even in geometry and data handling. This revision guide explains how to solve linear equations and inequalities step by step, with clear examples, common mistakes, and exam-style practice. By mastering these techniques, you will build a strong foundation for higher-level topics such as simultaneous equations, quadratic functions, and calculus preparation.
线性方程与不等式是 IGCSE 数学教学大纲的核心内容，广泛出现在代数题、应用题、图像题甚至几何和数据题中。这份复习指南将逐步讲解如何解线性方程和不等式，并提供清晰的例题、常见错误和考试风格练习。掌握这些技巧后，你将为进一步学习联立方程、二次函数以及微积分打下坚实基础。
1. Introduction to Linear Equations | 线性方程简介
tics, you are expected to solve linear equations efficiently and to use them in real-life problems. The solution of a linear equation is the value of the variable that makes the equation true.
线性方程是指未知数的最高次数为 1 的方程。例如 2x + 3 = 11 是线性方程，而 x² + 2 = 6 不是。在 IGCSE 数学中，你需要熟练地解线性方程，并能用它们解决实际应用问题。线性方程的解就是使方程成立的未知数的值。
nstance, in 2x + 3 = 11, substituting x = 4 gives 2(4) + 3 = 11, so x = 4 is the solution.
要检查某个值是否为解，可以把它代入原方程。如果两边相等，这个值就是正确的。例如在 2x + 3 = 11 中，代入 x = 4 可得 2(4) + 3 = 11，因此 x = 4 是方程的解。
2. Solving One-Step Linear Equations | 解一步线性方程
es by 3 to get x = 6. The key idea is to isolate the variable on one side of the equation.
解一步方程时，使用逆运算。若 x + 5 = 12，则两边同时减去 5，得 x = 7。若 3x = 18，则两边同时除以 3，得 x = 6。关键思路是把未知数单独留在方程的一边。
The table below summarises the most common inverse operations used in IGCSE algebra.
下表总结了 IGCSE 代数中最常用的逆运算。
| Operation | Inverse operation |
|---|---|
| Addition | Subtraction |
| Subtraction | Addition |
| Multiplication | Division |
| Division | Multiplication |
Always apply the inverse operation to both sides of the equation. This keeps the equation balanced and preserves the equality.
一定要对等式两边同时进行逆运算。这样可以保持方程平衡，等量关系不变。
3. Solving Two-Step and Multi-Step Equations | 解两步与多步方程
= 11. Subtract 3 from both sides to get 2x = 8. Then divide both sides by 2 to get x = 4.
解两步方程时，先消去加法或减法，再消去乘法或除法。例如解 2x + 3 = 11。两边先减 3，得 2x = 8，再两边除以 2，得 x = 4。
2x + 3 = 11 → 2x = 8 → x = 4
btract 3x from both sides to get 2x − 7 = 9. Add 7 to both sides to get 2x = 16, so x = 8.
在多步方程中，同样遵循先化简两边、合并同类项、再分离未知数的原则。例如 5x − 7 = 3x + 9。两边同时减去 3x，得 2x − 7 = 9；两边加 7，得 2x = 16，因此 x = 8。
5x − 7 = 3x + 9 → 2x − 7 = 9 → 2x = 16 → x = 8
Remember that the order of inverse operations matters. Always undo addition or subtraction before multiplication or division unless brackets force a different order.
请注意逆运算的顺序很重要。除非出现括号要求不同顺序，否则通常先消去加法或减法，再消去乘法或除法。
4. Equations with Brackets | 含括号的方程
the left side: 3x − 6 = 12. Then add 6 to both sides: 3x = 18. Finally divide by 3: x = 6.
解含括号的方程时，通常先展开括号。例如解 3(x − 2) = 12。展开左边得 3x − 6 = 12，两边加 6，得 3x = 18，最后除以 3，得 x = 6。
3(x − 2) = 12 → 3x − 6 = 12 → 3x = 18 → x = 6
(x + 3) = 8 becomes −2x − 6 = 8. Adding 6 gives −2x = 14, and dividing by −2 gives x = −7.
当括号外是负数时，要特别注意符号。例如 −2(x + 3) = 8 展开后为 −2x − 6 = 8。两边加 6 得 −2x = 14，再除以 −2 得 x = −7。
If there are two brackets or brackets on both sides, expand each one carefully, collect like terms, and then solve as usual.
如果出现两个括号或等号两边都有括号，要分别仔细展开，合并同类项后按常规方法求解。
5. Equations with Fractions | 含分数的方程
. The LCD of 3, 2, and 6 is 6. Multiply each term by 6: 2x + 3 = 5. Then 2x = 2, so x = 1.
为了消去分母，可以把每一项都乘以最小公分母，通常记作 LCD。例如解 x/3 + 1/2 = 5/6。3、2 和 6 的最小公分母是 6。每项乘以 6，得 2x + 3 = 5，于是 2x = 2，x = 1。
x/3 + 1/2 = 5/6 → 2x + 3 = 5 → 2x = 2 → x = 1
4(x − 2). Expand: 3x + 3 = 4x − 8. Subtract 3x from both sides: 3 = x − 8. Add 8: x = 11.
当等号两边各只有一个分式时，也可以使用交叉相乘。例如 (x + 1)/4 = (x − 2)/3。交叉相乘得 3(x + 1) = 4(x − 2)。展开为 3x + 3 = 4x − 8。两边减去 3x，得 3 = x − 8；再加 8，得 x = 11。
Always check that the denominator is not zero. If a solution makes any original denominator equal to zero, it must be rejected as an invalid solution.
务必检查分母是否为零。如果某个解使原方程中任何一个分母为零，那么这个解必须舍去，因为它不是有效解。
6. Introduction to Inequalities | 不等式简介
. For example, x > 4 means x is greater than 4. Unlike equations, inequalities usually have a whole range of solutions rather than a single value.
不等式使用 <、>、≤ 或 ≥ 来比较两个表达式。例如 x > 4 表示 x 大于 4。与方程不同，不等式的解通常是一个范围，而不是一个单一的数值。
Here are the four inequality symbols and their meanings:
下面是不等式的四种符号及其含义：
- x < a means x is less than a | x < a 表示 x 小于 a
- x > a means x is greater than a | x > a 表示 x 大于 a
- x ≤ a means x is less than or equal to a | x ≤ a 表示 x 小于或等于 a
- x ≥ a means x is greater than or equal to a | x ≥ a 表示 x 大于或等于 a
The solution set of an inequality can be described using set notation, interval notation, or a number line. In IGCSE exams, you must be comfortable with all three forms.
不等式的解集可以用集合表示法、区间表示法或数轴来表示。在 IGCSE 考试中，你需要熟练运用这三种表示方式。
7. Solving Linear Inequalities | 解线性不等式
Solve inequalities in the same way as equations, but remember to reverse the inequality sign when multiplying or dividing both sides by a negative number. For example, solve −2x > 6. Divide both sides by −2 and flip the sign: x < −3.
解不等式的方法与方程类似，但当两边同时乘以或除以一个负数时，必须改变不等号的方向。例如解 −2x > 6。两边同时除以 −2 并翻转不等号，得 x < −3。
−2x > 6 → x < −3
This rule exists because multiplying or dividing by a negative number reverses the order of numbers on the number line. If you forget to flip the sign, your solution will be incorrect.
这条规则存在的原因是，乘以或除以负数会反转数轴上数字的顺序。如果忘记翻转不等号，你的解就会出错。
ld give −2(−2) = 4, which is not greater than 6, confirming that x = −2 is not a solution.
要检验不等式的解，可以在解的范围内取一个值代入原不等式。对于 −2x > 6，取 x = −4，得 −2(−4) = 8，而 8 > 6 成立。若取 x = −2，则 −2(−2) = 4，4 不大于 6，这证明 x = −2 不是解。
8. Compound Inequalities | 复合不等式
by 2: −1 < x ≤ 2. The solution says that x is greater than −1 and less than or equal to 2.
复合不等式将两个不等式合并成一个式子。例如 −1 < 2x + 1 ≤ 5。三边同时减去 1，得 −2 < 2x ≤ 4；再同时除以 2，得 −1 < x ≤ 2。这个解表示 x 大于 −1 且小于或等于 2。
−1 < 2x + 1 ≤ 5 → −2 < 2x ≤ 4 → −1 < x ≤ 2
When solving compound inequalities, keep the same operation on all three parts at each step. This maintains the relationship between the middle expression and both bounds.
解复合不等式时，每一步都要对三部分进行同样的运算。这样才能保持中间表达式与左右两个边界之间的关系。
Sometimes you may need to solve two separate inequalities and then combine their solution sets using 'and' or 'or'. The phrase 'and' gives the intersection of two ranges, while 'or' gives the union.
有时你可能需要分别解两个不等式，然后用 'and' 或 'or' 来合并它们的解集。'and' 表示两个范围的交集，'or' 表示两个范围的并集。
9. Representing Inequalities on a Number Line | 用数轴表示不等式
. For −1 < x ≤ 2, use an open circle at −1 and a closed circle at 2, shading between them.
在数轴上表示不等式时，< 或 > 用空心圆圈，≤ 或 ≥ 用实心圆圈，并涂出满足不等式的区域。对于 −1 < x ≤ 2，在 −1 处画空心圆，在 2 处画实心圆，中间涂上阴影。
hade to the left of 2. A compound inequality shades the overlapping region between the two bounds.
如果不等式向右延伸，例如 x > −1，就在 −1 的右侧涂阴影；如果向左延伸，例如 x ≤ 2，就在 2 的左侧涂阴影。复合不等式则在两个边界之间涂出重叠区域。
In an exam, always label the required numbers clearly on the number line and make the direction of shading obvious. Use a ruler if necessary, and check whether the circle should be open or closed.
在考试中，务必在数轴上清楚地标出所需的数字，并明确表示阴影方向。必要时使用直尺，同时检查圆圈应该是空心还是实心。
10. Word Problems and Applications | 应用题与实际应用
Translate word problems into equations or inequalities by identifying the unknown quantity and the relationships between quantities. Common phrases include 'more than', 'less than', 'at least', and 'at most'. Example: A taxi charges $2 plus $3 per mile. If the total cost is less than $20, write 2 + 3m < 20. Solve: 3m < 18, so m < 6. The journey must be less than 6 miles.
解应用题时，要先找出未知量和数量之间的关系，然后把文字翻译成方程或不等式。常见的表述包括 'more than'、'less than'、'at least' 和 'at most'。例如：出租车起步价 2 美元，每英里加收 3 美元。如果总费用少于 20 美元，可列出 2 + 3m < 20。解得 3m < 18，所以 m < 6。行程必须少于 6 英里。
2 + 3m < 20 → 3m < 18 → m < 6
When the word problem says 'at least' or 'no more than' kg, the inequality is 0.8b ≤ 12, so b ≤ 15. This means the maximum number of books is 15.
当应用题中出现 'at least' 或 'no more than' 时，应使用 ≤ 或 ≥ 而不是严格不等号。例如一个箱子最多能装 12 kg，如果每本书重 0.8 kg，则不等式为 0.8b ≤ 12，解得 b ≤ 15。这意味着最多可以放 15 本书。
Always define your variable clearly, write the equation or inequality step by step, and state the final answer in the context of the problem, including units if needed.
务必清晰地定义变量，逐步写出方程或不等式，并结合题意
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