📚 Algebra and Functions for CIE IGCSE Mathematics | CIE IGCSE 数学：代数与函数考点精讲
In CIE IGCSE Mathematics, Algebra and Functions form a core part of the syllabus, testing your ability to manipulate expressions, solve equations, and understand relationships between variables. This revision guide covers the essential topics, with step-by-step explanations and key formulae.
在 CIE IGCSE 数学中，代数与函数是教学大纲的核心部分，考查你处理表达式、解方程以及理解变量之间关系的能力。本复习指南涵盖了基本知识点，并提供了逐步解释和关键公式。
1. Core Notation and Vocabulary | 核心符号与术语
a fixed number; a term is a single number, a single letter, or a product of them; and a coefficient is the number multiplying a variable. An expression such as 4x + 7 contains no equals sign, whereas an equation such as 4x + 7 = 19 makes a statement that must be solved. A formula relates two or more variables, and an identity is true for every value of the variable, which is why it is written with the symbol ≡.
在进行代数运算之前，你需要掌握准确的术语。变量是表示未知量的字母；常数是固定不变的数；项是单个数字、单个字母或它们的乘积；系数则是乘以变量的那个数。像 4x + 7 这样的表达式没有等号，而像 4x + 7 = 19 这样的方程是一个需要求解的陈述。公式描述两个或多个变量之间的关系，而恒等式对变量的所有取值都成立，因此用符号 ≡ 表示。
| Notation | Meaning | 中文含义 |
| 3x | coefficient 3 multiplied by variable x | 系数 3 乘以变量 x |
| x² | index (power) 2 applied to x | x 的指数（幂）为 2 |
| f(x) | function f acting on input x | 函数 f 作用于输入 x |
| ≡ | identity, true for all values | 恒等号，对所有取值成立 |
2. Simplifying Algebraic Expressions | 代数表达式的化简
Algebraic expressions are built from terms involving variables (like x and y) and constants. Simplification means rewriting an expression in its shortest equivalent form, and the key skill is collecting like terms - terms that have exactly the same variable part. Numbers collect with numbers, x terms collect with x terms, and x² terms collect with x² terms. Terms with different letter parts can never be added together.
代数表达式由包含变量（如 x、y）和常数的项构成。化简就是把表达式改写成最短的等价形式，核心技巧是合并同类项——也就是字母部分完全相同的项。数字与数字合并，含 x 的项与含 x 的项合并，含 x² 的项与含 x² 的项合并；字母部分不同的项永远不能相加。
tiplication works differently: 4a × 3b = 12ab, a × a = a², and a ÷ a = 1 (provided a ≠ 0).
例如，3x + 2y - x + 5y 化简为 2x + 7y。注意 2x 与 7y 必须保持分开，因为 x 和 y 不是同类项。乘法则不同：4a × 3b = 12ab，a × a = a²，a ÷ a = 1（要求 a ≠ 0）。
3x + 2y − x + 5y = 2x + 7y
t, as in 5a - (2a - 3b), requires you to change every sign inside: 5a - 2a + 3b = 3a + 3b.
务必检查每一项前面的符号。减去一个括号时，例如 5a - (2a - 3b)，括号内每一项都要变号：5a - 2a + 3b = 3a + 3b。
3. Laws of Indices | 指数法则
The laws of indices allow you to combine powers quickly, and they appear constantly in IGCSE papers. When you multiply powers of the same base you add the indices; when you divide you subtract them; when you raise a power to a power you multiply them. Any non-zero number raised to the power 0 equals 1.
指数法则让你能够快速合并幂，在 IGCSE 试卷中频繁出现。同底数幂相乘，指数相加；相除，指数相减；幂的乘方，指数相乘。任何非零数的 0 次幂都等于 1。
x² × x³ = x⁵ x⁵ ÷ x² = x³ (x²)³ = x⁶
so a⁰ = 1 for a ≠ 0. Negative and fractional indices are also examinable: a⁻ⁿ = 1 ÷ aⁿ, a1/2 = √a, and am/n = (ⁿ√a)ᵐ.
记住任何非零数的 0 次幂都是 1，即当 a ≠ 0 时 a⁰ = 1。负指数与分数指数同样属于考查范围：a⁻ⁿ = 1 ÷ aⁿ，a1/2 = √a，am/n = (ⁿ√a)ᵐ。
| Rule | Example | 中文说明 |
| Multiplying: add indices | x² × x³ = x⁵ | 相乘：指数相加 |
| Dividing: subtract indices | x⁵ ÷ x² = x³ | 相除：指数相减 |
| Power of a power: multiply | (x²)³ = x⁶ | 幂的乘方：指数相乘 |
| Zero index | a⁰ = 1, a ≠ 0 | 零指数：结果为 1 |
| Negative index | a⁻² = 1 ÷ a² | 负指数：取倒数 |
4. Expanding Brackets and Factorising | 展开括号与因式分解
Expanding removes brackets by multiplying every term in the first bracket by every term in the second. Factorising is the reverse process: you rewrite an expression as a product of factors. These two skills are the backbone of almost every later algebra question.
展开括号就是把第一个括号中的每一项与第二个括号中的每一项相乘，从而去掉括号。因式分解则是逆过程：把表达式写成因式的乘积。这两项技能几乎是后续所有代数题的基础。
9x = 3x(2x + 3). Then look for special patterns, especially the difference of two squares.
以 (x + 3)(x - 5) 为例，展开得到 x² - 5x + 3x - 15 = x² - 2x - 15。因式分解时先找公因式：6x² + 9x = 3x(2x + 3)；再找特殊结构，尤其是平方差。
a² − b² = (a + b)(a − b) (a ± b)² = a² ± 2ab + b²
4 = 7. Check your answer by expanding it again - it takes seconds and protects easy marks.
对于形如 x² + bx + c 的二次式，找两个数使它们的积为 c、和为 b。所以 x² + 7x + 12 = (x + 3)(x + 4)，因为 3 × 4 = 12 且 3 + 4 = 7。把答案再展开验证一遍只需几秒，却能保住容易丢的分数。
5. Linear Equations and Rearranging Formulae | 一次方程与公式变形
ractions appear, multiply every term by the lowest common denominator first to clear them.
解一次方程时，对方程两边做相同的运算，直到变量单独留在一边。例如 5x - 7 = 3x + 9 化为 2x = 16，所以 x = 8。若出现分数，先给每一项乘以最小公分母以消去分母。
h sides to get at = v - u, then divide by a, giving t = (v - u) ÷ a. Treat every letter except the one you want as if it were a number.
公式变形使用完全相同的思路，只不过答案是字母而不是数字。要把 v = u + at 变形为 t 为主项，两边同时减去 u 得 at = v - u，再除以 a，得到 t = (v - u) ÷ a。除了你要解出的那个字母，其他字母都当作数字处理。
When an equation has brackets, expand first. When the variable appears on both sides with a negative coefficient, collect all variable terms on the side that keeps the coefficient positive - it reduces sign errors.
方程含括号时先展开。当变量出现在两边且系数为负时，把所有含变量的项移到能使其系数为正的一侧，这样可以减少符号错误。
6. Inequalities on a Number Line | 不等式与数轴
Linear inequalities are solved in the same way as linear equations, with one crucial exception: if you multiply or divide both sides by a negative number, the inequality sign reverses. So from -2x > 6 you get x < -3, not x > -3.
一次不等式的解法与一次方程相同，但有一个关键例外：当两边同时乘以或除以负数时，不等号方向要反转。因此由 -2x > 6 得到 x < -3，而不是 x > -3。
4 means x is greater than -3 and at the same time less than or equal to 4, so the solution is drawn as a joined line segment.
在数轴上，空心圆圈表示严格不等式（< 或 >），实心圆圈表示包含端点（≤ 或 ≥）。像 -3 < x ≤ 4 这样的双重不等式表示 x 大于 -3 且同时小于或等于 4，因此画成一条连起来的线段。
−2x > 6 ⟹ x < −3
x ≤ 4, the integer values are -2, -1, 0, 1, 2, 3 and 4 - a very common one-mark follow-up.
题目要求写出整数解时一定要写。若解为 -3 < x ≤ 4，整数解是 -2、-1、0、1、2、3、4——这是非常常见的后续一分小题。
7. Simultaneous Linear Equations | 联立一次方程
A pair of simultaneous equations has one pair of values that satisfies both equations at the same time. The elimination method is usually fastest: multiply one or both equations so that the coefficients of one variable match, then add or subtract the equations to remove that variable.
一对联立方程只有一对取值能同时满足两个方程。消元法通常最快：把一个或两个方程乘以适当的数，使某个变量的系数相同，再把两式相加或相减以消去该变量。
e equations eliminates y: 5x = 20, so x = 4. Substituting back gives 8 + y = 11, so y = 3.
以 2x + y = 11 和 3x - y = 9 为例。由于 y 的系数分别为 +1 和 -1，两式相加即可消去 y：5x = 20，所以 x = 4。代回得 8 + y = 11，故 y = 3。
ker. Always substitute your answer into the equation you did not use for the check - that way an arithmetic slip cannot hide.
如果某个方程
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