📚 Mastering Algebra and Functions: AQA GCSE Maths | AQA GCSE 数学：代数和函数考点精讲
Algebra and functions form the backbone of the AQA GCSE Mathematics syllabus. From simplifying basic expressions to manipulating complex functions and solving equations, these topics test your ability to think logically and work with abstract symbols. Mastering them is essential not only for exam success but also for further study in maths and sciences. This guide breaks down every key concept, giving you clear explanations and examples in both English and Chinese so you can revise confidently and efficiently.
代数和函数是 AQA GCSE 数学教学大纲的核心支柱。从简化基本表达式到操作复杂函数和解方程，这些主题考查你的逻辑思维和抽象符号运用能力。掌握它们不仅是考试成功的关键，也是进一步学习数学和科学的基础。本指南将每个关键概念分解，用中英双语提供清晰的解释和示例，让你能够自信高效地复习。
1. Algebraic Terminology and Simplification | 代数术语与化简
An algebraic expression is a combination of numbers, variables (letters), and operations such as addition and multiplication. A term is a single number, variable, or product of numbers and variables, for example 3x or -5y². The coefficient is the numeric factor of a term, so in 4a the coefficient is 4. Like terms contain exactly the same variable part raised to the same power, such as 2x and 7x, and can be combined by adding or subtracting their coefficients.
代数表达式是数字、变量（字母）和运算（如加法和乘法）的组合。项是单个数字、变量或数字与变量的乘积，例如 3x 或 -5y²。系数是项中的数字因子，因此在 4a 中系数为 4。同类项含有完全相同的变量部分且指数相同，如 2x 和 7x，可以通过加上或减去它们的系数来合并。
To simplify an expression, identify and collect like terms. For instance, simplify 3a + 5b – 2a + 7b by grouping: (3a – 2a) + (5b + 7b) = a + 12b. Always pay attention to signs and the index laws when multiplying terms, such as x² × x³ = x⁵.
要化简表达式，识别并合并同类项。例如，化简 3a + 5b – 2a + 7b，分组：(3a – 2a) + (5b + 7b) = a + 12b。合并时注意符号，并在项相乘时运用指数律，如 x² × x³ = x⁵。
2. Expanding Brackets | 展开括号
Expanding brackets means multiplying each term inside the bracket by the term outside. For a single bracket, use the distributive law: a(b + c) = ab + ac. Be careful with negative signs: -2(x – 3) = -2x + 6. When expanding two brackets, multiply every term in the first bracket by every term in the second, often using the FOIL method (First, Outer, Inner, Last). For example, (x + 4)(x – 2) expands to x² – 2x + 4x – 8, which simplifies to x² + 2x – 8.
展开括号是指将括号外的项与括号内的每一项相乘。对于单个括号，使用分配律：a(b + c) = ab + ac。注意负号：-2(x – 3) = -2x + 6。展开两个括号时，将第一个括号中的每一项乘以第二个括号中的每一项，常用 FOIL 法则（首项、外项、内项、末项）。例如，(x + 4)(x – 2) 展开得 x² – 2x + 4x – 8，化简为 x² + 2x – 8。
For expressions like (a + b)², remember this means (a + b)(a + b). A common mistake is to write a² + b²; the correct expansion is a² + 2ab + b². Similarly, (a – b)² = a² – 2ab + b² and (a + b)(a – b) = a² – b². These special products save time and should be memorised.
对于 (a + b)² 这类表达式，记住它表示 (a + b)(a + b)。一个常见错误是写成 a² + b²；正确的展开是 a² + 2ab + b²。类似地，(a – b)² = a² – 2ab + b² 且 (a + b)(a – b) = a² – b²。这些特殊乘积可节省时间，应熟记。
3. Factorising Expressions | 因式分解
Factorising is the reverse of expanding; it writes an expression as a product of its factors. Start by looking for the highest common factor (HCF) of all terms. For example, factorise 4x² + 6x by taking out the HCF of 2x, giving 2x(2x + 3). When the expression is a quadratic in the form x² + bx + c, find two numbers that multiply to c and add to b. So x² + 7x + 10 factorises to (x + 2)(x + 5) because 2 and 5 multiply to 10 and add to 7.
因式分解是展开的逆运算，它将表达式写成因子的乘积。首先寻找所有项的最高公因数（HCF）。例如，分解 4x² + 6x，提取 HCF 2x，得到 2x(2x + 3)。当表达式是形如 x² + bx + c 的二次式时，找出两个数，它们相乘得 c 且相加得 b。因此 x² + 7x + 10 分解为 (x + 2)(x + 5)，因为 2 和 5 相乘得 10，相加得 7。
For quadratics with a leading coefficient not equal to 1, such as 2x² + 7x + 3, you need to split the middle term or use trial and error. Split 7x into 6x + x because 2×3=6 and we need factors of 6 that add to 7. Then factor by grouping: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Always check by expanding the brackets.
对于首项系数不是 1 的二次式，如 2x² + 7x + 3，需要拆分中间项或使用试错法。将 7x 拆分为 6x + x，因为 2×3=6，我们需要 6 的两个因数之和为 7。然后分组分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。务必通过展开括号来检验。
4. Solving Linear Equations | 解线性方程
A linear equation is an equation of the first degree, meaning the variable appears to the power of 1 only. The goal is to isolate the variable on one side using inverse operations. For example, to solve 3x + 5 = 20, subtract 5 from both sides to get 3x = 15, then divide both sides by 3 to obtain x = 5. Always do the same operation to both sides to maintain the equality.
线性方程是一次方程，意味着变量仅以 1 次幂出现。目标是利用逆运算将变量孤立在一边。例如，解方程 3x + 5 = 20，两边减去 5 得 3x = 15，然后两边除以 3 得 x = 5。始终对等式两边进行相同的运算以保持相等。
When the equation contains fractions, multiply every term by the lowest common denominator to clear them. For (2x)/3 + 1 = 5, multiply every term by 3: 2x + 3 = 15, so 2x = 12 and x = 6. If there are brackets, expand them first. For example, 2(x + 3) = 14 becomes 2x + 6 = 14, then 2x = 8 and x = 4.
当方程含有分数时，将每一项乘以最小公分母以消除分母。对 (2x)/3 + 1 = 5，每项乘 3：2x + 3 = 15，故 2x = 12，x = 6。若有括号，先展开。例如，2(x + 3) = 14 变为 2x + 6 = 14，然后 2x = 8，x = 4。
5. Solving Quadratic Equations | 解二次方程
A quadratic equation takes the form ax² + bx + c = 0, with a ≠ 0. There are three main methods: factorising, completing the square, and using the quadratic formula. If the quadratic factorises easily, set each factor equal to zero. For instance, solve x² – 5x + 6 = 0 which factorises to (x – 2)(x – 3) = 0, giving solutions x = 2 and x = 3.
二次方程的形式为 ax² + bx + c = 0，其中 a ≠ 0。主要有三种解法：因式分解法、配方法和二次公式法。若二次式容易分解，令每个因子等于零。例如，解 x² – 5x + 6 = 0，分解为 (x – 2)(x – 3) = 0，得解 x = 2 和 x = 3。
When factorising is not straightforward, use the quadratic formula:
x = (-b ± √(b² – 4ac)) / 2a
Remember to express solutions in exact form unless asked otherwise. The discriminant b² – 4ac tells you about the nature of the roots: if positive, two distinct real roots; if zero, one repeated root; if negative, no real roots. Completing the square is especially useful for sketching graphs and finding the vertex.
当因式分解不易时，使用二次公式。除非题目另有要求，解应表示为精确形式。判别式 b² – 4ac 可说明根的性质：若为正，有两个不等实根；若为零，有一个重根；若为负，无实根。配方法对于绘制图像和求顶点十分有用。
6. Simultaneous Equations | 联立方程
Simultaneous equations are sets of equations with multiple unknowns. The elimination method involves adding or subtracting equations to eliminate one variable. For the system 2x + y = 10 and x – y = 2, adding gives 3x = 12, so x = 4, and substitute back to find y = 2. You can also scale equations before elimination to match coefficients.
联立方程是含有多个未知数的方程组。消元法通过将方程相加或相减来消去一个变量。对于方程组 2x + y = 10 和 x – y = 2，相加得 3x = 12，故 x = 4，代回求得 y = 2。你也可以在消元前对方程进行缩放以匹配系数。
The substitution method is ideal when one equation is easily rearranged for a variable. For instance, y = 3x – 1 and 2x + 3y = 16: substitute the first into the second to get 2x + 3(3x – 1) = 16, solve for x, then y. When one equation is linear and the other quadratic, substitute the linear expression into the quadratic. This yields a quadratic in one variable, which you then solve, giving up to two solutions.
当其中一个方程容易变形为某一变量的表达式时，代入法非常理想。例如，y = 3x – 1 和 2x + 3y = 16：将第一个代入第二个得 2x + 3(3x – 1) = 16，解出 x 再求 y。当一个方程是线性而另一个是二次方程时，将线性表达式代入二次式，得到关于一个变量的二次方程，求解后最多可得两组解。
7. Inequalities | 不等式
An inequality compares expressions using symbols <, >, ≤ or ≥. Solving linear inequalities follows similar steps to solving equations, but remember: if you multiply or divide both sides by a negative number, you must reverse the inequality sign. For example, solve -2x < 8 by dividing both sides by -2, giving x > -4.
不等式使用 <、>、≤ 或 ≥ 符号比较表达式。解线性不等式的步骤与解方程类似，但要记住：若两边乘以或除以一个负数，必须反转不等式符号。例如，解 -2x < 8，两边除以 -2，得 x > -4。
To represent the solution set on a number line, use an open circle for < or > (value not included) and a closed circle for ≤ or ≥ (value included). For double inequalities like 3 < 2x + 1 ≤ 7, break them into two separate inequalities or solve simultaneously: subtract 1, then divide by 2 to get 1 < x ≤ 3. Quadratic inequalities require sketching the graph or using a sign table; for x² – 4 > 0, the solution is x < -2 or x > 2.
在数轴上表示解集时，用空心圆圈表示 < 或 >（不含该值），用实心圆点表示 ≤ 或 ≥（含该值）。对于 3 < 2x + 1 ≤ 7 这样的双重不等式，可拆分为两个单独的不等式，或同时求解：减 1 再除以 2 得 1 < x ≤ 3。二次不等式需绘制图像或使用符号表；对于 x² – 4 > 0，解为 x < -2 或 x > 2。
8. Sequences and the nth Term | 数列与第 n 项
A sequence is an ordered list of numbers following a rule. An arithmetic sequence has a constant difference between terms. To find the nth term of a linear sequence, use the formula aₙ = a + (n – 1)d, where a is the first term and d is the common difference. For the sequence 5, 8, 11, 14, …, the nth term is 3n + 2, because the difference is 3 and adjusting gives the first term.
数列是按照某种规则排列的一列数。等差数列的相邻两项之差为常数。要求线性数列的第 n 项，使用公式 aₙ = a + (n – 1)d，其中 a 为首项，d 为公差。对于数列 5, 8, 11, 14, …，第 n 项为 3n + 2，因为公差为 3，调整后得首项。
Quadratic sequences have a constant second difference. The nth term takes the form an² + bn + c. To find it, first work out the second difference, halve it to get a, then use simultaneous equations to find b and c. For example, the sequence 3, 6, 13, 24, … has a second difference of 4, so a = 2, and eventually yields 2n² – 3n + 4. Recognising patterns like square numbers and triangle numbers also helps.
二次数列的二阶差分为常数。第 n 项的形式为 an² + bn + c。要求出它，先计算二阶差分，将其除以 2 得到 a，然后用联立方程求出 b 和 c。例如数列 3, 6, 13, 24, … 的二阶差分为 4，故 a = 2，最终得到 2n² – 3n + 4。识别平方数、三角形数等常见模式也有帮助。
9. Introduction to Functions | 函数入门
A function is a rule that maps each input to exactly one output. It is often written as f(x) = … where x is the input. For instance, f(x) = 2x + 3 means that for any x, the output is twice the input plus three. To evaluate a function, substitute the given value: f(4) = 2(4) + 3 = 11.
Published by TutorHao | GCSE Mathematics Revision Series | aleveler.com更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导