📚 Mastering Algebraic Expressions and Linear Equations: KS3 Cambridge p104_2 Skills | 掌握代数式与线性方程：KS3 剑桥 p104_2 核心技能
olidates the key methods, worked examples and common mistakes so you can approach p104_2-style questions with confidence.
在许多 KS3 剑桥数学教材中，第 104 页第 2 题考查的是化简代数式、展开括号以及解一元一次方程的能力。这些技能是代数的核心，贯穿剑桥初中数学的整个学习过程。本复习指南总结了关键方法、典型例题和常见错误，帮助你自信应对 p104_2 类型的题目。
1. Understanding Variables, Terms and Coefficients | 理解变量、项与系数
In algebra, letters such as x, y or n stand for unknown numbers or quantities that can change. These letters are called variables. An algebraic expression combines numbers, variables and operations. In the expression 5x + 3y - 7, the terms are 5x, 3y and -7; the coefficients are 5 and 3; and -7 is the constant term.
在代数中，x、y 或 n 等字母表示未知数或可以变化的量，这些字母称为变量。代数式由数字、变量和运算组合而成。在表达式 5x + 3y - 7 中，项分别是 5x、3y 和 -7；系数是 5 和 3；-7 是常数项。
A term is a single number, a variable, or a product of numbers and variables. For example, 4x, -2y, 9 and x are all terms. Identifying terms correctly helps you decide which parts can be combined when simplifying an expression.
项是一个单独的数字、变量或数字与变量的乘积。例如 4x、-2y、9 和 x 都是项。正确识别项有助于你在化简表达式时判断哪些部分可以合并。
- Term: a single number, variable or product (项：单个数字、变量或乘积)
- Coefficient: the number factor in a term (系数：项中的数字因数)
- Constant: a term with no variable (常数项：不含变量的项)
2. Collecting Like Terms | 合并同类项
Like terms contain exactly the same variable part, such as 3x and 7x, or 2y and -5y. You can add or subtract like terms by combining their coefficients. Unlike terms, such as 3x and 4y, cannot be combined into a single term.
同类项含有完全相同的变量部分，例如 3x 和 7x，或 2y 和 -5y。你可以通过合并系数来加减同类项。不同类项，例如 3x 和 4y，不能合并为一个项。
he result is 6x. In a longer expression, group like terms first, then simplify each group.
例如，化简 3x + 5x - 2x 时，先把系数相加：3 + 5 - 2 = 6，因此结果是 6x。在较长的表达式中，先分组同类项，再分别化简每一组。
4a + 3b - 2a + 5b = (4a - 2a) + (3b + 5b) = 2a + 8b
Always keep the sign in front of each term with the term itself. This prevents mistakes when rearranging terms.
始终把每一项前面的符号与该连在一起。这样在重新排列各项时可以避免错误。
3. Expanding Brackets: The Distributive Law | 展开括号：分配律
c) = ab + ac. For example, 3(x + 4) expands to 3x + 12 because 3 x x = 3x and 3 x 4 = 12.
展开括号就是用括号外面的项乘以括号内的每一项。这使用了分配律：a(b + c) = ab + ac。例如，3(x + 4) 展开后得到 3x + 12，因为 3 x x = 3x，3 x 4 = 12。
term, be careful with signs: -2(x - 5) = -2x + 10 because -2 x x = -2x and -2 x -5 = +10.
当括号前是负数时，需要特别注意符号：-2(x - 5) = -2x + 10，因为 -2 x x = -2x，而 -2 x -5 = +10。
2(3y - 4) + 5 = 6y - 8 + 5 = 6y - 3
After expanding, collect any like terms to simplify the expression fully. This is often required in KS3 Cambridge p104-style questions.
展开后，合并所有同类项以完全化简表达式。这通常是 KS3 剑桥 p104 类型题目中要求的步骤。
4. Factorising Expressions | 因式分解
Factorising is the reverse of expanding. You rewrite an expression as a product of a common factor and a bracket. Find the highest common factor of the terms, write it outside the bracket, and divide each term by that factor inside the bracket.
因式分解是展开的逆运算。你将一个表达式改写为一个公因式与括号的乘积。先找出各项的最高公因式，把它写在括号外，再将括号内每一项除以该公因式。
or of 6 and 9 is 3, so 6x + 9 = 3(2x + 3). You can check by expanding: 3(2x + 3) = 6x + 9.
例如，对 6x + 9 进行因式分解。6 和 9 的最高公因数是 3，所以 6x + 9 = 3(2x + 3)。你可以通过展开来检查：3(2x + 3) = 6x + 9。
8x² + 4x = 4x(2x + 1)
x = 2x and 4x ÷ 4x = 1. Factorising is important for solving more complex equations later.
这里两项的最高公因式是 4x，因为 8x² ÷ 4x = 2x，4x ÷ 4x = 1。因式分解对以后解决更复杂的方程非常重要。
5. Solving One-Step Equations | 解一步方程
A linear equation contains a variable raised only to the power of 1. To solve a one-step equation, perform the inverse operation on both sides of the equation to isolate the variable. This keeps the equation balanced.
一元一次方程中的变量指数仅为 1。解一步方程时，在方程两边执行逆运算以分离变量。这可以保持方程平衡。
- 7 = 15 - 7, so x = 8. For the equation 3x = 21, divide both sides by 3: x = 21 ÷ 3 = 7.
例如，解 x + 7 = 15。两边同时减去 7：x + 7 - 7 = 15 - 7，所以 x = 8。对于方程 3x = 21，两边同时除以 3：x = 21 ÷ 3 = 7。
x - 4 = 10 → x = 10 + 4 = 14
Always write each step clearly and keep the equals signs aligned. This helps you avoid arithmetic slips and makes your method easy to check in an exam.
始终清晰地书写每一步，并保持等号对齐。这有助于避免计算错误，也便于考试时检查解题过程。
6. Solving Two-Step Equations | 解两步方程
Two-step equations involve two operations on the variable, such as multiplication and addition. Undo the operations in reverse order: first solve any addition or subtraction, then solve any multiplication or division.
两步方程中的变量涉及两种运算，例如乘法和加法。按相反顺序逐步解除运算：先处理加法或减法，再处理乘法或除法。
by 2 to get x = 6. Always check by substituting the value back into the original equation.
解方程 2x + 5 = 17。第一步：两边减去 5，得到 2x = 12。第二步：两边除以 2，得到 x = 6。始终将值代回原方程进行检查。
x/3 - 2 = 4 → x/3 = 6 → x = 18
variable, multiply or divide by -1 as the final step. For example, if -x = 7, then x = -7.
对于变量为负数的方程，最后一步乘以或除以 -1。例如，如果 -x = 7，那么 x = -7。
7. Solving Equations with Brackets | 解含括号的方程
ten appears in p104_2-style questions where you must combine expanding and solving skills.
当方程含有括号时，首先展开括号，然后化简两边，再分离变量。这类题目经常出现在 p104_2 类型的练习中，需要综合运用展开和求解技能。
and the left side: 3x + 12 = 27. Subtract 12 from both sides: 3x = 15. Divide by 3: x = 5.
解方程 3(x + 4) = 27。展开左边：3x + 12 = 27。两边减去 12：3x = 15。除以 3：x = 5。
2(x - 5) = x + 3 → 2x - 10 = x + 3 → x = 13
After expanding, collect like terms on each side. If the variable appears on both sides, move the smaller variable term first to keep the coefficient positive.
展开后，先合并各自边的同类项。如果两边都含有变量，先移走较小的变量项，以使系数保持为正。
8. Writing Equations from Word Problems | 根据应用题建立方程
Word problems require you to translate a real-life situation into an algebraic equation. Start by defining the unknown with a variable, then build an equation using the given relationships and operations.
应用题要求你将实际情境转化为代数方程。先设未知数为变量，再利用题目给出的关系和运算建立方程。
more, and he now has 20. Write and solve the equation. Translation: x + 6 = 20, so x = 14.
例如：Sam 有 x 颗糖。Emma 又给了他 6 颗，现在他有 20 颗。写出并求解方程。转化：x + 6 = 20，因此 x = 14。
ngth 8 cm and width w cm, and perimeter 26 cm, then 2(8 + w) = 26. Solving gives w = 5 cm.
对于周长问题，使用公式并代入已知值。如果一个矩形的长为 8 cm，宽为 w cm，周长为 26 cm，则 2(8 + w) = 26。解得 w = 5 cm。
Always define the variable clearly and answer the original question. Include units if the problem involves measurements.
始终明确定义变量并回答原题所问。如果题目涉及测量，答案中要包含单位。
9. Common Errors and How to Avoid Them | 常见错误与如何避免
One frequent mistake is forgetting to multiply all terms inside a bracket. For example, 2(x + 3) is sometimes written incorrectly as 2x + 3 instead of 2x + 6. Check that every term inside the bracket has been multiplied by the outside term.
一个常见错误是忘记乘以括号内的所有项。例如，2(x + 3) 有时被错误地写成 2x + 3，而正确答案是 2x + 6。请检查括号内每一项是否都乘以外面的项。
= 0, not 5x - 5x = 0 incorrectly skipping the signs. Keep every sign attached to its term.
另一个错误是合并同类项时丢失负号。表达式 5x - 2x - 3x 应化简为 (5 - 2 - 3)x = 0x = 0，而不是错误地忽略符号。要把每个符号与对应项连在一起。
In equations, avoid applying operations to only one side. Whatever you do to one side, you must do to the other. For example, if you subtract 4 from the left, subtract 4 from the right as well.
在方程中，避免只对一边进行运算。对一边做什么，对另一边也必须做同样的运算。例如，如果左边减去 4，右边也要减去 4。
| Common error | Correct method |
| 3(y - 2) = 3y - 2 | 3(y - 2) = 3y - 6 |
| -2(x + 4) = -2x + 8 | -2(x + 4) = -2x - 8 |
| x + 5 = 12 → x = 17 | x + 5 = 12 → x = 7 |
10. Exam-Style Practice Questions | 考试题型练习
idge p104_2 tasks. Complete all steps, then compare your answers with the solutions below.
尝试以下三道与 KS3 剑桥 p104_2 题型结构相似的练习题。写出完整步骤，然后与下面的答案对照。
Question 1: Simplify 7x - 3y + 2x + 5y - 4x.
问题 1：化简 7x - 3y + 2x + 5y - 4x。
Question 2: Solve 5(n - 3) + 2 = 27.
问题 2：解方程 5(n - 3) + 2 = 27。
Question 3: A rectangle has length (x + 6) cm and width 4 cm. Its perimeter is 32 cm. Form an equation and find x.
问题 3：一个矩形的长为 (x + 6) cm，宽为 4 cm，周长为 32 cm。建立方程并求 x。
Solutions:
答案：
1. 7x - 3y + 2x + 5y - 4x = (7x + 2x - 4x) + (-3y + 5y) = 5x + 2y
2. 5(n - 3) + 2 = 27 → 5n - 15 + 2 = 27 → 5n - 13 = 27 → 5n = 40 → n = 8
3. 2((x + 6) + 4) = 32 → 2(x + 10) = 32 → x + 10 = 16 → x = 6
Regular practice of expanding, factorising and solving equations will build speed and accuracy for Cambridge KS3 assessments. Review each step and always check your final answer by substitution.
经常练习展开、因式分解和解方程，有助于提高剑桥 KS3 考试的解题速度和准确率。复习每一步，并始终通过代入检验最终答案。
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