📚 Collecting Like Terms in IGCSE Algebra | IGCSE 代数：合并同类项
Collecting like terms is the process of simplifying an algebraic expression by adding or subtracting terms that have exactly the same variable parts. It is the first real algebra skill that IGCSE students meet, and it never disappears: you use it again when you solve equations, expand brackets, factorise, rearrange formulae and work with sequences. Getting it right at the start makes every later topic easier and faster.
合并同类项是通过对变量部分完全相同的项进行加减来化简代数表达式的过程。这是 IGCSE 学生接触到的第一个真正的代数技能，而且它从未消失：解方程、展开括号、因式分解、公式变形和处理数列时你都会用到它。一开始就把它做对，会让后面每一个知识点都更简单、更迅速。
In this guide we build the skill from the definition of a term all the way up to exam-style questions, with the signs, powers, brackets and fractions that Cambridge 0580 / 0980 and Edexcel 4MA1 papers actually test.
在本指南中，我们从"项"的定义出发，一路讲到考试真题风格的问题，涵盖剑桥 0580 / 0980 与爱德思 4MA1 试卷真正会考的正负号、幂、括号和分数。掌握这个主题有助于你写出最简形式的答案，并减少在较长的代数题目中出错的可能性。
1. What Are Like Terms? | 什么是同类项
An algebraic term is a single number, a single variable, or a product of numbers and variables. A term is never separated by a plus or minus sign in the middle. Terms are called like terms when they have exactly the same variable factors raised to exactly the same powers. The coefficients, that is the numbers written in front, may be completely different.
代数项是一个单独的数、一个单独的变量，或数与变量的乘积。项的内部不会被加号或减号分开。如果项具有完全相同的变量因式并且次数完全相同，就称为同类项。系数，也就是写在最前面的数，完全可以不同。
- 3x and -7x are like terms — 3x 与 -7x 是同类项。
- 4x² and 2x² are like terms — 4x² 与 2x² 是同类项。
- 5xy and -2xy are like terms — 5xy 与 -2xy 是同类项。
- 8 and -3 are like terms, because both are constants — 8 与 -3 是同类项，因为两者都是常数项。
- 3x and 3y are not like terms — 3x 与 3y 不是同类项。
- x² and x are not like terms, because the powers differ — x² 与 x 不是同类项，因为次数不同。
The order of the letters does not matter: xy and yx contain the same variables with the same powers, so they are like terms. Only the powers matter, not the order of writing.
字母的顺序无关紧要：xy 与 yx 含有相同的变量且次数相同，因此它们是同类项。只有次数重要，书写顺序不重要。
2. Terms, Coefficients and the Variable Part | 项、系数与变量部分
Every term can be split into two parts. The coefficient is the numerical factor, and the variable part is everything else, including the powers. For example, in the term -5x²y the coefficient is -5 and the variable part is x²y. If a term looks like x, the coefficient is understood to be 1; if it looks like -x, the coefficient is -1.
每一项都可以拆成两部分。系数是数字因式，变量部分是其余的一切，包括幂。例如在项 -5x²y 中，系数是 -5，变量部分是 x²y。如果一项写成 x，系数就默认为 1；如果写成 -x，系数就是 -1。
Two terms can only be collected when their variable parts match exactly, letter by letter and power by power. This single idea decides every question in the topic.
只有当两项的变量部分完全一致——字母相同、次数也相同——它们才能被合并。这唯一的一条原则决定了本主题中的每一道题。
- Term 7ab → coefficient 7, variable part ab — 项 7ab → 系数 7，变量部分 ab。
- Term -x² → coefficient -1, variable part x² — 项 -x² → 系数 -1，变量部分 x²。
- Term 12 → coefficient 12, no variable part — 项 12 → 系数 12，没有变量部分。
3. The Golden Rule of Collecting Like Terms | 合并同类项的基本法则
To collect like terms, add or subtract the coefficients and keep the variable part unchanged. The variable part is the "label" of the group: it tells you which terms belong together, and it must appear in the answer exactly as it appeared in the question.
合并同类项的方法是：把系数相加或相减，变量部分保持不变。变量部分是这个组的"标签"：它告诉你哪些项属于同一组，并且在答案中必须与题目中出现时完全一致。
3x + 5x = (3 + 5)x = 8x
ause of the distributive law read backwards: 3x + 5x = x(3 + 5) = 8x. You are not adding x's together as objects; you are counting how many x's you have.
这是成立的，因为分配律反过来读就是：3x + 5x = x(3 + 5) = 8x。你并不是把 x 当作物品相加，而是数一数你一共有多少个 x。
7a − 2a = (7 − 2)a = 5a
4x² + 2x² = (4 + 2)x² = 6x²
4. Handling Positive and Negative Signs | 正负号的处理
The most common source of lost marks is the sign in front of a term. A plus or minus sign belongs to the term that follows it, not to the term before it. In the expression 9y − 4y + y, the terms are +9y, −4y and +y, so the coefficients are 9, −4 and 1.
失分最常见的原因就是项前面的符号。加号或减号属于它后面的那一项，而不是前面的那一项。在表达式 9y − 4y + y 中，各项是 +9y、−4y 和 +y，所以系数是 9、−4 和 1。
- 9y − 4y + y = 6y — 9y − 4y + y = 6y。
- −3m − 5m = −8m, because the two negative coefficients add — −3m − 5m = −8m，因为两个负系数相加。
- 2p − 7p = −5p, and the answer may be negative — 2p − 7p = −5p，答案可以是负数。
- 6 − 10 = −4 for constant terms — 对常数项而言 6 − 10 = −4。
If a term has no sign written in front, it is positive. If a term is written as just a minus sign followed by a letter, its coefficient is −1.
如果一项前面没有写符号，它就是正的。如果一项只写了一个减号后面跟字母，那么它的系数就是 −1。
5. Expressions with More Than One Variable | 含多个变量的表达式
When an expression contains several different variables, sort the terms into groups first, then collect each group separately. A tidy method is to underline or tick each family in a different way, or to physically rewrite the expression with the groups next to each other.
当一个表达式含有几个不同的变量时，先把各项分组，再分别合并每一组。一个整洁的方法是：用不同的方式给每个"家族"划线或打勾，或者干脆把表达式重写，让同组的项挨在一起。
x + 5y. The x terms are 3x − x = 2x. The y terms are 2y + 5y = 7y. The answer is 2x + 7y.
化简：3x + 2y − x + 5y。含 x 的项是 3x − x = 2x。含 y 的项是 2y + 5y = 7y。答案是 2x + 7y。
- 6ab − 2ba = 4ab, because ab and ba are the same variable part — 6ab − 2ba = 4ab，因为 ab 与 ba 的变量部分相同。
- 2x²y and 3xy² can never be collected, since the powers on x and y are swapped — 2x²y 与 3xy² 永远不能合并，因为 x 与 y 上的次数被调换了。
- 5mn + 4nm − 9mn = 0, and a zero result is a perfectly good answer — 5mn + 4nm − 9mn = 0，结果为零是完全合理的答案。
6. Powers and Indices: x, x² and x³ | 幂与指数：x、x² 与 x³
Powers are part of the variable part, so terms with different powers are not like terms. x and x² are as different as apples and oranges; x² and x³ are also different. Only identical powers may be collected.
幂是变量部分的一部分，所以次数不同的项不是同类项。x 和 x² 就像苹果和橘子一样不同；x² 和 x³ 也是不同的。只有完全相同的次数才能合并。
x² + x² = 2x² but x² + x³ cannot be simplified
wers: x² × x³ = x⁵. When you collect, you add the coefficients and leave the powers alone.
注意不要把合并同类项与乘法的指数法则混为一谈。做乘法时，你把次数相加：x² × x³ = x⁵。做合并时，你把系数相加，次数保持不变。
- 5x² + 3x² − x² = 7x² — 5x² + 3x² − x² = 7x²。
- 4x³ + 2x² + x³ = 5x³ + 2x², and it stops there — 4x³ + 2x² + x³ = 5x³ + 2x²，到此为止不能再化。
- 3x × 4x = 12x², which is a product, not a collection — 3x × 4x = 12x²，这是乘积，不是合并。
7. Collecting Constant Terms | 合并常数项
Numbers without variables are constants, and all constants are like terms with each other. Keep them in their own group and combine them at the end. A common exam instruction is "simplify", which means the final answer should have no repeated variable parts and no uncombined numbers.
不含变量的数是常数项，而所有常数项彼此都是同类项。把它们单独成组，最后再合并。考试中常见的指令是 "simplify"（化简），意思是最终答案中不应有重复的变量部分，也不应有未合并的数字。
3x + 5 + 2x − 8. The x terms give 5x. The constants give 5 − 8 = −3. The answer is 5x − 3.
化简：3x + 5 + 2x − 8。含 x 的项得到 5x。常数项得到 5 − 8 = −3。答案是 5x − 3。
- 7 + 4a − 2 + a = 5a + 5 — 7 + 4a − 2 + a = 5a + 5。
- 2x² − 6 + x² + 10 = 3x² + 4 — 2x² − 6 + x² + 10 = 3x² + 4。
- Written as 5 + 5a or 5a + 5, both are accepted, but a consistent order looks neater — 写成 5 + 5a 或 5a + 5 都被接受，但保持统一顺序会更整齐。
8. Collecting Like Terms After Expanding Brackets | 展开括号后合并同类项
Many IGCSE questions combine two skills: expand first, then collect. Multiply out every bracket carefully, watching the sign in front of each bracket, and only then group the like terms.
许多 IGCSE 题目把两个技能组合在一起：先展开，再合并。仔细地把每个括号乘开，注意每个括号前面的符号，然后才分组合并同类项。
Example: 2(x + 3) + 3(x − 1). Expanding gives 2x + 6 + 3x − 3. Collecting gives 5x + 3.
例如：2(x + 3) + 3(x − 1)。展开得到 2x + 6 + 3x − 3。合并得到 5x + 3。
Example with a minus sign in front: 4(x + 2) − (x + 5). Expanding gives 4x + 8 − x − 5, because the minus sign flips both terms inside the second bracket. Collecting gives 3x + 3.
前面带减号的例子：4(x + 2) − (x + 5)。展开得到 4x + 8 − x − 5，因为减号会把第二个括号里的两项都变号。合并得到 3x + 3。
- 3(2x − 1) + 4(x + 5) = 6x − 3 + 4x + 20 = 10x + 17 — 3(2x − 1) + 4(x + 5) = 6x − 3 + 4x + 20 = 10x + 17。
- 5(2x − 3) − 2(4x − 7) = 10x − 15 − 8x + 14 = 2x − 1 — 5(2x − 3) − 2(4x − 7) = 10x − 15 − 8x + 14 = 2x − 1。
- x(x + 4) − x² = x² + 4x − x² = 4x — x(x + 4) − x² = x² + 4x − x² = 4x。
9. Fractional and Decimal Coefficients | 分数与小数系数
Coefficients do not have to be whole numbers. Fractions and decimals are collected in exactly the same way: add or subtract the coefficients, keep the variable part. With fractions, use a common denominator; with decimals, line up the decimal points.
系数不一定是整数。分数和小数的合并方式完全相同：系数相加或相减，变量部分保持不变。遇到分数就通分；遇到小数就把小数点对齐。
½x + ⅓x = ⅚x and 0.5x + 1.25x = 1.75x
For mixed forms it is usually safest to convert everything to fractions or everything to decimals before you start, so that you do not make a comparison error halfway through.
遇到混合形式时，最安全的做法通常是在开始之前把所有数都换成分数，或全部换成小数，以免在中途出现比较错误。
- ¾y − ¼y = ½y — ¾y − ¼y = ½y。
- 2.4a + 3.6a − a = 5a — 2.4a + 3.6a − a = 5a。
- ⅔m + ⅙m = ⅚m, using denominator 6 — ⅔m + ⅙m = ⅚m，取分母 6。
10. Common Mistakes to Avoid | 常见错误
Examiners report the same handful of errors year after year. Knowing them in advance is the cheapest way to protect your marks, because most of these mistakes can be spotted by reading your own answer back.
考官年复一年地报告同样几种错误。提前了解它们，是保住分数成本最低的方式，因为其中大多数错误只要回头读一遍自己的答案就能发现。
- Adding unlike terms, e.g. writing 3x + 2y = 5xy — 把非同类项相加，例如写成 3x + 2y = 5xy。
- Changing the power when collecting, e.g. writing 3x + 4x = 7x² — 合并时改变了次数，例如写成 3x + 4x = 7x²。
- Dropping the sign of a negative term, e.g. writing 5a − 8a = 3a — 丢掉负项的符号，例如写成 5a − 8a = 3a。
- Forgetting to change both signs when subtracting a bracket, e.g. −(x − 4) written as −x − 4 — 减去括号时忘记两项都变号，例如把 −(x − 4) 写成 −x − 4。
- Leaving the answer unfinished, e.g. stopping at 4x + 2 + 3x instead of 7x + 2 — 答案没有化到底，例如停在 4x + 2 + 3x 而不是 7x + 2。
- Multiplying instead of adding, e.g. writing 2p + 3p = 6p — 把相加做成了相乘，例如写成 2p + 3p = 6p。
11. Worked Exam-Style Examples | 考试风格例题精讲
The following examples show the working standard expected in an IGCSE answer. Write one line per step: identify, group, collect, and state the simplified expression clearly on its own line.
下面的例题展示了 IGCSE 答案所要求的书写标准。每一步写一行：识别、分组、合并，并把化简后的表达式单独清晰地写在最后一行。
| Question 题目 | Working 过程 | Answer 答案 |
| Simplify 8x + 3 − 5x + 7 | x terms: 8x − 5x = 3x; constants: 3 + 7 = 10 | 3x + 10 |
| Simplify 4a² + 3a − a² + 6a | a² terms: 4a² − a² = 3a²; a terms: 3a + 6a = 9a | 3a² + 9a |
| Simplify 3(2x − 1) + 4(x + 5) | Expand: 6x − 3 + 4x + 20; collect: 10x + 17 | 10x + 17 |
| Simplify ½(4x − 6) + 3(x + 1) | Expand: 2x − 3 + 3x + 3; collect: 5x + 0 | 5x |
| Simplify 7mn − 4nm + 2mn | All three share variable part mn: 7 − 4 + 2 = 5 | 5mn |
Notice how the final answer never contains two terms with the same variable part, and never contains a coefficient of 0 written out. An answer of 5x is cleaner than 5x + 0x.
注意最终答案中绝不会出现两个变量部分相同的项，也绝不会把系数 0 写出来。答案 5x 比 5x + 0x 更简洁。
12. Exam Tips and Practice Questions | 应试技巧与练习
In a written paper, show one clear line of working before the final answer, because method marks are awarded separately from accuracy marks. If a question says "simplify fully" or "write in its simplest form", any uncollected term left behind can cost the final mark.
在笔试中，请在最终答案之前写出一行清晰的过程，因为方法分与准确分是分开给分的。如果题目写着 "simplify fully" 或 "write in its simplest form"（化到最简），任何未合并完的项都可能让你丢掉最后一分。
- Always scan the expression once and mark each group before you calculate — 计算前先扫一遍表达式，给每一组做上标记。
- Underline the sign together with the term it belongs to — 把符号和它所属的项一起划线。
- Check whether brackets must be expanded first — 检查是否需要先展开括号。
- Keep the variable part identical to the question, including letter order if it helps you — 变量部分与题目保持一致，如果有助于判断就保留字母顺序。
- Do a quick numerical check: let x = 2 and test your simplified answer against the original — 做一次快速数值检验：令 x = 2，用化简后的答案与原始表达式对照。
Practice set: simplify each of the following. (1) 7x + 3x − 2x. (2) 5a − 8a + a. (3) 4x² + 3x − x² + 2x. (4) 3(2x − 1) + 4(x + 5). (5) 6ab − 2ba.
练习题：化简下列各式。(1) 7x + 3x − 2x。(2) 5a − 8a + a。(3) 4x² + 3x − x² + 2x。(4) 3(2x − 1) + 4(x + 5)。(5) 6ab − 2ba。
Answers: (1) 8x. (2) −2a. (3) 3x² + 5x. (4) 10x + 17. (5) 4ab. If you reached all five, you have the core of collecting like terms under control and you are ready to use it inside equations, inequalities and formulae.
答案：(1) 8x。(2) −2a。(3) 3x² + 5x。(4) 10x + 17。(5) 4ab。如果这五题你都做对了，说明你已经掌握了合并同类项的核心，可以把它用到方程、不等式和公式中去了。
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