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Combining Like Terms | 合并同类项

2026-08-30 05:01 · 本文部分内容由AI辅助创作

📚 Combining Like Terms | 合并同类项

In algebra, one of the most fundamental skills is simplifying expressions by combining like terms. This process reduces the complexity of an expression, making it easier to evaluate, substitute, and solve.

在代数学中,合并同类项是最基本的技能之一。通过合并同类项,我们可以化简代数表达式,使其更易于求值、代入和解方程。


1. What Are Like Terms? | 什么是同类项?

Like terms are terms that have exactly the same variable part — the same variables raised to the same powers. Only the numerical coefficients may differ.

同类项是指变量部分完全相同的项——即所含变量及其指数都相同。只有数字系数可以不同。

In contrast, unlike terms have different variable parts — for example, x and x², or x and y, are unlike terms.

相反,不同类项的变量部分不同——例如 x 与 x²、x 与 y 都是不同类项。


2. Identifying the Parts of a Term | 识别项的组成

A term consists of three main parts: the coefficient (the numerical factor), the variable (such as x, y, or a), and the exponent (the power to which the variable is raised).

一个项由三部分组成:系数(数字因数)、变量(如 x、y 或 a)以及指数(变量的幂次)。

TermCoefficientVariableExponent
6x³6x3
-2y-2y1
44(none)0

A term with no visible coefficient has a coefficient of 1 or -1. For example, x means 1x, and -y means -1y.

如果项中没有明确写出系数,则其系数为 1 或 -1。例如,x 就是 1x,-y 就是 -1y。


3. Collecting Like Terms — Basic Rule | 合并同类项——基本规则

To combine like terms, simply add or subtract their coefficients, keeping the variable part unchanged.

合并同类项时,只需将它们的系数相加或相减,变量部分保持不变。

ax + bx = (a + b)x

For example, 3x + 5x = 8x, and 7y - 2y = 5y.

例如,3x + 5x = 8x,7y - 2y = 5y。

This rule works because multiplication is distributive: (3 + 5)x = 8x.

这个规则之所以成立,是因为乘法分配律:(3 + 5)x = 8x。


4. Step-by-Step Simplification | 逐步化简

When simplifying an expression with several terms, follow these steps:

化简包含多个项的表达式时,请遵循以下步骤:

Example: Simplify 4a + 3b - 2a + b.

示例:化简 4a + 3b - 2a + b。

Group the a-terms and b-terms: (4a - 2a) + (3b + b) = 2a + 4b.

将含 a 的项和含 b 的项分组:(4a - 2a) + (3b + b) = 2a + 4b。


5. Working with Subtraction | 处理减法符号

When subtraction appears, treat the minus sign as part of the coefficient. This prevents sign errors.

当出现减法时,将负号视为系数的一部分,这样可以避免符号错误。

Consider: 6x - 2x - 3x + x

考虑:6x - 2x - 3x + x

6x, -2x, -3x, +x. Now add the coefficients: 6 + (-2) + (-3) + 1 = 2. So the result is 2x.

写出每一项及其符号:6x、-2x、-3x、+x。然后将系数相加:6 + (-2) + (-3) + 1 = 2。因此结果为 2x。

Always be careful with double negatives: 5a - (2a - 3a) requires distributing the negative sign first.

始终注意双重负号:5a - (2a - 3a) 需要先分配负号。


6. Combining Terms with Different Powers | 合并不同幂次的项

Terms with different exponents on the same variable are not like terms and cannot be combined.

同一变量但指数不同的项不是同类项,不能合并。

mbine: x² + 2x² = 3x². The x terms can combine: 3x - x = 2x. The final result is 3x² + 2x.

例如,在表达式 x² + 3x + 2x² - x 中,x² 项可以合并:x² + 2x² = 3x²;x 项可以合并:3x - x = 2x。最终结果为 3x² + 2x。

We cannot combine 3x² and 2x further because they have different powers.

不能再合并 3x² 和 2x,因为它们的幂次不同。


7. Multiplying a Bracket First | 先去括号再合并

Often, a question involves brackets. In such cases, expand the brackets first, then combine like terms.

许多题目包含括号。遇到括号时,首先展开括号,然后再合并同类项。

Example: Simplify 3(2x + 4) + 2(5x - 1).

示例:化简 3(2x + 4) + 2(5x - 1)。

Expand: 3 × 2x + 3 × 4 + 2 × 5x - 2 × 1 = 6x + 12 + 10x - 2.

展开:3 × 2x + 3 × 4 + 2 × 5x - 2 × 1 = 6x + 12 + 10x - 2。

Combine like terms: (6x + 10x) + (12 - 2) = 16x + 10.

合并同类项:(6x + 10x) + (12 - 2) = 16x + 10。


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8. Applying the Distributive Law with Negative Coefficients | 用负系数应用分配律

When a negative number multiplies a bracket, every term inside changes sign.

当负数乘以括号时,括号内的每一项都要变号。

Example: Simplify 4(x - 3) - (2x + 5).

示例:化简 4(x - 3) - (2x + 5)。

Note that -(2x + 5) is equivalent to -1(2x + 5). So:

注意 -(2x + 5) 等价于 -1(2x + 5)。因此:

4x - 12 - 2x - 5 = (4x - 2x) + (-12 - 5) = 2x - 17

This common mistake is often tested in IGCSE exams — forgetting to change the sign of the terms inside a subtracted bracket.

这是 IGCSE 考试中常见的一个易错点——学生往往会忘记改变被减去括号内各项的符号。

9. Combining Like Terms in Linear Equations | 在线性方程中合并同类项

Combining like terms is a crucial step when solving linear equations. It simplifies both sides before applying inverse operations.

解线性方程时,合并同类项是关键步骤。它能在应用逆运算之前简化方程两边。

Example: Solve 5x + 3 + 2x = 24.

示例:解方程 5x + 3 + 2x = 24。

left: 7x + 3 = 24. Then subtract 3 from both sides: 7x = 21. Finally, divide by 7: x = 3.

首先,合并左边的 x 项:7x + 3 = 24。然后两边减去 3:7x = 21。最后两边除以 7:x = 3。

Always check your answer by substituting the value back into the original equation.

务必通过将答案代回原方程来检查结果是否正确。


10. Common Mistakes and How to Avoid Them | 常见错误及如何避免

A good habit is to underline like terms in the same way before adding them together.

一个好习惯是先用同一种标记划出同类项,然后再相加。


11. Worked Exercises | 例题精讲

Exercise 1: Simplify 8m + 3n - 5m + 2n + 4.

练习 1:化简 8m + 3n - 5m + 2n + 4。

erms: 8m - 5m = 3m. Group n-terms: 3n + 2n = 5n. Keep the constant 4. Answer: 3m + 5n + 4.

合并 m 项:8m - 5m = 3m;合并 n 项:3n + 2n = 5n;保留常数 4。答案:3m + 5n + 4。

Exercise 2: Simplify 2(3x + 1) - 3(x - 2) + 5.

练习 2:化简 2(3x + 1) - 3(x - 2) + 5。

each bracket: 6x + 2 - 3x + 6 + 5. Combine: 6x - 3x = 3x; 2 + 6 + 5 = 13. Answer: 3x + 13.

展开每个括号:6x + 2 - 3x + 6 + 5。合并:6x - 3x = 3x;2 + 6 + 5 = 13。答案:3x + 13。

Exercise 3: Solve 4(2x - 1) = 20.

练习 3:解方程 4(2x - 1) = 20。

Expand: 8x - 4 = 20. Add 4 to both sides: 8x = 24. Divide by 8: x = 3.

展开:8x - 4 = 20。两边加 4:8x = 24。两边除以 8:x = 3。


12. Quick Revision Checklist | 快速复习清单

Use this checklist before your exam:

考试前请对照以下清单进行自查:

✔Checklist Item中文要点
✔I can identify like terms correctly.我能正确识别同类项。
✔I can combine coefficients with correct signs.我能带着正确符号合并系数。
✔I can expand brackets before collecting terms.我会先展开括号再合并同类项。
✔I can apply this skill to solve linear equations.我能运用该技能解线性方程。

Mastering the skill of combining like terms is essential for success in IGCSE Mathematics. It is not only tested directly in questions but also forms the foundation for solving equations, simplifying algebraic fractions, and exploring functions.

掌握合并同类项的技能对于 IGCSE 数学的成功至关重要。它不仅会直接出现在考题中,也是解方程、化简代数分式和探索函数的基础。

Keep practising — algebra becomes easier and more intuitive with consistent effort.

坚持练习——只要勤奋不断,代数就会变得更容易、更熟练。


Published by TutorHao | Mathematics Revision Series | aleveler.com

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