Page 118 Exercise 1: Solving Linear Equations with Brackets | 第118页练习1：解含括号的线性方程

📚 Page 118 Exercise 1: Solving Linear Equations with Brackets | 第118页练习1：解含括号的线性方程

Welcome to this targeted revision guide based on Exercise 1 from page 118 of your Cambridge KS3 Mathematics textbook. This exercise challenges you to solve linear equations that involve brackets, fractions, and variables on both sides. Mastering these problems will build a rock-solid foundation for IGCSE and beyond. In this article, we break down every concept, step by step, so you can tackle similar questions with confidence and accuracy.

欢迎阅读这篇基于剑桥 KS3 数学教材第118页练习1的针对性复习指南。该练习要求你求解含有括号、分数和两边都有变量的线性方程。掌握这些题目将为 IGCSE 及更高层次的数学打下坚实基础。本文会将每个概念逐层拆解，让你能信心十足、准确无误地应对同类题目。

1. What Is the Exercise About? | 练习内容概览

Exercise 1 on page 118 presents a series of linear equations designed to test your understanding of algebraic manipulation. You will find simple one-step equations, two-step equations, and more complex examples where you must expand brackets before isolating the variable. Some questions also include fractions and negative coefficients, which are typical of the Cambridge KS3 syllabus.

第118页练习1列出了一系列线性方程，旨在检验你的代数运算能力。你会遇到简单的一步方程、两步方程，以及需要先展开括号再分离变量的复杂例子。部分题目还包含分数和负系数，这些正是剑桥 KS3 大纲的典型内容。

The key skills tested here are: using the distributive property, collecting like terms, balancing equations, and checking solutions. By the end of this guide, you will have a clear strategy for every equation type.

这里考查的核心技能包括：运用分配律、合并同类项、保持等式平衡以及验证解。读完本指南，你将对每一种方程类型形成清晰的解题策略。

2. Core Rule: Keep the Equation Balanced | 核心法则：保持等式平衡

Every linear equation is like a set of scales. If you add, subtract, multiply, or divide one side by a number, you must do exactly the same to the other side. This golden rule ensures that the equation remains true while you work towards isolating the unknown variable, often denoted by x.

每个线性方程都像一架天平。如果你在等式的一边进行加、减、乘、除，就必须在另一边进行完全相同的操作。这条黄金法则能够确保在分离未知数（通常用 x 表示）的过程中等式始终成立。

For example, to solve x + 5 = 12, subtract 5 from both sides: x = 7. The same principle applies when brackets or fractions are involved, you simply need to take one logical step at a time.

例如，要解 x + 5 = 12，两边同时减去 5 即可得到 x = 7。同样的原则也适用于包含括号或分数的方程，你只需要有计划地逐次操作。

3. Tackling One-Step and Two-Step Equations | 处理一步方程和两步方程

The simplest equations in Exercise 1 require only one operation. For instance, x/4 = 9 is solved by multiplying both sides by 4, giving x = 36. If the equation is x – 7 = 2, add 7 to both sides to obtain x = 9. These straightforward examples remind you of the inverse operations: addition reverses subtraction, multiplication reverses division, and vice versa.

练习1中最简单的方程只需一步运算。例如，x/4 = 9 可通过两边同时乘以 4 求解，得到 x = 36。如果是 x – 7 = 2，两边同时加 7 便得 x = 9。这些直截了当的例子帮助你回忆逆运算：加法与减法互为逆运算，乘法与除法互为逆运算。

Two-step equations, such as 2x + 3 = 11, need two moves: first subtract 3 from both sides (2x = 8), then divide by 2 (x = 4). Always perform addition/subtraction before multiplication/division when peeling away layers around the variable.

像 2x + 3 = 11 这样的两步方程需要两次操作：先从两边减去 3 得到 2x = 8，再除以 2 得出 x = 4。在逐层剥离变量周围的运算时，请始终先做加减，再做乘除。

4. Expanding Brackets Using the Distributive Property | 运用分配律展开括号

Many equations on page 118 feature brackets, such as 3(x + 4) = 21. The distributive property tells us to multiply the term outside the bracket by each term inside: 3 × x and 3 × 4, giving 3x + 12 = 21. After expansion, the equation becomes a standard two-step problem: subtract 12 from both sides (3x = 9), then divide by 3 to find x = 3.

第118页上的许多方程含有括号，比如 3(x + 4) = 21。分配律告诉我们，用括号外的项乘以括号内的每一项：3 × x 和 3 × 4，得到 3x + 12 = 21。展开后，方程就变成一个标准的两步问题：两边减12得出 3x = 9，再除以3求得 x = 3。

3(x + 4) = 21 → 3x + 12 = 21 → x = 3

Be careful when a minus sign appears, for example 2(x – 5) = 8 expands to 2x – 10 = 8. Adding 10 to both sides yields 2x = 18, so x = 9. Always write the expanded terms with their correct signs.

当出现减号时要格外小心，例如 2(x – 5) = 8 展开后变为 2x – 10 = 8。两边加10得到 2x = 18，所以 x = 9。务必正确书写展开后各项的符号。

5. Dealing with Negative Coefficients | 处理负系数

Some equations in the exercise include a negative coefficient outside the bracket, like -2(x – 3) = 10. Expand carefully: -2 × x = -2x, and -2 × (-3) = +6, giving -2x + 6 = 10. Subtract 6 from both sides to get -2x = 4, then divide by -2 to obtain x = -2. Remember that dividing two negative numbers gives a positive result.

练习中的部分方程在括号外带有负系数，例如 -2(x – 3) = 10。展开时需细心：-2 × x = -2x，-2 × (-3) = +6，得到 -2x + 6 = 10。两边减去6得 -2x = 4，再除以 -2 得到 x = -2。请记住，两个负数相除结果为正。

If the negative sign is inside the bracket, like 4(3 – x) = 20, expand to 12 – 4x = 20. Subtract 12 from both sides: -4x = 8, then divide by -4 to find x = -2. Keeping track of signs is a key accuracy skill for Cambridge assessments.

如果负号在括号内部，例如 4(3 – x) = 20，展开得到 12 – 4x = 20。两边减12：-4x = 8，再除以 -4 得 x = -2。准确记录符号是剑桥评估中的一项关键技能。

6. Collecting Like Terms Before Solving | 先合并同类项再求解

Some equations require you to collect like terms before expanding or balancing. For example, 2(x + 1) + 3x = 13 first expands to 2x + 2 + 3x = 13. Then combine the x-terms: 5x + 2 = 13. Subtract 2: 5x = 11, so x = 11/5 or 2.2. This approach stops you from making careless errors with multiple x-terms.

部分方程需要你先合并同类项，再进行展开或平衡。比如 2(x + 1) + 3x = 13 首先展开为 2x + 2 + 3x = 13，接着合并含 x 的项：5x + 2 = 13。减去2：5x = 11，因此 x = 11/5 或 2.2。这种方法能防止在出现多个含 x 项时粗心犯错。

When terms appear on both sides, such as 5x – 3 = 2x + 9, the aim is to gather variables on one side and constants on the other. Subtract 2x from both sides: 3x – 3 = 9. Add 3: 3x = 12, so x = 4. Always choose the side that keeps the variable coefficient positive if possible.

当等式两边都有含 x 的项时，例如 5x – 3 = 2x + 9，目标是将变量集中在一边，常数集中在另一边。两边减去 2x：3x – 3 = 9。加上3：3x = 12，所以 x = 4。尽量让变量系数保持在正的一边。

7. Equations with Fractions | 含分数的方程

Page 118 includes equations like (x+2)/3 = 5. The fraction bar indicates division, so to undo it, multiply both sides by 3: x + 2 = 15, then x = 13. When the numerator itself contains multiple terms, treat the fraction as a group by protecting it with brackets during mental steps.

第118页包含如 (x+2)/3 = 5 的方程。分数线表示除法，因此要解除它，两边同乘以 3：x + 2 = 15，然后 x = 13。如果分子包含多项，在脑海中应用括号将其视为一个整体，以方便操作。

More challenging fractions appear in the form (2x – 1)/4 = (x + 3)/2. The efficient method is to cross-multiply: 2(2x – 1) = 4(x + 3). Expand: 4x – 2 = 4x + 12. Subtract 4x from both sides: -2 = 12, which is a contradiction, meaning no solution. Recognizing such cases is part of higher-level algebraic thinking.

更具挑战性的分数形式如 (2x – 1)/4 = (x + 3)/2。高效的方法是交叉相乘：2(2x – 1) = 4(x + 3)。展开得 4x – 2 = 4x + 12。两边减去 4x：-2 = 12，这是一个矛盾式，表明无解。识别这类情况是高阶代数思维的一部分。

When a solution arises like 0 = 0, the equation has infinitely many solutions, meaning any value of x satisfies it. Keep an eye out for these special outcomes in the exercise.

若得出 0 = 0 这样的结果，则方程有无穷多解，意味着任何 x 值都能使等式成立。在练习中留意这些特殊情况。

8. Checking Your Answer | 验证你的答案

After finding a value for x, always substitute it back into the original equation to verify correctness. For 3(x + 4) = 21, we found x = 3. Plugging in: 3(3 + 4) = 3 × 7 = 21, which matches. This habit catches arithmetic errors and is a requirement in Cambridge exams to ensure complete marks.

求出 x 的值后，务必代回原方程验证正确性。以 3(x + 4) = 21 为例，我们解得 x = 3。代回得 3(3 + 4) = 3 × 7 = 21，与原式吻合。这一习惯能发现计算错误，同时也是剑桥考试中确保拿满分的必备步骤。

If the left and right sides differ, retrace your steps. Common errors include misapplying the distributive property, forgetting to change signs, or adding instead of subtracting. Checking allows you to correct these before finalising your answer.

如果左右两边不相等，回溯你的解题步骤。常见错误包括误用分配律、忘记变号，或者应当减法时做了加法。验证能让你在最终确定答案之前纠正这些错误。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Learners frequently lose marks by mishandling negative signs. When expanding -2(x – 3), it is easy to write -2x – 6 instead of the correct -2x + 6. A useful trick is to think of the minus sign as ‘multiply by -2’ and apply it to every term inside the brackets individually.

学生们常常因误用负号而失分。展开 -2(x – 3) 时，很容易错写成 -2x – 6，而正确结果是 -2x + 6。一个实用的技巧是把负号看作“乘以 -2”，并逐一应用到括号内的每一项。

Another pitfall involves adding or subtracting terms without maintaining balance. Forgetting to subtract the same amount from both sides results in an incorrect solution. Writing each step line by line, with clear ‘do to both sides’ annotations, dramatically reduces such errors.

另一个陷阱是进行加减时没有保持等式平衡。忘记在等式两边同时减去同一个数会导致错误的解。逐行书写每一步，并清楚标注“对两边操作”，可以大幅减少此类错误。

Finally, when dealing with fractions, students sometimes multiply only part of an expression. For (x/2) + 3 = 7, you must either multiply every term by 2 (x + 6 = 14) or subtract 3 first. Rushing leads to x/2 = 10, which gives a different answer.

最后，处理分数时，学生有时只乘表达式的某一部分。对于 (x/2) + 3 = 7，要么先将每一项乘以2得到 x + 6 = 14，要么先减去3。操之过急会变成 x/2 = 10，导致不同答案。

10. Step-by-Step Worked Example | 逐步完整示例

Let us work through a challenging equation that combines several skills: 2(3x – 1) – 4 = x + 5(2 – x). First, expand the brackets on the left: 6x – 2 – 4, which simplifies to 6x – 6. On the right, expand: x + 10 – 5x, which simplifies to 10 – 4x. The equation now is 6x – 6 = 10 – 4x.

下面让我们解一道综合多种技能的挑战性方程：2(3x – 1) – 4 = x + 5(2 – x)。首先，展开左边的括号：6x – 2 – 4，化简得 6x – 6。右边展开为：x + 10 – 5x，化简得 10 – 4x。现在方程变为 6x – 6 = 10 – 4x。

Add 4x to both sides to gather variables: 10x – 6 = 10. Add 6 to both sides: 10x = 16. Divide by 10: x = 16/10, which reduces to 8/5 or 1.6. Check by substituting back: left side 2(3(1.6) – 1) – 4 = 2(4.8 – 1) – 4 = 2(3.8) – 4 = 7.6 – 4 = 3.6. Right side: 1.6 + 5(2 – 1.6) = 1.6 + 5(0.4) = 1.6 + 2 = 3.6. Both sides match, confirming the solution.

两边同时加 4x 以集中变量：10x – 6 = 10。两边同时加 6：10x = 16。除以 10 得 x = 16/10，约简为 8/5 或 1.6。代回检验：左边 2(3(1.6) – 1) – 4 = 2(4.8 – 1) – 4 = 2(3.8) – 4 = 7.6 – 4 = 3.6。右边：1.6 + 5(2 – 1.6) = 1.6 + 5(0.4) = 1.6 + 2 = 3.6。两边相等，确认解正确。

11. Practice Questions from Page 118 Style | 第118页风格的练习题

Here are three additional questions styled like those in Exercise 1. Try solving them, then check your answers below.

以下三道练习题的风格与练习1相似。请尝试求解，然后与下方答案核对。

• a) 4(x – 3) + 2 = 2x + 8
• b) (5x + 1)/2 = 3x – 4
• c) 3(2 – x) – 5 = -2(x + 1)

Solutions: a) Expand 4x – 12 + 2 = 2x + 8 → 4x – 10 = 2x + 8 → 2x = 18 → x = 9. b) Multiply both sides by 2: 5x + 1 = 6x – 8 → 1 + 8 = 6x – 5x → x = 9. c) Expand 6 – 3x – 5 = -2x – 2 → 1 – 3x = -2x – 2 → Add 3x to both sides: 1 = x – 2 → x = 3. Substitute to verify each.

答案：a) 展开 4x – 12 + 2 = 2x + 8 → 4x – 10 = 2x + 8 → 2x = 18 → x = 9。b) 两边乘2：5x + 1 = 6x – 8 → 1 + 8 = 6x – 5x → x = 9。c) 展开 6 – 3x – 5 = -2x – 2 → 1 – 3x = -2x – 2 → 两边加 3x：1 = x – 2 → x = 3。通过代回验证每一个解。

12. Exam Tips and Summary | 考试技巧与总结

In the Cambridge KS3 progression test or checkpoint exam, you must show all working clearly. Examiners award method marks even if the final answer is wrong, so never skip steps. Always write the equation you are solving, expand brackets methodically, and note each balancing operation.

在剑桥 KS3 阶段考试或 Checkpoint 考试中，你需要清晰展示所有步骤。即使最终答案错误，考官也会给步骤分，因此绝不要跳过解题过程。务必写出你要解的方程，有条不紊地展开括号，并标注每个平衡操作。

Time management is crucial. Start with the easier one-step equations to build confidence, then move to the brackets and fractions. Use checking strategically, perhaps after every two or three equations, to catch mistakes early.

时间管理至关重要。先做简单的一步方程以建立信心，然后再攻克带括号和分数的题目。有策略地进行验证，例如每做完两三道题就检查一次，以便尽早发现错误。

Remember the core tools: distributive property, collecting like terms, inverse operations, and balancing. With consistent practice, you will solve page 118 Exercise 1 fluently and be ready for more advanced algebraic challenges ahead.

记住这些核心工具：分配律、合并同类项、逆运算以及平衡。通过持续练习，你将能顺利攻克第118页练习1，并为未来更高阶的代数挑战做好准备。

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