📚 Solving Linear Equations with Brackets | 去括号解一元一次方程
In Cambridge KS3 Mathematics, one of the most important algebra skills is solving linear equations where brackets appear on one or both sides. This topic builds directly on two-step equations and prepares you for more demanding questions in Checkpoint tests. In this article, we will break the method into clear steps, practise with worked examples, and highlight the common errors that cost marks.
在 Cambridge KS3 数学中，最重要的代数技能之一是解一元一次方程，尤其是方程的一边或两边含有括号的情况。这个主题直接建立在两步方程的基础上，并为 Checkpoint 考试中更难的题目做好准备。本文将把解题方法拆分为清晰步骤，通过例题进行练习，并指出容易丢分的常见错误。
Before you start, remember that an equation is a statement that two expressions are equal. You can think of it as a balanced scale: whatever you do to one side, you must also do to the other side. This balance idea is the foundation of every method below.
开始之前，请记住：方程表示两个表达式相等。你可以把它想象成一个平衡的天平——你对一边做什么，另一边也必须做同样的操作。这个平衡思想是下面所有方法的基础。
1. Understanding Equations and Balance | 理解方程与天平平衡
s that the expression on the left has exactly the same value as the number on the right. The letter x stands for an unknown number that we need to find. To find it, we perform inverse operations while keeping both sides balanced.
像 3(x + 2) = 21 这样的方程告诉我们，左边的表达式与右边的数完全相等。字母 x 代表我们需要找到的未知数。为了求出 x，我们使用逆运算，同时保持两边平衡。
When a bracket is multiplied by a number, the multiplication applies to every term inside the bracket. This is called the distributive law. For example, 3(x + 2) means 3 times x plus 3 times 2, which gives 3x + 6.
当一个括号乘以一个数时，这个乘法作用于括号内的每一项。这叫做分配律。例如，3(x + 2) 表示 3 乘 x 加 3 乘 2，得到 3x + 6。
- Keep the equals sign lined up in your working.
- Always do the same operation to both sides.
- Write each new equation on a new line.
- 在书写过程中对齐等号。
- 始终对两边进行相同的运算。
- 每一步新方程另起一行书写。
2. Reading the Question: What Does p182_2 Mean? | 读题：p182_2 在考什么
In many Cambridge Checkpoint books, page 182 question 2 focuses on solving equations that involve brackets, fractions, or variables on both sides. The question usually asks you to find the value of x and then check your answer. The wording may be short, but the marks are awarded for clear method, not just the final answer.
在许多 Cambridge Checkpoint 教材中，第 182 页第 2 题通常涉及解含有括号、分数或两边都有未知数的方程。题目通常会要求你求出 x 的值，并检验答案。题干可能很短，但分数是给清晰的解题过程，而不仅仅是最终答案。
ion tests the distributive law, collecting like terms, and the balance method at the same time. You should read it carefully and identify the operations around x step by step.
这类典型题目可能是：解方程 2(3x − 1) = 4x + 10。这道题同时考查分配律、合并同类项和平衡法。你应该仔细读题，并一步步识别围绕 x 的运算。
| Clue in the question | What it tells you |
| bracket | expand first |
| x on both sides | collect x terms on one side |
| fraction | multiply by the denominator |
| 题目线索 | 说明 |
| 括号 | 先展开括号 |
| 两边都有 x | 把含 x 的项移到一边 |
| 分数 | 乘以分母 |
3. Step 1: Expand the Brackets | 第一步：展开括号
When you see a number directly outside a bracket, multiply each term inside the bracket by that number. Pay attention to signs. For 2(3x − 1), multiplying gives 6x − 2. The minus sign belongs to the term after it, so the second product is −2, not +2.
当你看到括号外直接有一个数时，用这个数乘以括号内的每一项。注意符号。对于 2(3x − 1)，相乘得到 6x − 2。减号属于后面的项，所以第二项是 −2，而不是 +2。
If there are two brackets, expand them separately before combining terms. For example, in 3(x + 4) − 2(x − 1), the first bracket gives 3x + 12, and the second gives −2x + 2 because −2 times −1 is positive 2.
如果有两个括号，先分别展开，然后再合并各项。例如，在 3(x + 4) − 2(x − 1) 中，第一个括号得到 3x + 12，第二个括号得到 −2x + 2，因为 −2 乘 −1 等于正 2。
a(b + c) = ab + ac
2(3x − 1) = 6x − 2
4. Step 2: Simplify Both Sides | 第二步：化简两边
y simplified on each side, but if you had 6x + 3 − 2, you would combine 3 and −2 to get 6x + 1.
展开后，合并方程每一边的同类项。同类项是指变量部分完全相同的项。例如，6x − 2 = 4x + 10 两边已经是最简形式，但如果你有 6x + 3 − 2，就需要把 3 和 −2 合并成 6x + 1。
Never move terms across the equals sign before simplifying each side. This keeps your work tidy and reduces sign errors. Write the x terms first, then the constant terms, so the next step is easier.
在把各项移到等号另一边之前，一定要先化简每一边。这样可以使书写整洁，减少符号错误。先写含 x 的项，再写常数项，这样下一步会更容易。
6x − 2 = 4x + 10
5. Step 3: Move Variable Terms to One Side | 第三步：将含未知数的项移到一边
, subtract 4x from both sides to remove the x term from the right. This gives 2x − 2 = 10.
选择你想要保留未知数的一边，通常是未知数系数较大的一边。在 6x − 2 = 4x + 10 中，两边同时减去 4x，把右边的 x 项去掉，得到 2x − 2 = 10。
Use inverse operations: addition undoes subtraction, and subtraction undoes addition. Always show the same operation on both sides. This makes your method clear and earns method marks even if you make a small arithmetic error later.
使用逆运算：加法抵消减法，减法抵消加法。始终在两边写出相同的运算。这样你的解题过程清晰，即使后面有小计算错误，也能获得方法分。
6x − 2 − 4x = 4x + 10 − 4x
2x − 2 = 10
6. Step 4: Isolate and Solve for x | 第四步：解出 x
x − 2 = 10, add 2 to both sides to get 2x = 12. Then divide both sides by 2 to find x = 6.
接下来，把常数项从含未知数的一边移走。在 2x − 2 = 10 中，两边同时加 2，得到 2x = 12。然后两边同时除以 2，得到 x = 6。
The order of inverse operations is opposite to the order of operations in the original expression. We first removed the addition or subtraction, then the multiplication or division. This mirrors how you would undo a sequence: reverse the last step first.
逆运算的顺序与原表达式中的运算顺序相反。我们先去掉加法或减法，再去掉乘法或除法。这就像倒序撤销一系列操作：先撤销最后一步。
2x − 2 + 2 = 10 + 2
2x = 12
x = 6
7. Handling Fractions in Equations | 处理方程中的分数
integer equation. For example, in x/3 + 1 = 5, multiply all terms by 3 to get x + 3 = 15.
如果方程中出现分数，将两边每一项都乘以最小公分母。这样可以去掉分数，把方程变成整数方程。例如，在 x/3 + 1 = 5 中，所有项乘以 3，得到 x + 3 = 15。
Do not multiply only the fraction term. The multiplier must apply to every term in the equation. Missing a term breaks the balance and gives a wrong answer.
不要只乘含有分数的那一项。乘数必须应用于方程中的每一项。漏乘某一项会破坏平衡，导致答案错误。
(x/3) × 3 + 1 × 3 = 5 × 3
x + 3 = 15
8. Checking Your Answer by Substitution | 代入检验答案
t side becomes 4 × 6 + 10 = 24 + 10 = 34. Both sides equal 34, so the solution is correct.
求出 x = 6 后，把它代回原方程 2(3x − 1) = 4x + 10。左边变为 2(3 × 6 − 1) = 2(18 − 1) = 2 × 17 = 34。右边变为 4 × 6 + 10 = 24 + 10 = 34。两边都等于 34，因此解是正确的。
Checking takes only a few seconds but can catch sign errors, expansion mistakes, or arithmetic slips. In Checkpoint papers, many questions explicitly ask you to verify your answer, so make this a habit.
检验只需要几秒钟，但可以发现符号错误、展开错误或计算错误。在 Checkpoint 考试中，许多题目明确要求验证答案，所以要养成这个习惯。
- Write the original equation.
- Replace x with your value.
- Work out each side separately.
- Compare the two results.
- 写出原方程。
- 用求出的值替换 x。
- 分别计算两边。
- 比较两个结果。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
One common mistake is forgetting to multiply every term inside the bracket. For 3(x + 2), some learners write 3x + 2 instead of 3x + 6. To avoid this, draw arrows from the outside number to each term inside the bracket before writing the expanded form.
一个常见错误是忘记乘括号内的每一项。对于 3(x + 2)，有些学生会写成 3x + 2，而不是 3x + 6。为了避免这种错误，在写出展开式之前，先从括号外的数向括号内每一项画箭头。
Another frequent error is dropping the minus sign when the outside multiplier is negative. In −2(x − 1), the correct expansion is −2x + 2, not −2x − 2. Always be careful with negative signs: a negative times a negative gives a positive.
另一个常见错误是当括号外的乘数为负数时丢失负号。在 −2(x − 1) 中，正确展开是 −2x + 2，而不是 −2x − 2。一定要小心负号：负数乘负数得正数。
Students also sometimes move terms incorrectly. When you subtract a term from one side, you must subtract the same term from the other side. Changing the sign of one term without doing the same to both sides breaks the equation.
学生有时还会错误地移项。当你从一边减去一项时，必须从另一边减去同样的项。只改变某一项的符号而不对两边进行相同操作，会破坏方程。
| Mistake | Correction |
| 3(x + 2) = 3x + 2 | 3x + 6 |
| −2(x − 1) = −2x − 2 | −2x + 2 |
| 4x + 3 = 7x | 4x and 3 are not like terms |
| 错误 | 正确做法 |
| 3(x + 2) = 3x + 2 | 3x + 6 |
| −2(x − 1) = −2x − 2 | −2x + 2 |
| 4x + 3 = 7x | 4x 和 3 不是同类项 |
10. Worked Example: Two Brackets | 例题：含有两个括号
side by combining like terms: x + 8 = 12. Finally subtract 8 from both sides to get x = 4.
我们来逐步解方程 3(x + 2) − 2(x − 1) = 12。首先展开两个括号：3x + 6 − 2x + 2 = 12。然后合并左边的同类项：x + 8 = 12。最后两边同时减去 8，得到 x = 4。
2(4 − 1) = 3 × 6 − 2 × 3 = 18 − 6 = 12. The right side is 12, so the solution is correct.
代入检验：左边为 3(4 + 2) − 2(4 − 1) = 3 × 6 − 2 × 3 = 18 − 6 = 12。右边是 12，所以解是正确的。
3(x + 2) − 2(x − 1) = 12
3x + 6 − 2x + 2 = 12
x + 8 = 12
x = 4
11. Practice Questions | 练习题
Try these questions without looking at the solutions. For each one, show all the steps: expand brackets, simplify, move terms, then solve. Finally check your answer in the original equation.
尝试不看答案做以下题目。每题都要写出所有步骤：展开括号、化简、移项、求出解。最后代回原方程检验答案。
- Solve 2(x + 5) = 18
- Solve 4(x − 3) = 2x + 10
- Solve 5(2x + 1) = 3(x + 7)
- Solve x/2 + 4 = 9
- Solve 3(x + 4) − 2(x + 1) = 20
- 解方程 2(x + 5) = 18
- 解方程 4(x − 3) = 2x + 10
- 解方程 5(2x + 1) = 3(x + 7)
- 解方程 x/2 + 4 = 9
- 解方程 3(x + 4) − 2(x + 1) = 20
The answers are: 4, 11, 16/7, 10, and 10. If you made an error, go back to the relevant section and rework the question. Writing each step clearly will help you find where the mistake happened.
答案分别是：4、11、16/7、10 和 10。如果你做错了，回到相关小节重新做一遍。把每一步写清楚，可以帮助你找到错误发生的位置。
12. Exam Tips for Cambridge KS3 | Cambridge KS3 考试技巧
In Checkpoint tests, many algebra marks come from showing your method, not just writing the final answer. Always write the equation, then show one operation on both sides per line. This demonstrates that you understand the balance method.
在 Checkpoint 考试中，许多代数分数来自展示解题方法，而不仅仅是写出最终答案。一定要写出方程，然后每行展示对两边进行的一次运算。这可以证明你理解平衡法。
When you finish, read the question again. If it asks for the value of x, give a number. If it asks to check your answer, show the substitution clearly. If you have time, recheck signs and arithmetic in each line.
完成后，重新读一遍题目。如果题目要求 x 的值，就写出一个数。如果题目要求检验答案，就要清楚地展示代入过程。如果有时间，重新检查每一行的符号和计算。
- Expand brackets first.
- Simplify each side before moving terms.
- Use inverse operations in the correct order.
- Show your working line by line.
- Substitute to check your final answer.
- 先展开括号。
- 移项前先化简每一边。
- 按正确顺序使用逆运算。
- 一行一行展示解题过程。
- 代入检验最终答案。
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