Solving Linear Equations (Exercise P136_1) | 解线性方程（练习P136_1）

📚 Solving Linear Equations (Exercise P136_1) | 解线性方程（练习P136_1）

Welcome to this focused revision guide for KS3 Cambridge Mathematics. The problems on page 136, Exercise 1, are designed to strengthen your ability to solve linear equations. Whether you are tackling one-step puzzles or equations with unknowns on both sides, a systematic approach will help you avoid common slips and build confidence. This article explains the balancing method step by step, shows how to expand brackets correctly, and provides plenty of worked examples that mirror the style of your practice sheet.

欢迎阅读这份针对KS3剑桥数学的专项复习指南。第136页练习1中的题目旨在强化你解线性方程的能力。无论你面对的是简单的一步方程，还是未知数出现在等式两边的复杂情况，有条不紊的方法都能帮你避开常见失误，树立解题信心。本文会逐步讲解天平平衡法，演示如何正确展开括号，并提供大量与你练习页风格一致的详细例题。

1. What Is an Equation? | 什么是方程？

An equation is a mathematical statement that shows two expressions are equal, using the equals sign ‘=’. For example, x + 5 = 12 means that when you add 5 to an unknown number x, the result is 12. Solving the equation is the process of finding the value of x that makes this statement true. In KS3 work, we only deal with linear equations, where the variable has no exponent higher than 1.

方程是一个数学陈述，它用等号“=”表明两个表达式相等。例如，x + 5 = 12 表示当你把未知数 x 加上 5 后，结果是 12。解方程就是找出能让这个陈述成立的 x 值。在 KS3 的学习中，我们只处理线性方程，也就是变量的指数不超过 1。

2. The Balancing Method | 天平平衡法

Think of an equation as a perfectly balanced scale. Whatever you do to one side, you must do exactly the same to the other side to keep it balanced. If you add 3 to the left, add 3 to the right. If you divide the left by 2, divide the right by 2. This ‘golden rule’ is the foundation of all algebraic manipulation at this level. Visualising the scale helps you remember never to perform an operation on only one side.

你可以把方程想象成一架完全平衡的天平。你对其中的一边做什么操作，就必须对另一边做完全相同的操作，才能保持平衡。如果左边加上 3，右边也要加上 3；如果左边除以 2，右边也要除以 2。这条“黄金法则”是本阶段所有代数变形的基础。把方程想象成天平，有助于你记住绝不能只对单边进行运算。

For instance, with x + 4 = 9, subtract 4 from both sides to isolate x. The process can be shown in a simple table.

例如对于 x + 4 = 9，两边同时减去 4 就能分离出 x。这个过程可以用一个简单的表格展示。

Equation	Operation
x + 4 = 9	−4 on both sides
x = 5	Simplified

3. Solving One-Step Equations | 解一步方程

One-step equations require exactly one inverse operation to find the unknown. If the equation shows addition, we subtract; if it shows multiplication, we divide. Consider y − 7 = 3: here the inverse of subtracting 7 is adding 7. Applying this to both sides gives y = 10. The same logic applies to 4a = 24, which yields a = 6 after dividing both sides by 4.

一步方程只需要进行一次逆运算就能求出未知数。如果方程中出现加法，我们就用减法；如果出现乘法，我们就用除法。来看 y − 7 = 3：这里减去 7 的逆运算是加上 7。对两边同时加 7 就得到 y = 10。同样的逻辑也适用于 4a = 24，两边同除以 4 即可得出 a = 6。

Always write the operation neatly beside the equation. With m ÷ 5 = 3, the inverse is multiplication by 5, so m = 15. Getting these quick wins right builds the fluency needed for more complicated problems later in Exercise 136.

始终保持运算步骤整洁地写在方程旁边。对于 m ÷ 5 = 3，逆运算是乘以 5，因此 m = 15。把这些基础题做对，可以积累流畅的运算能力，为练习 136 后面更复杂的题目做好准备。

4. Two-Step Equations | 两步方程

Two-step equations combine two operations, such as multiplication and addition. The key is to undo the addition or subtraction first, then the multiplication or division. Take 2x + 3 = 11. Subtract 3 from both sides to obtain 2x = 8, then divide both sides by 2 to find x = 4. Breaking the process into layers prevents confusion.

两步方程结合了两种运算，例如乘法和加法。关键在于先撤销加法或减法，再处理乘法或除法。以 2x + 3 = 11 为例，两边先减去 3 得到 2x = 8，然后两边同除以 2 求出 x = 4。把这个过程分层处理可以避免混乱。

2x + 3 = 11 → 2x = 8 → x = 4

Another common type is 5 − 2y = 1. Here subtracting 5 from both sides gives −2y = −4. Dividing by −2 then yields y = 2. Notice how careful handling of negative signs is essential; many learners lose marks by forgetting that dividing two negatives gives a positive.

另一种常见类型是 5 − 2y = 1。两边同时减去 5 得到 −2y = −4。再除以 −2 便得出 y = 2。注意，这里必须小心翼翼处理负号；很多学生因为忘记“负负得正”而丢分。

5. Equations with Brackets | 含括号的方程

When an equation includes brackets, your first job is to expand them using the distributive property. For 3(x + 4) = 21, multiply both x and 4 by 3 to get 3x + 12 = 21. Then proceed as with a standard two-step equation: subtract 12, then divide by 3, arriving at x = 3. Brackets simply hide a multiplication that must be unpacked before isolating the variable.

当方程里出现括号时，你首先要运用分配律把括号展开。对于 3(x + 4) = 21，把 x 和 4 都乘以 3，得到 3x + 12 = 21。然后按照标准的两步方程来解：先减去 12，再除以 3，得出 x = 3。括号只是把乘法隐藏起来了，你必须在分离变量之前先把它展开。

Be especially careful with a negative sign outside the bracket, such as −2(3x − 1) = 10. Expanding gives −6x + 2 = 10. Many students mistakenly write −6x − 2. Remember that −2 multiplied by −1 gives +2. From there, subtract 2 from both sides to get −6x = 8, and divide by −6 to obtain x = −4/3. Fractions are perfectly acceptable solutions in KS3.

当括号外面带负号时尤其要小心，例如 −2(3x − 1) = 10。展开后得到 −6x + 2 = 10。很多学生会错误地写成 −6x − 2。请记住：−2 乘以 −1 结果为正 2。之后两边减去 2 得到 −6x = 8，再除以 −6 得出 x = −4/3。在 KS3 阶段，分数解完全是可接受的答案。

6. Unknowns on Both Sides | 未知数在等式两边

Some of the trickier problems on page 136 place the variable on both sides of the equals sign, like 5x + 2 = 3x + 10. The strategy is to collect the x terms on one side and the constants on the other. Subtract 3x from both sides, giving 2x + 2 = 10. Then subtract 2, and finally divide by 2 to reach x = 4. Always aim to have a positive coefficient for x when possible.

第 136 页中一些较为棘手的题目会把变量放在等号两边，比如 5x + 2 = 3x + 10。解题策略是把含 x 的项集中到一边，常数项集中到另一边。两边同时减去 3x，得到 2x + 2 = 10。再减去 2，最后除以 2，便可得出 x = 4。只要有可能，你应当尽量让 x 的系数保持为正数。

With an equation like 7 − x = 2x − 5, you can add x to both sides to get 7 = 3x − 5, then add 5 and divide by 3, obtaining x = 4. Avoid moving the variable to the side that makes the coefficient negative, as it increases the chance of sign errors. Practising this type regularly will make the page 136 questions feel straightforward.

遇到像 7 − x = 2x − 5 这样的方程，你可以在两边同时加上 x，得到 7 = 3x − 5，然后再加 5 并除以 3，求得 x = 4。不要贸然把变量移到会导致系数为负的那一边，那样只会增加符号出错的概率。经常练习这一类题目，第 136 页的练习就会变得很顺手。

7. Checking Your Solution | 检验你的解

Substituting your answer back into the original equation is the best way to catch mistakes. For 4(x − 2) = 2x + 6, if you found x = 7, check by replacing every x with 7: left side becomes 4(7 − 2) = 20, right side becomes 2(7) + 6 = 20. Both sides match, so the solution is correct. This habit will save you marks in assessments and gives you instant feedback during revision.

把你的答案代回原方程是发现错误的最佳途径。对于 4(x − 2) = 2x + 6，如果你求出 x = 7，那就把每个 x 替换成 7 来检验：左边变成 4(7 − 2) = 20，右边变成 2(7) + 6 = 20。两边相等，说明解是正确的。养成这个习惯，不仅能在考试中为你保住分数，还能在复习时即时反馈问题所在。

When a solution involves a fraction or a negative number, checking is even more important. Take 2x/3 = 8, giving x = 12. Plugging back: (2 × 12) ÷ 3 = 8, which simplifies to 8 = 8. Without this step, a simple arithmetic slip could go unnoticed. Treat checking as a non-negotiable final stage of every equation problem.

当解含有分数或负数时，检验就更加重要。以 2x/3 = 8 为例，解出 x = 12。代回去：(2 × 12) ÷ 3 = 8，化简后为 8 = 8。如果没有这一步，一个简单的计算疏忽可能就会被忽略。把检验当作每道方程题不可省略的最后一步。

8. Word Problems Leading to Equations | 由应用题引出方程

Exercise 136 likely includes scenarios where you must first form an equation from a written description. For example: ‘I think of a number, multiply it by 3, add 10, and the result is 31.’ Let the unknown number be n; the equation is 3n + 10 = 31. Solving gives n = 7. Always define your variable clearly at the start, as good mathematical communication is part of the Cambridge marking criteria.

练习 136 中很可能包含需要你先从文字描述中建立方程的应用题。例如：“我想一个数，把它乘以 3，再加 10，结果是 31。”设未知数为 n，对应方程就是 3n + 10 = 31。解出 n = 7。一开始就要清楚地定义你的变量，因为良好的数学表达习惯是剑桥评分标准的一部分。

Another typical KS3 question reads: ‘The perimeter of a rectangle is 50 cm. The length is 4 cm longer than the width. Find the dimensions.’ Let width = w, then length = w + 4. Perimeter = 2(w + w + 4) = 2(2w + 4) = 4w + 8. Set this equal to 50: 4w + 8 = 50, so w = 10.5 cm. The length is 14.5 cm. Writing the equation correctly is often half the battle.

另一个典型的 KS3 题目是：“一个矩形的周长是 50 厘米，长比宽多 4 厘米，求矩形的长和宽。”设宽为 w，则长为 w + 4。周长 = 2(w + w + 4) = 2(2w + 4) = 4w + 8。令它等于 50：4w + 8 = 50，因此 w = 10.5 厘米，长为 14.5 厘米。正确地列出方程往往就成功了一大半。

9. Common Mistakes and How to Avoid Them | 常见错误及其避免方法

One frequent error is applying an operation to only part of one side. For example, when solving x/2 + 3 = 7, some learners subtract 3 from the +3 and the 7 but forget to subtract 3 from the x/2 term properly — although in this case you subtract 3 from the whole left side, which is fine. The danger arises when they multiply only the x/2 by 2 instead of the entire expression. Remember: the whole side must be treated as a single quantity. Use brackets if necessary.

一个常见的错误是只对单边的某一部分进行运算。例如，在解 x/2 + 3 = 7 时，有些学生从 +3 和 7 那里减去了 3，却忽略了 x/2 ——尽管这里两边减去 3 是正确的。真正危险的是，当他们接下来只把 x/2 乘以 2，而没有把左边整个表达式乘以 2。请记住：必须将左（或右）侧视为一个整体。必要时用括号把表达式括起来。

Misplacing negative signs is another pitfall. In −2x = 8, the solution is x = −4, not x = 4. A quick mental check can help: −2 × (−4) does give 8. Similarly, when expanding brackets, always multiply each term inside by the factor outside, including the sign. Writing a plus sign as a raised ‘+’ and a minus as ‘−’ clearly will reduce slip-ups. Slow down and write every line completely; rushing is the main cause of these errors.

负号错位是另一个陷阱。对于 −2x = 8，解是 x = −4，而不是 x = 4。快速心算检验一下：−2 × (−4) 确实等于 8。同样地，在展开括号时，务必将括号内每一项都乘以外面的因数，包括符号。把加号和减号写清楚也能减少失误。放慢速度，把每一步都完整地写出来；仓促行事是这类错误的主因。

10. Tips for Tackling Page 136 Exercise 1 | 应对第 136 页练习 1 的技巧

Begin by scanning the exercise to identify the equation types: one-step, two-step, brackets, or unknowns on both sides. Jot down the inverse operations you will need next to each question. Work through the problems in order, but if you get stuck, move on and return later. Pencil in checking steps at the bottom of the page; this reinforces accuracy and aligns with Cambridge’s emphasis on reasoning and verification.

开始做题前，先快速浏览一遍练习，判断方程的类型：一步、两步、含括号，还是未知数在两边。在每道题旁边简要写下你需要的逆运算。按照顺序逐一完成，如果遇到卡住的题目，先跳过去，之后再回头。在页脚处用铅笔写下检验步骤；这既能提升准确率，也契合剑桥对推理和验证的重视。

Finally, use the examples in this article as models. If you face a question like 5(2x − 1) = 3x + 9, expand the bracket first to obtain 10x − 5 = 3x + 9, then collect x terms: subtract 3x, add 5, and divide. With disciplined practice, the methods become automatic, and you will tackle Exercise P136_1 with confidence and speed.

最后，把本文中的例题当作模板。如果你遇到类似 5(2x − 1) = 3x + 9 的题目，先展开括号得到 10x − 5 = 3x + 9，然后集中 x 项：减去 3x，加 5，再除以系数。经过有针对性的训练，这些方法会成为你的本能，你也就能自信而迅速地完成练习 P136_1 了。

10x − 5 = 3x + 9 → 7x = 14 → x = 2

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Solving Linear Equations (Exercise P136_1) | 解线性方程（练习P136_1）