Tag: KS3

📚 KS3 Cambridge Maths: Solving Linear Equations with Brackets and Fractions | KS3 剑桥数学：解含括号和分数的线性方程

Welcome to this Cambridge KS3 Maths revision guide on solving linear equations. This article focuses on the skills you need when an equation includes brackets, fractions, or unknowns on both sides – exactly the type of question often found in the p247_2.pdf worksheet. Work through the methods step by step, then test yourself with the practice questions.

欢迎阅读本剑桥 KS3 数学复习指南，主题是解线性方程。本文重点讲解方程中带括号、分数或两边都有未知数时所需的解题技巧，这正是 p247_2.pdf 练习中常见的题型。请逐步学习解题方法，然后用练习题进行自测。

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1. What Is a Linear Equation? | 什么是线性方程？

A linear equation is a statement that two expressions are equal, and the unknown variable is only raised to the power 1. You will see equations such as x + 7 = 12, 3x – 4 = 11, 2(x + 5) = 18, and x/3 + 1 = 6. The main task is to find the value of the unknown that makes the statement true.

线性方程是指两个表达式相等，且未知数的次数仅为 1 的等式。你会遇到如 x + 7 = 12、3x – 4 = 11、2(x + 5) = 18 和 x/3 + 1 = 6 这样的方程。主要任务是求出使等式成立的未知数的值。

3x + 5 = 20

2(x + 4) = 18

x/5 + 2 = 6

In Cambridge KS3 Mathematics, linear equations often appear in non-calculator papers, in word problems, and in sequences. You must be able to solve them accurately and show clear working.

在剑桥 KS3 数学中，线性方程经常出现在非计算器试卷、应用题和数列题中。你必须能够准确求解，并写出清晰的解题过程。

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2. The Balance Method and Inverse Operations | 天平法与逆运算

Think of an equation as a balance scale. To keep the equation true, you must do exactly the same operation to both sides. The opposite operations used to undo each other are addition and subtraction, and multiplication and division.

把方程想象成一台天平。为了保持方程成立，你必须对方程两边做完全相同的运算。互为逆运算的操作包括：加法和减法、乘法和除法。

• Inverse operations: + and – undo each other; × and ÷ undo each other. — 逆运算：加和减互相抵消；乘和除互相抵消。

For example, to solve x + 8 = 15, subtract 8 from both sides to get x = 7. To solve 4x = 20, divide both sides by 4 to get x = 5. Always apply the inverse operation in the reverse order of the order of operations.

例如，解 x + 8 = 15 时，两边同时减去 8，得到 x = 7。解 4x = 20 时，两边同时除以 4，得到 x = 5。始终按照运算顺序的相反顺序来使用逆运算。

x + 8 = 15 → x = 15 – 8 → x = 7

4x = 20 → x = 20 ÷ 4 → x = 5

Operation | 运算	Inverse operation | 逆运算
+ 6	– 6
– 9	+ 9
× 3	÷ 3
÷ 5	× 5

Keep your work tidy by writing each new equation directly below the previous one. This makes checking much easier.

书写要整齐，把每个新的方程直接写在上一个方程下面。这样检查起来会容易得多。

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3. Solving Equations with Brackets | 解含括号的方程

When an equation has brackets, you can expand the brackets first using the distributive law. For example, 3(x + 4) means 3 × x + 3 × 4, which equals 3x + 12. Then solve the resulting equation.

当方程中含有括号时，你可以先用分配律展开括号。例如，3(x + 4) 表示 3 × x + 3 × 4，等于 3x + 12。然后求解所得的方程。

3(x + 4) = 27

3x + 12 = 27

3x = 27 – 12

3x = 15 → x = 5

Another example: solve 4(x – 2) = 20. Expand to get 4x – 8 = 20. Add 8 to both sides: 4x = 28. Divide by 4: x = 7. You can also divide first if the number outside the bracket divides the right side evenly: x – 2 = 5, so x = 7.

另一个例子：解 4(x – 2) = 20。展开得到 4x – 8 = 20。两边加 8：4x = 28。除以 4：x = 7。你也可以先除以括号外的数，如果它能整除右边：x – 2 = 5，所以 x = 7。

When a bracket is multiplied by a negative number, be careful with signs. For example, -2(x + 5) = -2x – 10, not -2x + 10.

当括号乘以负数时，要注意符号。例如，-2(x + 5) = -2x – 10，而不是 -2x + 10。

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4. Solving Equations with Fractions | 解含分数的方程

Equations with fractions usually come in two forms. The first form has a coefficient that is a fraction, such as x/5 or 2x/3. The second form has an expression as the numerator over a denominator, such as (2x + 4)/3.

含分数的方程通常有两种形式。第一种形式是系数为分数，如 x/5 或 2x/3。第二种形式是分子为表达式、分母为数字，如 (2x + 4)/3。

For the first form, use multiplication as the inverse operation. To solve x/5 + 2 = 6, subtract 2 first: x/5 = 4. Then multiply both sides by 5: x = 20.

对于第一种形式，用乘法作为逆运算。解 x/5 + 2 = 6 时，先减 2：x/5 = 4。然后两边同时乘以 5：x = 20。

x/5 + 2 = 6 → x/5 = 4 → x = 20

For the second form, clear the denominator first by multiplying both sides by the denominator. Solve (2x + 4)/3 = 6 by multiplying both sides by 3: 2x + 4 = 18. Then subtract 4: 2x = 14, and divide by 2: x = 7.

对于第二种形式，先通过两边乘以分母来去掉分母。解 (2x + 4)/3 = 6 时，两边同时乘以 3：2x + 4 = 18。然后减 4：2x = 14，再除以 2：x = 7。

(2x + 4)/3 = 6 → 2x + 4 = 18 → 2x = 14 → x = 7

When the coefficient is a fraction such as 2x/3, think of it as 2x divided by 3. To solve 2x/3 = 8, multiply both sides by 3 to get 2x = 24, then divide by 2 to get x = 12.

当系数是分数如 2x/3 时，可以把它看作 2x 除以 3。解 2x/3 = 8 时，两边同时乘以 3，得到 2x = 24，再除以 2，得到 x = 12。

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5. Unknowns on Both Sides | 两边都有未知数

When an equation has the unknown on both sides, collect the variable terms on one side first. A good habit is to remove the smaller unknown term so that the coefficient of the variable stays positive.

当方程两边都有未知数时，首先把含未知数的项移到同一边。一个良好的习惯是消去较小的未知数项，这样未知数的系数会保持为正。

Solve 7x + 2 = 3x + 18. Subtract 3x from both sides to get 4x + 2 = 18. Subtract 2: 4x = 16. Divide by 4: x = 4.

解 7x + 2 = 3x + 18。两边同时减去 3x，得到 4x + 2 = 18。再减 2：4x = 16。除以 4：x = 4。

7x + 2 = 3x + 18 → 4x + 2 = 18 → 4x = 16 → x = 4

Solve 2x + 9 = 5x – 6. Subtract 2x from both sides: 9 = 3x – 6. Add 6 to both sides: 15 = 3x. Divide by 3: x = 5.

解 2x + 9 = 5x – 6。两边同时减去 2x：9 = 3x – 6。两边同时加 6：15 = 3x。除以 3：x = 5。

2x + 9 = 5x – 6 → 9 = 3x – 6 → 15 = 3x → x = 5

Always check that you have not reversed a sign when moving terms. Write each step clearly to avoid losing marks.

移项时始终要注意没有搞错符号。把每一步写清楚，以免丢分。

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6. Combining Brackets and Fractions | 括号与分数综合

Some Cambridge questions combine brackets and fractions in one equation. For example, solve (2(x + 3))/5 = 4. Clear the denominator first by multiplying both sides by 5: 2(x + 3) = 20. Divide by 2: x + 3 = 10. Subtract 3: x = 7.

一些剑桥考试题会把括号和分数结合在同一个方程中。例如，解 (2(x + 3))/5 = 4。先通过两边乘以 5 去分母：2(x + 3) = 20。再除以 2：x + 3 = 10。减 3：x = 7。

(2(x + 3))/5 = 4 → 2(x + 3) = 20 → x + 3 = 10 → x = 7

You can also expand first: 2x + 6 = 20, then solve to get 2x = 14, x = 7. Both methods are valid. Choose the one that gives easier numbers.

你也可以先展开：2x + 6 = 20，然后求解得到 2x = 14，x = 7。两种方法都有效。选择数字更容易处理的那一种。

Another example: solve 3(x – 2)/4 = 6. Multiply by 4: 3(x – 2) = 24. Divide by 3: x – 2 = 8. Add 2: x = 10.

另一个例子：解 3(x – 2)/4 = 6。两边乘以 4：3(x – 2) = 24。除以 3：x – 2 = 8。加 2：x = 10。

3(x – 2)/4 = 6 → 3(x – 2) = 24 → x – 2 = 8 → x = 10

When you see a negative fraction in front of a bracket, such as -(x + 5)/2 = 3, remember that the negative sign applies to the whole numerator after you multiply.

当你看到括号前有负分数时，如 -(x + 5)/2 = 3，请记住，乘以分母后，

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📚 Cambridge KS3 Mathematics: Linear Equations from p261 Exercise 2 | 剑桥 KS3 数学：从 p261 练习 2 看线性方程

Linear equations are one of the most important topics in the Cambridge KS3 mathematics syllabus. They appear in almost every test, checkpoint paper and later IGCSE module. This article is based on the types of questions often found on page 261, exercise 2 of the Cambridge KS3 practice book. You will learn how to solve one-step, two-step and bracket equations, and how to check your answers confidently.

线性方程是剑桥 KS3 数学大纲中最重要的主题之一。它们几乎出现在每一次测试、Checkpoint 考试以及后续 IGCSE 模块中。本文基于剑桥 KS3 练习册第 261 页练习 2 中常见的题型编写。你将学会如何解一步方程、两步方程和含括号的方程，并能自信地检验答案。

1. What is a Linear Equation? | 什么是线性方程？

A linear equation is an equation in which the unknown variable, usually written as x, is only raised to the power of 1. This means there are no x² terms, no x³ terms and no variables in the denominator. The general form of a one-variable linear equation is ax + b = c, where a, b and c are numbers and a is not zero.

线性方程是指未知数（通常写作 x）的指数仅为 1 的方程。这意味着方程中没有 x² 项、没有 x³ 项，分母中也没有变量。一元一次线性方程的一般形式是 ax + b = c，其中 a、b 和 c 是数字，且 a 不等于零。

For example, the equation below is a linear equation because the highest power of x is 1.

例如，下面的方程是线性方程，因为 x 的最高次数是 1。

2x + 3 = 11

Solving a linear equation means finding the value of x that makes the left side equal to the right side. In this case, x = 4 because 2 × 4 + 3 = 11.

解线性方程意味着求出使左边等于右边的 x 的值。在这个例子中，x = 4，因为 2 × 4 + 3 = 11。

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2. The Balance Method | 天平法

Think of an equation as a balance scale. The left pan and the right pan must always have the same mass. If you add, subtract, multiply or divide one side, you must do exactly the same to the other side to keep the equation balanced.

把方程想象成一个天平。左盘和右盘必须始终保持相同的重量。如果你对一边进行加、减、乘、除，你必须对另一边做完全相同的操作，才能保持方程平衡。

This is called the balance method. It is the foundation for solving every linear equation at KS3 level. For example, to solve x + 4 = 10, subtract 4 from both sides.

这就是天平法。它是 KS3 阶段解每一个线性方程的基础。例如，要解 x + 4 = 10，需要在两边同时减去 4。

x + 4 − 4 = 10 − 4

x = 6

Using the balance method helps you avoid the common mistake of only changing one side of the equation. It also makes each step clear and easy to follow.

使用天平法可以帮助你避免只改变方程一边这一常见错误。它还能让每一步都清晰易循。

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3. Solving Equations with Addition and Subtraction | 含加减的方程

If the equation has a number added to the variable, subtract that number from both sides. For example, solve x + 7 = 15.

如果方程中变量加上了一个数，就从两边同时减去这个数。例如，解方程 x + 7 = 15。

x + 7 − 7 = 15 − 7

x = 8

If the equation has a number subtracted from the variable, add that number to both sides. For example, solve x − 5 = 9.

如果方程中变量减去了一个数，就在两边同时加上这个数。例如，解方程 x − 5 = 9。

x − 5 + 5 = 9 + 5

x = 14

Always write the operation on both sides before simplifying. This shows the examiner that you understand the balance method and not just guessing the answer.

一定要先写出两边的运算，然后再化简。这可以向阅卷人展示你理解天平法，而不是仅仅猜出答案。

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4. Solving Equations with Multiplication and Division | 含乘除的方程

When the variable is multiplied by a number, divide both sides by that number to find x. For example, solve 3x = 18.

当变量乘以一个数时，两边同时除以这个数即可求出 x。例如，解方程 3x = 18。

3x ÷ 3 = 18 ÷ 3

x = 6

When the variable is divided by a number, multiply both sides by that number. For example, solve x ÷ 4 = 5.

当变量除以一个数时，两边同时乘以这个数。例如，解方程 x ÷ 4 = 5。

x ÷ 4 × 4 = 5 × 4

x = 20

These are called one-step equations because only one operation is needed to isolate x. They are often the first step in longer multi-step questions.

这些被称为一步方程，因为只需要一步运算就可以把 x 单独分离出来。它们通常是更长的多步题中的第一步。

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5. Two-Step Equations | 两步方程

A two-step equation involves two operations, such as multiplication and addition. For example, solve 2x + 3 = 15. First remove the addition or subtraction, then remove the multiplication or division.

两步方程包含两种运算，例如乘法和加法。例如，解方程 2x + 3 = 15。先消去加法或减法，再消去乘法或除法。

2x + 3 − 3 = 15 − 3

2x = 12

2x ÷ 2 = 12 ÷ 2

x = 6

The order is important: do the opposite of the order of operations. Since BIDMAS says multiplication comes before addition, the inverse order is to undo addition first, then multiplication.

顺序很重要：要逆着运算顺序来做。因为 BIDMAS 规则说先乘后加，所以逆运算顺序是先消去加法，再消去乘法。

Always write the new equation after each step. This reduces errors and helps you check your solution later.

每一步之后都要写出新的方程。这样可以减少错误，也便于你之后检验解。

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6. Equations with Brackets | 含括号的方程

When an equation contains brackets, expand them first using the distributive law. For example, solve 3(x + 2) = 21.

当方程中含有括号时，首先使用分配律展开括号。例如，解方程 3(x + 2) = 21。

3 × x + 3 × 2 = 21

3x + 6 = 21

3x + 6 − 6 = 21 − 6

3x = 15

x = 5

Alternatively, you can divide both sides by 3 first, as long as the whole left side is divisible by 3. Both methods give the same answer, but expanding first is usually safer.

或者，你也可以先两边同时除以 3，只要左边整体都能被 3 整除。两种方法得到的答案相同，但先展开括号通常更稳妥。

Be careful with negative signs outside brackets: 2(x − 4) = 10 becomes 2x − 8 = 10, not 2x − 4 = 10.

注意括号外面的负号：2(x − 4) = 10 应变为 2x − 8 = 10，而不是 2x − 4 = 10。

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7. Equations with Fraction Coefficients | 分数系数方程

Some linear equations contain fractions. To solve them, multiply every term on both sides by the denominator. For example, solve x/3 + 2 = 7.

有些线性方程含有分数。要解这类方程，需要把两边每一项都乘以分母。例如，解方程 x/3 + 2 = 7。

3 × (x/3 + 2) = 3 × 7

x + 6 = 21

x = 15

If there are two different denominators, find the lowest common denominator and multiply every term by it. This removes all fractions in one step.

如果有两个不同的分母，先找出最小公分母，然后把每一项都乘以它。这样就可以一步消去所有分数。

Keep the fraction bar as a division sign if you find that clearer. For example, x/3 means x ÷ 3, so multiplying by 3 undoes the division.

如果你觉得更清楚，可以把分数线看作除号。例如，x/3 表示 x ÷ 3，因此乘以 3 就抵消了除法。

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8. Word Problems and Applications | 应用题与实际应用

Linear equations are often hidden inside word problems. The first job is to translate the words into an equation. Let x represent the unknown number, then build the equation step by step.

线性方程常常隐藏在文字题中。第一项任务是把文字转换成方程。设 x 表示未知数，然后逐步建立方程。

Example: A number is doubled, then 5 is added, and the result is 17. Find the number.

例子：一个数先加倍，再加上 5，结果是 17。求这个数。

2x + 5 = 17

2x = 12

x = 6

Common word clues: ‘added to’ means +, ‘subtracted from’ means −, ‘multiplied by’ means ×, ‘divided by’ means ÷, and ‘is’ or ‘equals’ means =.

常见的文字线索：“加上”表示 +，“减去”表示 −，“乘以”表示 ×，“除以”表示 ÷，“是”或“等于”表示 =。

Always reread the question after finding x to make sure your answer makes sense in the real situation.

求出 x 之后，一定要重新读一遍题目，确认你的答案在实际情境中是合理的。

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

One common error is only applying an operation to one side of the equation. For example, writing x + 4 = 10, then x = 10 − 4 is correct, but writing x + 4 = 10 and then x = 6 without showing −4 on both sides can lead to mistakes in harder equations.

一个常见错误是只对方程的一边进行运算。例如，写 x + 4 = 10，然后 x = 10 − 4 是正确的，但在更复杂的方程中，如果不在两边同时写出 −4，就容易出错。

Another common error is forgetting to multiply every term when clearing brackets or fractions. For example, 2(x + 3) = 10 should become 2x + 6 = 10, not 2x + 3 = 10.

另一个常见错误是在去括号或去分母时忘记乘以每一项。例如，2(x + 3) = 10 应变为 2x + 6 = 10，而不是 2x + 3 = 10。

Sign errors are also frequent. When subtracting a negative term, change it to addition. Write each step clearly and double-check your signs.

符号错误也很常见。减去一个负项时，要把它变为加法。每一步写清楚，并仔细检查符号。

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10. Check Your Answers | 检验你的答案

After solving an equation, always substitute your value of x back into the original equation. The left side and right side must be equal.

解完方程后，一定要把 x 的值代回原方程。左边和右边必须相等。

Example: Solve 5x − 4 = 26. We find x = 6. Check: 5 × 6 − 4 = 30 − 4 = 26, which matches the right side.

例子：解方程 5x − 4 = 26。我们得到 x = 6。检验：5 × 6 − 4 = 30 − 4 = 26，与右边相等。

Checking is especially important in Cambridge Checkpoint papers because it helps you catch small arithmetic errors and gain full marks.

在剑桥 Checkpoint 考试中，检验尤为重要，因为它能帮助你发现小的计算错误，从而获得满分。

Make checking a habit after every practice question, even if the question does not ask for it.

即使题目没有要求检验，也要养成每道练习题都检验的习惯。

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📚 Solving Linear Equations for KS3 Cambridge Maths | KS3剑桥数学：解一元一次方程

At KS3 Cambridge Mathematics, solving linear equations is a key algebraic skill. You learn how to find the value of an unknown letter, usually x or y, by using inverse operations and keeping an equation balanced. This guide walks through the essential methods, written examples and common exam-style tips for Cambridge Lower Secondary learners.

在KS3剑桥数学中，解一元一次方程是一项核心代数技能。你将学习如何利用逆运算并保持等式平衡，求未知字母（通常是x或y）的值。本指南为剑桥初中阶段学生梳理关键方法、书写示例和常见考试技巧。

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1. What Is a Linear Equation? | 什么是线性方程

A linear equation is an equation in which the unknown appears only to the power of 1. For example, 2x + 3 = 11 is linear because x is not squared or cubed. The word ‘linear’ tells you that the graph of the equation is a straight line.

线性方程是指未知数的次数仅为1的方程。例如，2x + 3 = 11 是线性的，因为 x 没有平方或立方。’线性’一词表示该方程的图像是一条直线。

The solution of an equation is the value of the unknown that makes the equation true. In 2x + 3 = 11, if x = 4, then 2 × 4 + 3 = 11, so the solution is x = 4.

方程的解是使等式成立的未知数的值。在 2x + 3 = 11 中，如果 x = 4，那么 2 × 4 + 3 = 11，因此解为 x = 4。

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2. The Balance Method | 天平法（等式性质）

Think of an equation as a balance scale. The equals sign means both sides have the same value. Whatever you do to one side, you must do to the other side to keep the balance.

把方程想象成一台天平。等号意味着两边数值相等。你对一边做的任何运算，都必须对另一边做同样的运算，以保持平衡。

The key operations are: add or subtract the same number on both sides, multiply both sides by the same non-zero number, and divide both sides by the same non-zero number.

关键运算是：两边同时加或减同一个数；两边同时乘以同一个非零数；两边同时除以同一个非零数。

Always write your working vertically, keeping the equals signs lined up. This shows the examiner that you have used balance correctly.

书写时始终竖向排列，保持等号对齐。这能让阅卷老师看出你正确使用了等式平衡。

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3. One-Step Equations | 一步方程

In a one-step equation, only one inverse operation is needed. For example, solve x + 5 = 12. Subtract 5 from both sides: x + 5 − 5 = 12 − 5, so x = 7.

一步方程只需进行一次逆运算。例如，解 x + 5 = 12。两边同时减去5：x + 5 − 5 = 12 − 5，所以 x = 7。

If the equation is 3x = 18, divide both sides by 3: 3x ÷ 3 = 18 ÷ 3, so x = 6.

如果方程是 3x = 18，两边同时除以3：3x ÷ 3 = 18 ÷ 3，所以 x = 6。

If the unknown is in the denominator, such as 20 ÷ x = 5, multiply both sides by x first, then divide: 20 = 5x, so x = 4.

如果未知数在分母中，例如 20 ÷ x = 5，先两边乘以 x，再除以：20 = 5x，所以 x = 4。

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4. Two-Step Equations | 两步方程

A two-step equation involves two operations. For example, solve 2x + 3 = 11. First undo the addition: subtract 3 from both sides to get 2x = 8. Then undo the multiplication: divide both sides by 2 to get x = 4.

两步方程包含两种运算。例如，解 2x + 3 = 11。先消去加法：两边同时减去3，得到 2x = 8。再消去乘法：两边同时除以2，得到 x = 4。

Remember to reverse the order of operations. Undo adding/subtracting before multiplying/dividing, because in the original expression multiplication happens before addition.

记住要逆用运算顺序。先消去加减法，再消去乘除法，因为原表达式中乘法先于加法。

Example: solve 7x − 9 = 26. Add 9 to both sides: 7x = 35. Then divide by 7: x = 5.

例如：解 7x − 9 = 26。两边加9：7x = 35。然后除以7：x = 5。

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5. Equations with Brackets | 带括号的方程

If an equation contains brackets, expand them first using the distributive law. For example, 3(x + 2) = 21. Expand: 3x + 6 = 21. Then subtract 6: 3x = 15. Divide by 3: x = 5.

如果方程含有括号，先利用乘法分配律展开。例如，3(x + 2) = 21。展开：3x + 6 = 21。然后减去6：3x = 15。除以3：x = 5。

Sometimes you can divide both sides by the factor first if it divides evenly. In 3(x + 2) = 21, divide both sides by 3 to get x + 2 = 7, then x = 5. This is often quicker.

如果因数能整除，也可以先两边除以这个因数。在 3(x + 2) = 21 中，两边除以3得到 x + 2 = 7，所以 x = 5。这通常更快捷。

Be careful with negative brackets: −2(x − 4) = 10 expands to −2x + 8 = 10, not −2x − 8.

注意负号括号：−2(x − 4) = 10 展开为 −2x + 8 = 10，而不是 −2x − 8。

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6. Unknowns on Both Sides | 两边都有未知数

When x appears on both sides, aim to collect all x terms on one side and numbers on the other. For example, solve 5x + 2 = 2x + 11. Subtract 2x

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📚 Bearings and Scale Drawings in KS3 Cambridge Maths | KS3 剑桥数学：方位角与比例绘图

In the Cambridge KS3 mathematics curriculum, bearings and scale drawings bring together geometry, ratio and real-world measurement. A bearing describes a direction as a three-figure angle measured clockwise from north, while a scale drawing shows a real object or distance at a fixed ratio. This topic is common in checkpoint tests because it tests practical skills such as using a protractor, converting units and applying map scales.

在剑桥 KS3 数学课程中，方位角与比例绘图把几何、比和实际测量结合在一起。方位角用一个从正北顺时针测量的三位数角度来表示方向，而比例绘图则按照固定比例展示真实物体或距离。这个主题在 Checkpoint 考试中很常见，因为它考查使用量角器、换算单位和应用地图比例尺等实用技能。

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1. What is a Bearing? | 什么是方位角？

A bearing is a way of describing the direction from one point to another. It is always measured from north in a clockwise direction and is written as a three-figure number. For example, due east is written as 090°, due south as 180°, due west as 270°, and due north as 000°.

方位角是描述从一个点到另一个点方向的一种方式。它总是从正北开始按顺时针方向测量，并写成三位数的形式。例如，正东写作 090°，正南写作 180°，正西写作 270°，正北写作 000°。

Bearings are useful in navigation, map reading and construction because they give an exact direction without needing words like ‘north-east’ or ‘south-west’. A bearing of 045° means exactly halfway between north and east, while 225° means halfway between south and west.

方位角在航海、地图阅读和建筑绘图中很有用，因为它能给出准确方向，而不必使用“东北”或“西南”这样的词语。方位角 045° 表示正北与正东的中间方向，而 225° 表示正南与正西的中间方向。

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2. Measuring Bearings with a Protractor | 用量角器测量方位角

To measure the bearing of B from A, first draw the north line at point A. Then place the centre of the protractor on A and line up 000° with the north line. Measure clockwise from the north line to the line segment AB. If the angle is less than 100°, add a leading zero so the bearing has three digits.

要测量从 A 到 B 的方位角，首先在点 A 处画出正北线。然后把量角器的中心放在点 A，并将 000° 与正北线对齐。从正北线顺时针测量到线段 AB。如果角度小于 100°，就在前面补一个零，使方位角成为三位数。

For example, if the angle from north to AB is 75° clockwise, the bearing is 075°. If the angle is 130°, the bearing is simply 130°. A common mistake is measuring anticlockwise or forgetting to use three figures.

例如，如果从正北到 AB 的顺时针角度是 75°，方位角就是 075°。如果角度是 130°，方位角就是 130°。常见的错误是逆时针测量，或者忘记使用三位数表示。

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3. Three-Figure Bearings and Direction Rules | 三位数方位角与方向规则

All bearings must be written as three-figure numbers, so 8° becomes 008° and 95° becomes 095°. The rule makes bearings easy to read in navigation and avoids confusion with angles measured in other ways. When a bearing is given, it does not tell you the distance; it only tells you the direction.

所有方位角都必须写成三位数，因此 8° 写成 008°，95° 写成 095°。这条规则使方位角在导航中易于读取，也避免与其他方式测量的角度混淆。给出方位角时，它并不告诉你距离，只告诉你方向。

Bearings can be reversed. The bearing of A from B is not usually the same as the bearing of B from A. For example, if the bearing of B from A is 060°, then the bearing of A from B is 240° because you turn around and continue clockwise from north at B. In general, reversing a bearing means adding 180° if the original is less than 180°, or subtracting 180° if it is 180° or more.

方位角可以反向。从 B 到 A 的方位角通常与从 A 到 B 的方位角不一样。例如，如果从 A 到 B 的方位角是 060°，那么从 B 到 A 的方位角就是 240°，因为你在 B 点掉头后继续从正北顺时针旋转。一般来说，反向方位角的方法是：如果原方位角小于 180°，就加上 180°；如果大于或等于 180°，就减去 180°。

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4. Drawing Bearings Accurately | 准确绘制方位角

To draw a bearing from a point, start by drawing a vertical north line through that point. Place your protractor with 000° pointing north and mark the required clockwise angle. Then draw a straight line through the mark. Use a ruler to make the line long enough to reach the target or another given point.

要从一个点绘制方位角，先通过该点画一条竖直的正北线。将量角器的 000° 对准正北，标出所需的顺时针角度，然后通过标记画一条直线。用直尺把线画得足够长，以到达目标或另一个已知点。

If a scale is also given, you can mark the actual distance along the bearing line. For instance, a ship travels 8 km on a bearing of 120°. Using a scale of 1 cm = 2 km, you would draw 4 cm in the direction 120° to represent the ship’s position.

如果还给出了比例尺，你可以沿着方位线标出实际距离。例如，一艘船以 120° 的方位角行驶了 8 km。使用 1 cm = 2 km 的比例尺，你需要在 120° 方向上画 4 cm 来表示船的位置。

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5. Understanding Scale Drawings | 理解比例绘图

A scale drawing is an accurate representation of a real object or area drawn smaller or larger than the original. The scale tells you how the drawing length relates to the real length. For example, a scale of 1 : 100 means that 1 unit on the drawing represents 100 units in real life.

比例绘图是按原物体或区域缩小或放大后绘制的精确表示。比例尺告诉你绘图长度与实际长度的关系。例如，比例尺 1 : 100 表示图纸上的 1 个单位代表实际中的 100 个单位。

Scales are often written as ratios, such as 1 : 50000 on a map. They can also be written as statements, such as ‘1 cm represents 5 km’. The key skill is converting between the drawing length and the real length by multiplying or dividing by the scale factor.

比例尺通常写成比，例如地图上的 1 : 50000。它也可以写成文字说明，例如“1 cm 代表 5 km”。关键技能是通过乘以或除以比例因子，在图纸长度和实际长度之间进行换算。

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6. Scale Factor and Map Scales | 比例因子与地图比例尺

The scale factor is the number you multiply a drawing length by to find the real length. If the scale is 1 : 25000, then 1 cm on the map is 25000 cm in real life. Since 25000 cm = 250 m, the map scale tells us that 1 cm represents 250 m.

比例因子是你将图纸长度乘以它来得到实际长度的数。如果比例尺是 1 : 25000，那么地图上的 1 cm 就是实际中的 25000 cm。因为 25000 cm = 250 m，所以这个地图比例尺告诉我们 1 cm 代表 250 m。

To convert a drawing length to a real length, use the formula:

要将图纸长度换算成实际长度，可使用以下公式：

Real distance = Drawing distance × Scale factor

To convert a real length to a drawing length, divide by the scale factor:

要将实际长度换算成图纸长度，则除以比例因子：

Drawing distance = Real distance ÷ Scale factor

Always convert units carefully. If the scale is 1 : 200000 and the map distance is 3 cm, the real distance is 3 × 200000 = 600000 cm = 6 km.

换算单位时要格外小心。如果比例尺为 1 : 200000，地图距离为 3 cm，则实际距离为 3 × 200000 = 600000 cm = 6 km。

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7. Converting Real Distances and Areas | 实际距离与面积换算

Distance conversions using map scales are straightforward because they involve lengths only. However, when working with area, the scale factor must be squared. If a map scale is 1 : 100, then 1 cm² on the map represents 100² = 10000 cm² in real life.

使用地图比例尺进行距离换算很简单，因为它只涉及长度。然而，处理面积时，比例因子必须平方。如果地图比例尺是 1 : 100，那么地图上的 1 cm² 表示实际中的 100² = 10000 cm²。

For example, a field has an area of 2 cm² on a 1 : 5000 map. The real area is 2 × 5000² = 2 × 25000000 = 50000000 cm². Converting to m², divide by 10000, so the area is 5000 m².

例如，一块田地在一张 1 : 5000 的地图上的面积是 2 cm²。实际面积是 2 × 5000² = 2 × 25000000 = 50000000 cm²。换算成 m² 时要除以 10000，所以面积是 5000 m²。

Remember that 1 m² = 10000 cm² because 1 m = 100 cm and area uses square units. This is a common source of error in checkpoint questions.

记住 1 m² = 10000 cm²，因为 1 m = 100 cm，而面积使用平方单位。这是 Checkpoint 题目中常见的错误来源。

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8. Combining Bearings with Scale Drawings | 方位角与比例绘图的结合

Many exam questions ask you to use a bearing and a distance to plot a point on a scale drawing, or to work out the bearing and distance between two points already marked. This combines accurate protractor work with ruler measurement and scale conversion.

许多考试题目要求你利用方位角和距离在比例图上标出一个点，或者计算已标记的两点之间的方位角和距离。这结合了准确的量角器作图、直尺测量和比例换算。

For example, a treasure map has a scale of 1 cm = 3 km. A rock is 9 km from a tower on a bearing of 210°. To plot the rock, draw the north line at the tower, mark 210° clockwise, then measure 3 cm along that line because 9 ÷ 3 = 3 cm.

例如，一张藏宝图的比例尺是 1 cm = 3 km。一块岩石在塔的 210° 方位角上，距离 9 km。要标出岩石，先在塔处画正北线，顺时针标出 210°，然后沿该线量出 3 cm，因为 9 ÷ 3 = 3 cm。

Similarly, if two points are shown on a scale drawing, measure the angle from north at the starting point and the length between them. Then convert the length using the scale to find the real distance.

同样，如果比例图上已经标出两个点，则测量从起点正北方向到两点连线的角度以及它们之间的长度。然后使用比例尺换算出实际距离。

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

Several mistakes appear regularly when students work with bearings and scale drawings. The most common is measuring a bearing anticlockwise from north instead of clockwise. Always follow the clockwise direction and double-check with the three-figure rule.

学生在处理方位角和比例绘图时，经常会出现几个错误。最常见的是从正北逆时针测量方位角，而不是顺时针。一定要始终沿顺时针方向测量，并用三位数规则进行双重检查。

Another mistake is forgetting to write a bearing as three figures, for example writing 60° instead of 060°. This may not change the direction but it loses method marks. Also, when reversing a bearing, many students simply add 180° to any value, which is incorrect when the result exceeds 360°.

另一个错误是忘记将方位角写成三位数，例如写作 60° 而不是 060°。这可能不会改变方向，但会丢掉方法分。另外，在反向方位角时，许多学生对任何数值都简单地加上 180°，当结果超过 360° 时，这是错误的。

With scale drawings, mixing up multiplication and division is very common. If you know the drawing length and want the real length, multiply. If you know the real length and want the drawing length, divide. Use the formulas to check your operation.

在比例绘图中，把乘法和除法搞混是非常常见的。如果你知道图纸长度而想求实际长度，就乘以比例因子。如果你知道实际长度而想求图纸长度，就除以比例因子。用公式来检查你的运算。

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10. Worked Example: Bearings and Distance | 例题：方位角与距离

A boat leaves a harbour H and travels 12 km on a bearing of 140° to point P. Using a scale of 1 cm = 2 km, draw the journey and find the bearing of H from P.

一艘船离开港口 H，沿 140° 方位角行驶 12 km 到达点 P。使用 1 cm = 2 km 的比例尺，绘制这次航行，并求从 P 到 H 的方位角。

First, the drawing distance is 12 ÷ 2 = 6 cm. Draw a north line at H, measure 140° clockwise, and draw a line 6 cm long. Label the end P.

首先，绘图距离是 12 ÷ 2 = 6 cm。在 H 处画正北线，顺时针量出 140°，画一条 6 cm 长的线段，末端标记为 P。

To find the bearing of H from P, draw a new north line at P. The bearing of H from P is the reverse bearing of 140°. Since 140° + 180° = 320°, the bearing of H from P is 320°.

要求从 P 到 H 的方位角，在 P 处画一条新的正北线。从 P 到 H 的方位角是 140° 的反向方位角。因为 140° + 180° = 320°，所以从 P 到 H 的方位角是 320°。

This example shows how a scale drawing and a bearing calculation work together. In a checkpoint question, you might need to measure the drawn angle rather than calculate it, but the principle is the same.

这个例子展示了比例绘图和方位角计算如何结合使用。在 Checkpoint 题目中，你可能需要测量图中角度而不是计算，但原理是相同的。

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11. Practice Skills for Checkpoint Success | 针对 Checkpoint 成功练习技能

To master this topic, practise drawing and measuring bearings with a protractor until you can do it quickly and accurately. Use squared paper or plain paper and check that your north lines are truly vertical and parallel to each other.

要掌握这个主题，请反复练习用量角器绘制和测量方位角，直到你能快速准确地完成。使用方格纸或白纸，并检查你的正北线是否真正竖直且彼此平行。

When working with scale drawings, always write the scale as a ratio or statement first. Then decide whether you need to multiply or divide. Underline the given units and convert them before calculating.

使用比例绘图时，首先要把比例尺写成比或文字说明。然后确定需要用乘法还是除法。在计算前给已知单位划线并先换算单位。

A useful revision exercise is to draw a simple journey with two or more legs, such as 5 km on a bearing of 060° followed by 8 km on a bearing of 150°. Plot both legs on a scale drawing and then measure the final bearing and distance from the starting point.

一个有用的复习练习是绘制包含两段或更多段的简单旅程，例如先沿 060° 方位角行驶 5 km，再沿 150° 方位角行驶 8 km。在比例图上画出两段旅程，然后测量从起点到终点的最终方位角和距离。

• Always write bearings as three digits.
• Always measure bearings clockwise from north.
• To reverse a bearing, add or subtract 180°.
• For lengths, multiply to find real distance and divide to find drawing distance.
• For areas, square the scale factor.

• 方位角始终写成三位数。
• 方位角始终从正北顺时针测量。
• 反向方位角加或减 180°。
• 长度换算：求实际距离用乘法，求图纸距离用除法。
• 面积换算：比例因子要平方。

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12. Key Summary and Exam Tips | 核心总结与考试技巧

Bearings and scale drawings test your ability to link measurement, direction and proportional reasoning. The key formulas to remember are the real distance and drawing distance conversions, and the reverse bearing rule.

方位角与比例绘图考查你将测量、方向和比例推理联系起来的能力。要记住的关键公式是实际距离与图纸距离的换算，以及反向方位角规则。

In an exam, show your working clearly. Draw north lines fine and parallel, write all bearings in three figures, and label your scale drawing. Even if your final answer is wrong, correct method marks can be earned.

考试时，要清晰地写出解题步骤。正北线画细且平行，所有方位角写成三位数，并标注你的比例图。即使最终答案错误，正确的解题方法也能获得步骤分。

Before moving to the next question, check whether the question asks for a bearing or a distance. Bearing answers need degrees and three figures; distance answers need units such as km, m or cm. This simple check can prevent losing marks.

在进入下一题之前，检查题目问的是方位角还是距离。方位角答案需要度数和三位数；距离答案需要单位，如 km、m 或 cm。这个简单的检查可以防止丢分。

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📚 Mastering Linear Equations for KS3 Maths | 掌握 KS3 数学中的一元一次方程

Linear equations are the foundation of algebra at KS3. In the Cambridge Lower Secondary Mathematics curriculum, you are expected to set up, solve and check one-variable equations with confidence. This article breaks down the key methods, common pitfalls and exam-style practice you need for success.

一元一次方程是 KS3 代数的基础。在剑桥初中数学课程中，你需要能够熟练地建立、求解并检验一元方程。本文拆解关键方法、常见错误和考试题型，帮助你在这一考点上取得成功。

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1. What Is a Linear Equation? | 什么是一元一次方程？

A linear equation is an equation in which the unknown, usually written as x, is only raised to the power of 1. It can be written in the form ax + b = c, where a, b and c are known numbers.

一元一次方程是指未知数（通常写作 x）的指数仅为 1 的方程。它可以写成 ax + b = c 的形式，其中 a、b 和 c 是已知数。

For example, 2x + 3 = 11 is linear because x appears without any square, cube or root. Equations like x² = 9 or 1/x = 4 are not linear at KS3 level.

例如，2x + 3 = 11 是一元一次方程，因为 x 没有平方、立方或根号。像 x² = 9 或 1/x = 4 这样的方程在 KS3 阶段不属于一元一次方程。

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2. The Balance Method | 天秤法

The balance method is the key idea behind solving linear equations. Think of an equation as a balanced set of scales: whatever you do to one side, you must do to the other side to keep it balanced.

天秤法是解一元一次方程的核心思想。把方程想象成一架平衡的天平：你对一边做什么操作，必须对另一边做同样的操作，才能保持平衡。

If you add, subtract, multiply or divide one side by a number, you must do exactly the same to the other side. This rule keeps the equation true while you isolate the unknown.

如果你对一边加、减、乘或除以一个数，你必须对另一边进行完全相同的操作。这一规则在分离未知数的过程中保持方程成立。

• Add the same value to both sides | 两边同时加上同一个数
• Subtract the same value from both sides | 两边同时减去同一个数
• Multiply or divide both sides by the same non-zero value | 两边同时乘以或除以同一个非零数

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3. Solving One-Step Equations | 解一步方程

A one-step equation needs only one inverse operation to solve. The inverse of addition is subtraction, and the inverse of multiplication is division.

一步方程只需要一次逆运算即可求解。加法的逆运算是减法，乘法的逆运算是除法。

To solve x + 7 = 15, subtract 7 from both sides:

要解 x + 7 = 15，两边同时减去 7：

x + 7 − 7 = 15 − 7

x = 8

To solve 4x = 28, divide both sides by 4:

要解 4x = 28，两边同时除以 4：

4x ÷ 4 = 28 ÷ 4

x = 7

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4. Solving Two-Step Equations | 解两步方程

A two-step equation combines two operations, such as 2x + 5 = 17. The golden rule is to undo addition or subtraction first, then undo multiplication or division.

两步方程包含两种运算，例如 2x + 5 = 17。黄金法则是先消去加法或减法，再消去乘法或除法。

Solve 2x + 5 = 17:

解 2x + 5 = 17：

2x + 5 − 5 = 17 − 5

2x = 12

2x ÷ 2 = 12 ÷ 2

x = 6

Another common form is 3x − 4 = 20. Add 4 to both sides first, then divide by 3:

另一种常见形式是 3x − 4 = 20。先两边加上 4，再除以 3：

3x − 4 + 4 = 20 + 4

3x = 24

x = 8

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5. Expanding Brackets First | 先去括号

When a linear equation contains brackets, expand them before using the balance method. This rule follows the order of operations in reverse.

当一元一次方程含有括号时，先展开括号，再使用天秤法。这一规则是逆用运算顺序。

Solve 3(x + 2) = 21:

解 3(x + 2) = 21：

3x + 6 = 21

3x = 15

x = 5

You can also solve by dividing both sides by the coefficient of the bracket first: 3(x + 2) ÷ 3 = 21 ÷ 3, giving x + 2 = 7, then x = 5. Both methods are valid.

你也可以先两边除以括号前的系数：3(x + 2) ÷ 3 = 21 ÷ 3，得到 x + 2 = 7，再得 x = 5。两种方法都正确。

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6. Equations with Variables on Both Sides | 两边含未知数的方程

Some equations have the unknown on both sides, such as 4x + 3 = 2x + 11. Start by collecting the x terms on one side and the numbers on the other.

有些方程两边都含有未知数，例如 4x + 3 = 2x + 11。先把含 x 的项移到一边，把数字移到另一边。

Subtract 2x from both sides:

两边同时减去 2x：

4x − 2x + 3 = 2x − 2x + 11

2x + 3 = 11

Then subtract 3 and divide by 2:

然后减去 3，再除以 2：

2x = 8

x = 4

Remember that you can move terms to either side as long as you apply the inverse operation correctly. Choose the side that keeps the coefficient positive when possible.

请记住，只要正确使用逆运算，你可以把项移到任意一边。尽量选择使系数为正的一边。

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7. Equations with Fractions | 含分数的方程

To solve an equation like x/3 + 2 = 7, first subtract 2 from both sides, then multiply by the denominator.

要解像 x/3 + 2 = 7 这样的方程，先两边减去 2，再乘以分母。

x/3 + 2 − 2 = 7 − 2

x/3 = 5

x/3 × 3 = 5 × 3

x = 15

For equations like (2x + 1)/3 = 5, multiply both sides by the denominator first, then solve the two-step equation:

对于像 (2x + 1)/3 = 5 这样的方程，先两边乘以分母，再解两步方程：

2x + 1 = 15

2x = 14

x = 7

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8. Writing Equations from Word Problems | 根据文字题建立方程

Many Cambridge exam questions ask you to form an equation from a real-life situation. Read the problem carefully and translate each phrase into algebra.

许多剑桥考试题要求你从实际情境中建立方程。仔细阅读题目，把每个短语转化为代数语言。

• “Three times a number” means 3x | “一个数的三倍” 表示 3x
• “Five more than a number” means x + 5 | “比一个数多五” 表示 x + 5
• “The sum of a number and 8 is 20” means x + 8 = 20 | “一个数与 8 的和是 20” 表示 x + 8 = 20

Example: I think of a number, multiply it by 4, then add 7. The answer is 31. Let the number be x:

例子：我想一个数，乘以 4，再加 7，结果是 31。设这个数为 x：

4x + 7 = 31

4x = 24

x = 6

Always define your variable clearly before forming the equation, then check that the answer makes sense in the original context.

在建立方程前，始终清晰地设出变量，然后检查答案在原始语境中是否合理。

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9. Checking Your Solution | 检验你的解

Always substitute your final x value back into the original equation to verify it works. This habit catches careless mistakes and shows full working in exams.

始终把你的最终 x 值代回原方程进行检验。这个习惯可以避免粗心错误，并在考试中展示完整步骤。

For 2x + 5 = 17, if you found x = 6, check:

对于 2x + 5 = 17，如果你得到 x = 6，检验：

2 × 6 + 5 = 12 + 5 = 17

Since the left side equals the right side, the solution is correct.

因为左边等于右边，所以解是正确的。

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10. Common Mistakes to Avoid | 常见错误

Many marks are lost through simple errors. The table below shows the most common mistakes and how to correct them.

很多分数因简单错误而丢失。下表列出最常见的错误及改正方法。

Mistake | 错误	Correct approach | 正确处理
Forgetting to operate on both sides | 忘记两边同时操作	Always apply the same operation to both sides | 始终对两边进行相同操作
Undoing operations in the wrong order | 逆运算顺序错误	Undo addition/subtraction before multiplication/division | 先逆加减，再逆乘除
Mis-handling negative signs | 负号处理错误	Use inverse operations carefully with negatives | 处理负数时谨慎使用逆运算
Expanding brackets incorrectly | 括号展开错误	Multiply every term inside the bracket | 乘以括号内的每一项

Pay particular attention to equations like 5x − 3 = 2x + 9. A common error is to subtract 2x from only one side or to subtract 3 incorrectly.

特别注意像 5x − 3 = 2x + 9 这样的方程。常见错误是只从一边减去 2x，或者错误地减去 3。

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11. Exam-Style Practice Questions | 考试题型练习

Try these questions to test your understanding. Solve each equation, showing every step.

尝试以下题目检验你的理解。解每个方程，展示每一步。

• Solve 5x + 2 = 37 | 解 5x + 2 = 37
• Solve 2(x − 3) = 14 | 解 2(x − 3) = 14
• Solve 3x + 4 = x + 16 | 解 3x + 4 = x + 16
• Solve x/4 − 1 = 6 | 解 x/4 − 1 = 6

Worked solutions:

参考解答：

For 5x + 2 = 37: subtract 2, then divide by 5, giving x = 7.

对于 5x + 2 = 37：先减 2，再除以 5，得到 x = 7。

For 2(x − 3) = 14: expand to 2x − 6 = 14, add 6, then divide by 2, giving x = 10.

对于 2(x − 3) = 14：展开得 2x − 6 = 14，加 6，再除以 2，得到 x = 10。

For 3x + 4 = x + 16: subtract x and subtract 4, giving 2x = 12, so x = 6.

对于 3x + 4 = x + 16：减去 x 和 4，得到 2x = 12，因此 x = 6。

For x/4 − 1 = 6: add 1, then multiply by 4, giving x = 28.

对于 x/4 − 1 = 6：加 1，再乘以 4，得到 x = 28。

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12. Summary and Key Tips | 总结与关键提示

To master linear equations, always use the balance method, undo operations in the correct order, and check your answer by substitution. Practice translating word problems into equations and be careful with negative numbers and brackets.

要掌握一元一次方程，始终使用天秤法，按正确顺序进行逆运算，并通过代入检验答案。练习将文字题转化为方程，并注意负数和括号。

Keep your working neat and write one operation per line. This makes it easier to spot mistakes and helps you gain method marks in Cambridge exams.

保持解题过程整洁，每行只写一步操作。这样更容易发现错误，并帮助你在剑桥考试中获得步骤分。

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📚 KS3 Cambridge Mathematics: Solving Linear Equations | 解一元一次方程

Linear equations are the foundation of algebra at Key Stage 3. In Cambridge Lower Secondary Mathematics, you are expected to solve equations using inverse operations and the balancing method, even when brackets or variables on both sides are involved. This article is suitable for KS3 students preparing for Cambridge Checkpoint or school assessments.

线性方程是 KS3 代数的基础。在剑桥初中数学中，你需要掌握使用逆运算和平衡法解方程，即使方程中含有括号或两边都有变量。本文适合准备剑桥 Checkpoint 或校内评估的 KS3 学生。

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1. What Is a Linear Equation? | 什么是线性方程？

A linear equation is an algebraic statement in which the highest power of the unknown is 1. It usually contains an equals sign and exactly one unknown, such as x or y. The solution is the value of the unknown that makes the statement true.

线性方程是指未知数的最高次数为 1 的代数等式。它通常包含一个等号和一个未知数，例如 x 或 y。方程的解就是使等式成立的未知数的值。

Examples of linear equations:

线性方程的例子：

• 2x + 3 = 11
• 5y – 7 = 2y + 8
• 3(x – 4) = 9
• x/4 + 1 = 6

Expressions like x² + 2 = 6 are not linear because the unknown is squared. Equations such as xy = 12 are also not linear because there are two unknowns multiplied together.

像 x² + 2 = 6 这样的表达式不是线性方程，因为未知数被平方了。像 xy = 12 这样的方程也不是线性方程，因为两个未知数相乘。

In Cambridge KS3, you will mainly work with one unknown and a variable power of 1. Recognising this helps you choose the correct solving method.

在剑桥 KS3 阶段，你主要会接触一个未知数且变量次数为 1 的方程。识别这一点有助于你选择正确的求解方法。

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2. The Balancing Method | 平衡法

An equation is like a balance scale. Whatever you do to one side, you must do to the other to keep it balanced. This is the core idea behind solving any linear equation and is heavily assessed in Cambridge Checkpoint.

方程就像一个天平。你对一边做的任何操作，都必须对另一边做同样的操作，以保持平衡。这是解任何线性方程的核心思想，也是剑桥 Checkpoint 中重点考查的内容。

For example, to solve x + 3 = 7, subtract 3 from both sides:

例如，要解 x + 3 = 7，两边同时减去 3：

x + 3 – 3 = 7 – 3 ⇒ x = 4

The symbol ⇒ means ‘leads to’ and shows the next step in your working. You should write one operation per line to make your method clear.

符号 ⇒ 表示“推出”，显示解题过程中的下一步。你应该每行只写一步运算，使解题过程更清晰。

You can check the balance by substituting x = 4 into x + 3 = 7. The left side becomes 4 + 3 = 7, which equals the right side.

你可以通过把 x = 4 代入 x + 3 = 7 来检验平衡。左边变成 4 + 3 = 7，与右边相等。

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3. Solving One-Step Equations | 解一步方程

One-step equations require only one inverse operation to isolate the unknown. Addition and subtraction are inverse operations, as are multiplication and division. Choosing the correct inverse is the key skill here.

一步方程只需要一次逆运算就能求出未知数。加法和减法是互逆运算，乘法和除法也是互逆运算。选择正确的逆运算这里的关键技能。

Type 1: Addition equation. Solve x + 5 = 12. Subtract 5 from both sides to get x = 7.

类型 1：加法方程。解 x + 5 = 12。两边同时减去 5，得到 x = 7。

x = 12 – 5 = 7

Type 2: Subtraction equation. Solve x – 6 = 9. Add 6 to both sides to get x = 15.

类型 2：减法方程。解 x – 6 = 9。两边同时加上 6，得到 x = 15。

x = 9 + 6 = 15

Type 3: Multiplication equation. Solve 4x = 20. Divide both sides by 4 to get x = 5.

类型 3：乘法方程。解 4x = 20。两边同时除以 4，得到 x = 5。

x = 20 ÷ 4 = 5

Type 4: Division equation. Solve y ÷ 3 = 9. Multiply both sides by 3 to get y = 27.

类型 4：除法方程。解 y ÷ 3 = 9。两边同时乘以 3，得到 y = 27。

y = 9 × 3 = 27

When the coefficient of x is negative, divide by the negative coefficient. For example, -2x = 8 gives x = 8 ÷ (-2) = -4.

当 x 的系数为负数时，要除以这个负系数。例如 -2x = 8 得到 x = 8 ÷ (-2) = -4。

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4. Solving Two-Step Equations | 解两步方程

A two-step equation has two operations acting on the unknown. You must undo them in reverse order, following the reverse of the order of operations. This means you handle addition or subtraction before multiplication or division.

两步方程中有两种运算作用于未知数。你必须按照运算顺序的逆序来逐步消去它们。也就是说，先处理加减法，再处理乘除法。

Example: Solve 3x + 2 = 14.

例子：解方程 3x + 2 = 14。

3x + 2 = 14

3x = 14 – 2 = 12

x = 12 ÷ 3 = 4

First subtract 2, then divide by 3. Always check by substituting x = 4 back into the original equation.

先减去 2，然后除以 3。始终要把 x = 4 代回原方程进行检验。

Another example: Solve x/5 – 2 = 3. First add 2 to both sides, then multiply by 5.

另一个例子：解方程 x/5 – 2 = 3。首先两边加 2，然后两边乘以 5。

x/5 = 3 + 2 = 5

x = 5 × 5 = 25

The reverse order is essential. If you divide before subtracting, you will often make a mistake with the constant term.

逆序处理非常重要。如果你先除后减，常常会在常数项上出错。

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5. Equations with Brackets | 带括号的方程

When an equation contains brackets, expand them first using the distributive law. Then collect like terms and solve using the balancing method. This topic appears frequently in Cambridge Checkpoint tests.

当方程中含有括号时，先用分配律展开括号。然后合并同类项，再用平衡法求解。这个主题经常出现在剑桥 Checkpoint 测试中。

Example: Solve 2(x + 3) = 16.

例子：解方程 2(x + 3) = 16。

2(x + 3) = 16

2x + 6 = 16

2x = 10

x = 5

Remember that a minus sign before a bracket changes every sign inside, for example -2(x – 3) = -2x + 6. This is a very common source of errors.

记住括号前有负号会改变括号内每一项的符号，例如 -2(x – 3) = -2x + 6。这是非常常见的错误来源。

Example with a negative sign: Solve -3(2x – 1) = 15.

带负号的例子：解方程 -3(2x – 1) = 15。

-6x + 3 = 15

-6x = 12

x = 12 ÷ (-6) = -2

If the equation has brackets on both sides, expand them all before collecting like terms.

如果方程两边都有括号，要先展开所有括号，再合并同类项。

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6. Equations with Variables on Both Sides | 两边都含未知数的方程

If the unknown appears on both sides of the equation, first collect all unknown terms on one side and all numbers on the other side. Then simplify and solve.

如果未知数出现在方程两边，首先把所有的未知项移到一边，把所有数字移到另一边。然后化简并求解。

Example: Solve 5x – 3 = 2x + 9.

例子：解方程 5x – 3 = 2x + 9。

5x – 2x = 9 + 3

3x = 12

x = 4

To decide which side to move the variable to, choose the side with the larger coefficient. This keeps the variable positive.

为了决定把变量移到哪一边，选择系数较大的一边。这样能让未知数保持为正。

• 5x – 2x = 3x, which is positive.
• If you moved 5x to the right, you would get -3x = 12, leading to x = -4 but with an extra negative step.

可补充：

5x – 2x = 3x，为正。如果把 5x 移到右边，会得到 -3x = 12，虽然也能得到 x = -4，但多了一个负号步骤。

Another example: Solve 7x + 2 = 3x – 6.

另一个例子：解方程 7x + 2 =

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📚 Solving Linear Equations for KS3 Cambridge Maths | KS3 剑桥数学：解一元一次方程

Linear equations are the foundation of algebra at Key Stage 3. Whether you are working through the Cambridge Lower Secondary Maths course or preparing for a Checkpoint test, being able to solve an equation confidently is one of the most important skills you can practise. This article explains the key methods step by step, with examples, common errors and practice questions.

一元一次方程是 KS3 代数的基础。无论你正在学习剑桥初中数学课程，还是准备 Checkpoint 测试，能够自信地解方程都是你可以练习的最重要技能之一。本文将逐步解释关键方法，配以例题、常见错误和练习题。

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1. What is a Linear Equation? | 什么是一元一次方程？

A linear equation is an equation where the unknown (usually written as x, y or n) is only raised to the power of 1. There are no x², x³, square roots or variables multiplied together. For example, 2x + 3 = 11 is a linear equation.

一元一次方程是指未知数（通常写作 x、y 或 n）的指数仅为 1 的方程。方程中不含 x²、x³、平方根，也不含变量相乘的形式。例如，2x + 3 = 11 就是一个一元一次方程。

The word ‘linear’ comes from ‘line’ because the graph of y = 2x + 3 is a straight line. Solving the equation tells us the value of the unknown that makes the statement true.

“线性” 一词来源于 “直线”，因为 y = 2x + 3 的图像是一条直线。解方程就是求出使等式成立的未知数的值。

2x + 3 = 11 is true when x = 4 because 2 × 4 + 3 = 11

2x + 3 = 11 在 x = 4 时成立，因为 2 × 4 + 3 = 11

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2. Key Vocabulary and Notation | 关键术语与符号

Before solving equations, you need to be confident with the language used in questions. The table below shows the most common terms you will meet in Cambridge KS3 maths.

在解方程之前，你需要熟悉题目中常用的数学语言。下表列出了剑桥 KS3 数学中最常遇到的术语。

English term	中文术语	Example
Variable	变量	x in 3x + 2
Coefficient	系数	3 in 3x + 2
Constant	常数	2 in 3x + 2
Left-hand side (LHS)	等号左边	2x + 3 in 2x + 3 = 11
Right-hand side (RHS)	等号右边	11 in 2x + 3 = 11
Solution	解	x = 4

An equation is different from an expression. An expression such as 3x + 2 does not have an equals sign, so it cannot be solved directly. An equation such as 3x + 2 = 14 has an equals sign and can be solved to find the value of x.

方程与表达式不同。表达式如 3x + 2 没有等号，因此不能直接求解。方程如 3x + 2 = 14 有等号，可以求出 x 的值。

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3. The Balance Method | 天平法原理

Think of an equation as a pair of balanced scales. The equals sign is the pivot. Whatever you do to one side, you must do exactly the same to the other side to keep the scales balanced.

把方程想象成一台平衡的天平。等号就是支点。无论你对一边做什么运算，都必须对另一边做完全相同的运算，才能保持天平平衡。

You can add, subtract, multiply or divide both sides by the same non-zero number. The aim is to isolate the unknown on one side of the equals sign.

你可以对两边同时加上、减去、乘以或除以同一个不为零的数。目标是把未知数单独留在等号的一边。

If x + 3 = 7, then x + 3 − 3 = 7 − 3, so x = 4

如果 x + 3 = 7，那么 x + 3 − 3 = 7 − 3，所以 x = 4

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4. Solving One-Step Equations | 解一步方程

One-step equations need just one operation to find the unknown. Start with addition and subtraction, then move on to multiplication and division.

一步方程只需要一次运算就能求出未知数。先练习加法和减法，再练习乘法和除法。

For x + 5 = 13, subtract 5 from both sides to get x = 8. For x − 7 = 20, add 7 to both sides to get x = 27.

对于 x + 5 = 13，两边同时减去 5，得到 x = 8。对于 x − 7 = 20，两边同时加上 7，得到 x = 27。

x + 5 = 13 → x = 13 − 5 → x = 8

x − 7 = 20 → x = 20 + 7 → x = 27

For 3x = 27, divide both sides by 3 to get x = 9. For x ÷ 4 = 6, multiply both sides by 4 to get x = 24.

对于 3x = 27，两边同时除以 3，得到 x = 9。对于 x ÷ 4 = 6，两边同时乘以 4，得到 x = 24。

3x = 27 → x = 27 ÷ 3 → x = 9

x ÷ 4 = 6 → x = 6 × 4 → x = 24

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5. Solving Two-Step Equations | 解两步方程

Two-step equations involve two operations. Always reverse the order of operations: undo addition or subtraction first, then undo multiplication or division.

两步方程包含两种运算。始终按运算的逆序求解：先消去加法或减法，再消去乘法或除法。

Solve 2x + 3 = 11. First subtract 3 from both sides to get 2x = 8. Then divide both sides by 2 to get x = 4.

解方程 2x + 3 = 11。首先两边同时减去 3，得到 2x = 8。然后两边同时除以 2，得到 x = 4。

2x + 3 = 11 → 2x = 11 − 3 → 2x = 8 → x = 8 ÷ 2 → x = 4

2x + 3 = 11 → 2x = 11 − 3 → 2x = 8 → x = 8 ÷ 2 → x = 4

Solve 5x − 4 = 21. Add 4 to both sides to get 5x = 25. Then divide both sides by 5 to get x = 5.

解方程 5x − 4 = 21。两边同时加上 4，得到 5x = 25。然后两边同时除以 5，得到 x = 5。

5x − 4 = 21 → 5x = 21 + 4 → 5x = 25 → x = 25 ÷ 5 → x = 5

5x − 4 = 21 → 5x = 21 + 4 → 5x = 25 → x = 25 ÷ 5 → x = 5

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6. Equations with Brackets | 含括号的方程

If an equation contains brackets, expand them first using the distributive law. Then solve the resulting equation in the usual way.

如果方程中含有括号，先用分配律展开括号。然后按通常方法解得到的方程。

Solve 3(x + 2) = 18. Expand the left-hand side to get 3x + 6 = 18. Subtract 6 from both sides to get 3x = 12. Divide by 3 to get x = 4.

解方程 3(x + 2) = 18。展开左边得到 3x + 6 = 18。两边同时减去 6，得到 3x = 12。再除以 3，得到 x = 4。

3(x + 2) = 18 → 3x + 6 = 18 → 3x = 12 → x = 4

3(x + 2) = 18 → 3x + 6 = 18 → 3x = 12 → x = 4

Solve 2(3x − 1) = 16. Expand to get 6x − 2 = 16. Add 2 to both sides to get 6x = 18. Divide by 6 to get x = 3.

解方程 2(3x − 1) = 16。展开得到 6x − 2 = 16。两边同时加上 2，得到 6x = 18。再除以 6，得到 x = 3。

2(3x − 1) = 16 → 6x − 2 = 16 → 6x = 18 → x = 3

2(3x − 1) = 16 → 6x − 2 = 16 → 6x = 18 → x = 3

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7. Equations with Unknowns on Both Sides | 未知数在等号两边的方程

When the unknown appears on both sides of the equals sign, collect the variable terms on one side and the constant terms on the other side. A good rule is to eliminate the smaller coefficient of x first.

当未知数出现在等号两边时，把含未知数的项移到一边，把常数项移到另一边。一个常用的方法是先消去系数较小的 x 项。

Solve 5x + 2 = 3x + 10. Subtract 3x from both sides to get 2x + 2 = 10. Subtract 2 from both sides to get 2x = 8. Divide by 2 to get x = 4.

解方程 5x + 2 = 3x + 10。两边同时减去 3x，得到 2x + 2 = 10。两边同时减去 2，得到 2x = 8。再除以 2，得到 x = 4。

5x + 2 = 3x + 10 → 2x + 2 = 10 → 2x = 8 → x = 4

5x + 2 = 3x + 10 → 2x + 2 = 10 → 2x = 8 → x = 4

Solve 7x − 5 = 2x + 20. Subtract 2x from both sides to get 5x − 5 = 20. Add 5 to both sides to get 5x = 25. Divide by 5 to get x = 5.

解方程 7x − 5 = 2x + 20。两边同时减去 2x，得到 5x − 5 = 20。两边同时加上 5，得到 5x = 25。再除以 5，得到 x = 5。

7x − 5 = 2x + 20 → 5x − 5 = 20 → 5x = 25 → x = 5

7x − 5 = 2x + 20 → 5x − 5 = 20 → 5x = 25 → x = 5

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8. Equations Involving Fractions | 涉及分数的方程

When an equation includes a fraction, you can remove the fraction by multiplying both sides by the denominator. If there is more than one fraction, multiply by the lowest common denominator.

当方程中含有分数时，可以将两边同时乘以分母来消去分数。如果方程中有多个分数，则乘以它们的最小公分母。

Solve x ÷ 3 + 2 = 5. Subtract 2 from both sides to get x ÷ 3 = 3. Multiply both sides by 3 to get x = 9.

解方程 x ÷ 3 + 2 = 5。两边同时减去 2，得到 x ÷ 3 = 3。两边同时乘以 3，得到 x = 9。

x ÷ 3 + 2 = 5 → x ÷ 3 = 3 → x = 3 × 3 → x = 9

x ÷ 3 + 2 = 5 → x ÷ 3 = 3 → x = 3 × 3 → x = 9

Solve (2x) ÷ 5 − 1 = 7. Add 1 to both sides to get (2x) ÷ 5 = 8. Multiply both sides by 5 to get 2x = 40. Divide by 2 to get x = 20.

解方程 (2x) ÷ 5 − 1 = 7。两边同时加上 1，得到 (2x) ÷ 5 = 8。两边同时乘以 5，得到 2x = 40。再除以 2，得到 x = 20。

(2x) ÷ 5 − 1 = 7 → (2x) ÷ 5 = 8 → 2x = 40 → x = 20

(2x) ÷ 5 − 1 = 7 → (2x) ÷ 5 = 8 → 2x = 40 → x = 20

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9. Checking Your Solution | 检验你的解

Always substitute your answer back into the original equation to check that it works. This is especially important in tests, because it can catch small mistakes quickly.

始终将你的答案代回原方程进行检验。这在考试中尤其重要，因为可以快速发现小的错误。

For 2x + 3 = 11, if x = 4, the left-hand side is 2 × 4 + 3 = 8 + 3 = 11. The right-hand side is 11. Both sides match, so the solution is correct.

对于 2x + 3 = 11，如果 x = 4，左边为 2 × 4 + 3 = 8 + 3 = 11。右边为 11。两边相等，因此解是正确的。

Check: 2 × 4 + 3 = 11 → 8 + 3 = 11 → 11 = 11 ✔

检验：2 × 4 + 3 = 11 → 8 + 3 = 11 → 11 = 11 ✔

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10. Common Mistakes to Avoid | 常见错误与避免方法

Many students lose marks on algebra not because they do not understand the topic, but because they make small avoidable errors. Watch out for the following common mistakes.

许多学生在代数上丢分，并不是因为不懂知识点

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📚 Pythagoras’ Theorem in Cambridge KS3 Maths | 剑桥 KS3 数学：勾股定理

Pythagoras’ theorem is one of the most important results in Cambridge KS3 mathematics. It connects the three sides of a right-angled triangle and gives you a reliable way to find a missing length. This article explains the theorem step by step, with worked examples, common errors and exam-style practice.

勾股定理是剑桥 KS3 数学中最重要的结论之一。它将直角三角形的三条边联系起来，为你提供了一种可靠的方法来求缺失的边长。本文逐步解释该定理，并配有例题、常见错误和考试风格练习。

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1. What Is Pythagoras’ Theorem? | 什么是勾股定理？

Pythagoras’ theorem only works in right-angled triangles. It says that the square of the longest side is equal to the sum of the squares of the other two sides.

勾股定理只适用于直角三角形。它表明最长边的平方等于另外两条边的平方之和。

a² + b² = c²

Here, c is the hypotenuse, the side opposite the right angle. The letters a and b are the two shorter sides.

这里 c 是斜边，也就是直角所对的边。字母 a 和 b 是两条较短的直角边。

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2. Identifying the Hypotenuse | 识别斜边

The hypotenuse is always the longest side of a right-angled triangle. It is also always opposite the right angle, never adjacent to it.

斜边始终是直角三角形中最长的边。它也总是与直角相对，从不与直角相邻。

For example, in a triangle with sides 3 cm, 4 cm and 5 cm, the side of 5 cm must be the hypotenuse because it is the largest value.

例如，在一个边长分别为 3 cm、4 cm 和 5 cm 的三角形中，5 cm 的边一定是斜边，因为它是最大的值。

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3. The Formula a² + b² = c² | 公式 a² + b² = c²

You can rearrange the formula depending on which side is missing. To find the hypotenuse, add the squares of the two shorter sides and then take the square root.

你可以根据缺失的是哪条边来调整公式。要求斜边，先求两条较短边的平方和，然后再开平方。

c = √(a² + b²)

To find a shorter side, subtract the square of the known shorter side from the square of the hypotenuse, then take the square root.

要求一条直角边，用斜边的平方减去已知直角边的平方，然后再开平方。

a = √(c² − b²)

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4. Worked Example: Finding the Hypotenuse | 例题：求斜边

A right-angled triangle has shorter sides of 6 cm and 8 cm. Find the length of the hypotenuse.

一个直角三角形的两条直角边分别为 6 cm 和 8 cm。求斜边的长度。

Substitute a = 6 and b = 8 into the formula:

将 a = 6 和 b = 8 代入公式：

c² = 6² + 8² = 36 + 64 = 100

c = √100 = 10 cm

Always remember to give units in your final answer.

请始终记得在最终答案中写出单位。

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5. Worked Example: Finding a Shorter Side | 例题：求直角边

In a right-angled triangle, the hypotenuse is 13 cm and one shorter side is 5 cm. Find the missing side.

在一个直角三角形中，斜边为 13 cm，一条直角边为 5 cm。求缺失的边。

Use the rearranged formula with c = 13 and b = 5:

使用调整后的公式，令 c = 13 和 b = 5：

a² = 13² − 5² = 169 − 25 = 144

a = √144 = 12 cm

This checks because 5, 12, 13 is a well-known Pythagorean triple.

这是成立的，因为 5、12、13 是一组著名的勾股数。

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6. Checking Right-Angled Triangles | 判断直角三角形

You can use the theorem in reverse. If the three sides satisfy a² + b² = c², then the triangle must be right-angled.

你可以反过来使用该定理。如果三条边满足 a² + b² = c²，那么这个三角形一定是直角三角形。

For example, check whether 4 cm, 5 cm and 6 cm form a right-angled triangle:

例如，判断 4 cm、5 cm 和 6 cm 是否构成直角三角形：

4² + 5² = 16 + 25 = 41, but 6² = 36

Since 41 ≠ 36, the triangle is not right-angled.

因为 41 ≠ 36，所以这个三角形不是直角三角形。

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7. Real-Life Applications | 实际应用

Pythagoras’ theorem is used in many real situations. For example, a ladder leaning against a wall forms a right-angled triangle with the ground.

勾股定理在许多实际情况中都有应用。例如，斜靠在墙上的梯子与地面构成一个直角三角形。

If a 5 m ladder is placed 2 m away from a wall, the height reached on the wall is:

如果一把 5 m 长的梯子底部离墙 2 m，那么它在墙上达到的高度为：

height = √(5² − 2²) = √(25 − 4) = √21 ≈ 4.58 m

This kind of calculation is common in construction, navigation and design.

这类计算在建筑、导航和设计中都很常见。

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8. Pythagorean Triples | 勾股数

A Pythagorean triple is a set of three whole numbers that satisfy the theorem exactly. These are useful for mental maths and quick checking.

勾股数是一组三个整数，能精确满足定理。它们对心算和快速检验很有用。

Shorter sides | 直角边	Hypotenuse | 斜边
3, 4	5
5, 12	13
7, 24	25
8, 15	17

Recognising these triples can save time in exams.

识别这些勾股数可以在考试中节省时间。

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9. Common Mistakes | 常见错误

One common mistake is adding or subtracting the sides before squaring them. Always square first, then add or subtract.

一个常见错误是先对边进行加减，然后再平方。一定要先平方，然后再加减。

• Wrong: (6 + 8)² = 196

  错误：(6 + 8)² = 196

• Right: 6² + 8² = 36 + 64 = 100

  正确：6² + 8² = 36 + 64 = 100

Another mistake is using the hypotenuse in the wrong position. The hypotenuse must be on its own on one side of the equation.

另一个错误是把斜边放在错误的位置。斜边必须单独位于方程的一边。

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10. Exam-Style Practice | 考试风格练习

Try these questions to test your understanding. Answers are given in the next section.

尝试以下问题来检验你的理解。答案在下一节中给出。

• A right-angled triangle has shorter sides 9 cm and 12 cm. Find the hypotenuse.

  一个直角三角形两条直角边分别为 9 cm 和 12 cm。求斜边。

• The hypotenuse is 17 cm and one shorter side is 8 cm. Find the other side.

  斜边为 17 cm，一条直角边为 8 cm。求另一条直角边。

• Do 6 cm, 7 cm and 9 cm form a right-angled triangle?

  6 cm、7 cm 和 9 cm 能否构成直角三角形？

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11. Answers and Explanations | 答案与解析

For the first question, compute c² = 9² + 12² = 81 + 144 = 225, so c = √225 = 15 cm.

第一题，计算 c² = 9² + 12² = 81 + 144 = 225，所以 c = √225 = 15 cm。

For the second question, compute b² = 17² − 8² = 289 − 64 = 225, so b = √225 = 15 cm.

第二题，计算 b² = 17² − 8² = 289 − 64 = 225，所以 b = √225 = 15 cm。

For the third question, check 6² + 7² = 36 + 49 = 85, but 9² = 81. Since 85 ≠ 81, it is not right-angled.

第三题，检验 6² + 7² = 36 + 49 = 85，而 9² = 81。因为 85 ≠ 81，所以不是直角三角形。

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12. Summary | 总结

Pythagoras’ theorem applies only to right-angled triangles. Remember that c is always the hypotenuse, and use a² + b² = c² to find a missing hypotenuse, or a² = c² − b² to find a missing shorter side.

勾股定理只适用于直角三角形。记住 c 始终是斜边，使用 a² + b² = c² 求缺失的斜边，或使用 a² = c² − b² 求缺失的直角边。

Practise recognising Pythagorean triples and always show your working clearly with units.

练习识别勾股数，并始终清晰展示计算过程和单位。

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📚 Solving Linear Equations for KS3 Cambridge Maths | 剑桥初中数学：解一元一次方程

Linear equations are at the heart of algebra in the Cambridge KS3 curriculum. Whether you are working through a textbook exercise or a Checkpoint test, being able to form and solve an equation such as x + 5 = 12 is an essential skill. This guide walks through the key ideas step by step, with examples and checks you can use in your own work.

线性方程是剑桥 KS3 数学课程中代数的核心内容。无论你是在做课本练习还是 Checkpoint 测试，能够建立并求解像 x + 5 = 12 这样的方程都是一项基本技能。本指南将逐步讲解关键概念，并给出例题和检验方法，帮助你在独立练习中使用。

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1. What Is a Linear Equation? | 什么是一元一次方程？

A linear equation is an equation where the unknown variable appears only to the first power (no squares, cubes, or roots). In KS3 work, the variable is usually written as x, but it can be any letter, such as y, n, or t. The word “linear” means that the graph of the equation is a straight line. For example, x + 2 = 7, 3x = 18 and 2x − 5 = 9 are all linear equations. An equation has two sides separated by an equals sign. The equals sign tells us that the left side and the right side have the same value. Solving the equation means finding the value of the unknown that makes this true.

一元一次方程是指未知数只出现一次幂（没有平方、立方或根号）的方程。在 KS3 阶段，未知数通常用 x 表示，但也可以是任何字母，如 y、n 或 t。”线性” 一词表示方程的图像是一条直线。例如，x + 2 = 7、3x = 18 和 2x − 5 = 9 都是一元一次方程。方程有等号隔开的两边。等号告诉我们左边和右边的值相等。解方程就是找出使等式成立的未知数的值。

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2. Key Terms and Symbols | 关键术语与符号

Before practising, it helps to know the language used in questions. The variable is the unknown letter, such as x. A coefficient is the number multiplying the variable, so in 4x the coefficient is 4. A constant is a number on its own, such as 5 in x + 5. An expression has no equals sign, for example 3x + 2. An equation has an equals sign, for example 3x + 2 = 11. We use symbols including +, −, ×, ÷ and = to build equations. Multiplication is often written without the × sign, so 3x means 3 × x.

在练习之前，了解题目中常用的语言很有帮助。变量是未知的字母，例如 x。系数是与变量相乘的数字，因此在 4x 中系数是 4。常数是单独出现的数字，例如 x + 5 中的 5。表达式没有等号，如 3x + 2。方程有等号，如 3x + 2 = 11。我们使用 +、−、×、÷ 和 = 等符号来建立方程。乘法通常省略 × 号，所以 3x 表示 3 × x。

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3. The Balancing Method | 平衡法原理

The balancing method is based on a simple idea: whatever you do to one side of an equation, you must also do to the other side. Imagine an old-fashioned balance scale. If the two pans are level, they stay level only when you add the same weight to both sides, remove the same weight from both sides, multiply both sides by the same number, or divide both sides by the same non-zero number. This keeps the equation true. For example, starting with x + 3 = 10, we can subtract 3 from both sides to remove the constant and isolate x.

平衡法基于一个简单的思想：对方程的一边做什么，对另一边也必须做同样的操作。想象一架老式天平。如果两个托盘平衡，只有当两边加上相同的重量、减去相同的重量、乘以相同的数或除以相同的非零数时，它们才能保持平衡。这样才能使等式仍然成立。例如，从 x + 3 = 10 出发，我们可以两边同时减去 3，以消去常数并单独留下 x。

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4. Solving One-Step Equations | 解一步方程

A one-step equation needs only one operation to isolate the variable. If the equation is x + 5 = 12, subtract 5 from both sides. The solution is x = 7. If the equation is x − 4 = 9, add 4 to both sides to get x = 13. If the equation is 3x = 18, divide both sides by 3 to get x = 6. If the equation is x ÷ 5 = 4, multiply both sides by 5 to get x = 20. In every case, use the inverse operation: addition is undone by subtraction, subtraction by addition, multiplication by division, and division by multiplication.

一步方程只需要一次运算就能单独求出未知数。如果方程是 x + 5 = 12，两边同时减去 5，解为 x = 7。如果方程是 x − 4 = 9，两边同时加上 4，得到 x = 13。如果方程是 3x = 18，两边同时除以 3，得到 x = 6。如果方程是 x ÷ 5 = 4，两边同时乘以 5，得到 x = 20。每种情况都要使用逆运算：加法由减法还原，减法由加法还原，乘法由除法还原，除法由乘法还原。

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5. Solving Two-Step Equations | 解两步方程

Two-step equations combine two operations. For example, in 2x + 3 = 13, the variable x is first multiplied by 2, then 3 is added. To solve, undo the operations in reverse order. First subtract 3 from both sides, then divide by 2. Step 1: 2x + 3 − 3 = 13 − 3 gives 2x = 10. Step 2: 2x ÷ 2 = 10 ÷ 2 gives x = 5. Always undo addition or subtraction before undoing multiplication or division when the equation has the form ax + b = c. This reverse order is important because it peels away the outer layer first.

两步方程包含两种运算。例如，在 2x + 3 = 13 中，未知数 x 先乘以 2，然后加上 3。求解时要按相反的顺序逆运算。首先两边同时减去 3，再除以 2。第一步：2x + 3 − 3 = 13 − 3，得到 2x = 10。第二步：2x ÷ 2 = 10 ÷ 2，得到 x = 5。当方程形如 ax + b = c 时，一定要先撤销加减法，再撤销乘除法。这个相反顺序很重要，因为它先剥去最外层的运算。

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6. Expanding Brackets First | 先展开括号

Some equations contain brackets, such as 3(x + 2) = 21. The bracket means 3 times the entire sum (x + 2). Before using the balancing method, expand the bracket by multiplying each term inside by the number outside. For 3(x + 2), multiply x by 3 and 2 by 3 to get 3x + 6. The equation becomes 3x + 6 = 21. Now it is a two-step equation. Subtract 6 from both sides to get 3x = 15, then divide by 3 to get x = 5. Always look for brackets early so you can simplify the equation first.

有些方程含有括号，例如 3(x + 2) = 21。括号表示 3 乘以整个和 (x + 2)。使用平衡法之前，要先用外面的数乘以括号内的每一项，把括号展开。对于 3(x + 2)，把 x 乘以 3，把 2 乘以 3，得到 3x + 6。方程变为 3x + 6 = 21。现在它是一个两步方程。两边同时减去 6，得到 3x = 15，再除以 3，得到 x = 5。一定要尽早寻找括号，以便先化简方程。

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7. Unknowns on Both Sides | 未知数在两侧

In equations like 5x + 2 = 3x + 8, the unknown x appears on both sides. The first goal is to collect like terms so that x appears only on one side. Subtract 3x from both sides: 5x − 3x + 2 = 3x − 3x + 8, which simplifies to 2x + 2 = 8. Then subtract 2 from both sides to get 2x = 6. Finally divide by 2: x = 3. You can also move the smaller variable term to the other side by using inverse operations. The key is to keep the equation balanced at every step so the equals sign remains true.

在像 5x + 2 = 3x + 8 这样的方程中，未知数 x 出现在两边。第一个目标是把同类项合并，让 x 只出现在一边。两边同时减去 3x：5x − 3x + 2 = 3x − 3x + 8，化简为 2x + 2 = 8。然后两边同时减去 2，得到 2x = 6。最后除以 2：x = 3。你也可以用逆运算把较小的变量项移到另一边。关键是在每一步都保持方程平衡，使等号始终成立。

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8. Equations with Fractions | 含分数的方程

Fractions can appear in linear equations, for example x/4 + 1 = 6. The fraction bar means division, so x/4 is the same as x ÷ 4. To solve, first subtract 1 from both sides: x/4 = 5. Then multiply both sides by 4 to undo the division: x = 20. If the equation is 2x/

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📚 Cambridge KS3 Mathematics: Fractions, Decimals and Percentages | 剑桥KS3数学：分数、小数和百分比

Welcome to this TutorHao revision guide for Cambridge KS3 Mathematics. This article focuses on fractions, decimals and percentages, one of the most important number topics at Key Stage 3. Many learners meet this topic in exercises such as page 226, question 2, and it appears frequently in Cambridge Checkpoint tests. Understanding how these three forms connect will help you solve a wide range of numerical and real-world problems.

欢迎阅读 TutorHao 为剑桥 KS3 数学准备的复习指南。本文聚焦分数、小数和百分比，这是第三学段最重要的数论主题之一。许多学习者在第226页第2题等练习中接触过这一内容，它也经常出现在剑桥 Checkpoint 考试中。理解这三种表达形式之间的联系将帮助你解决各种数值和现实问题。

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1. Understanding Fractions, Decimals and Percentages | 理解分数、小数和百分比

A fraction represents part of a whole. The numerator, which is the top number, tells us how many parts are taken. The denominator, which is the bottom number, tells us the total number of equal parts. For example, ¾ means three quarters of a whole, with 3 as the numerator and 4 as the denominator.

分数表示整体的一部分。分子是上面的数，表示取了多少份；分母是下面的数，表示整体被平均分成的份数。例如，¾ 表示整体的四分之三，其中3是分子，4是分母。

A decimal uses place value and a decimal point. The digits after the point show tenths, hundredths, thousandths, and so on. A percentage means ‘out of 100’, so 25% means 25 out of 100. This is the same as ¼ or 0.25. These three forms are just different ways of describing the same value.

小数使用位值和一个小数点。小数点后的数字分别表示十分位、百分位、千分位等。百分比表示每100份中的部分，所以25%表示100份中的25份。它等同于 ¼ 或 0.25。这三种形式只是描述同一个值的不同方式。

¾ = 0.75 = 75%

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2. Converting Between Fractions, Decimals and Percentages | 分数、小数和百分比之间的转换

To convert a fraction to a decimal, divide the numerator by the denominator. For example, ¾ = 3 ÷ 4 = 0.75. To convert a decimal to a percentage, multiply the decimal by 100. So 0.75 × 100 = 75%. To convert a percentage to a decimal, divide by 100, giving 75% = 0.75.

要将分数转换为小数，用分子除以分母。例如，¾ = 3 ÷ 4 = 0.75。要将小数转换为百分比，将小数乘以100。因此 0.75 × 100 = 75%。要将百分比转换为小数，除以100，得到75% = 0.75。

To convert a percentage to a fraction, write the percentage over 100 and then simplify if possible. For example, 40% = 40/100 = 2/5. To convert a fraction to a percentage, first convert the fraction to a decimal, then multiply by 100. For example, ⅝ = 5 ÷ 8 = 0.625, so ⅝ = 62.5%.

要将百分比转换为分数，先将百分比写在100上，然后尽可能化简。例如，40% = 40/100 = 2/5。要将分数转换为百分比，先将分数转换为小数，再乘以100。例如，⅝ = 5 ÷ 8 = 0.625，因此 ⅝ = 62.5%。

Fraction → Decimal → Percentage

Fraction	Decimal	Percentage
½	0.5	50%
¼	0.25	25%
¾	0.75	75%
⅕	0.2	20%
⅖	0.4	40%

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3. Ordering and Comparing Values | 数值的大小比较与排序

When you need to order fractions, decimals and percentages, the easiest method is to convert all values into decimals. For example, to compare ⅖, 0.3 and 35%, convert each value: ⅖ = 0.4, 35% = 0.35, and 0.3 is already a decimal. Then order the decimals: 0.3 < 0.35 < 0.4.

当你需要排列分数、小数和百分比的大小时，最简单的方法是把所有值都转换为小数。例如，比较 ⅖、0.3和35%，先转换每个值：⅖ = 0.4，35% = 0.35，0.3已经是小数。然后排列小数：0.3 < 0.35 < 0.4。

Another useful idea is that a larger denominator means smaller parts. Therefore ¼ is smaller than ⅓. For example, ½ is larger than ⅓ because halves are larger pieces than thirds. This concept helps you compare fractions without using a calculator.

另一个有用的概念是分母越大表示每一份越小。因此 ¼ 小于 ⅓。例如，½ 大于 ⅓，因为每一份更大。这个概念有助于你在不使用计算器的情况下比较分数。

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4. Adding and Subtracting Fractions | 分数的加减法

To add or subtract fractions, they must have the same denominator. You cannot add the numerators until the denominators are equal. For example, to add ½ + ⅓, first find the lowest common denominator of 2 and 3, which is 6. Convert: ½ = 3/6 and ⅓ = 2/6. Then add: 3

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📚 Scatter Graphs and Correlation for Cambridge KS3 Maths | 剑桥KS3数学：散点图与相关性

Scatter graphs are a key part of the Cambridge KS3 Mathematics curriculum. They help us display two sets of numerical data on the same diagram and show whether there is a relationship between the variables. In many exam papers, including page 221 question 2, you are asked to plot points, describe correlation, draw a line of best fit, and use it to make predictions. This article reviews the essential skills step by step.

散点图是剑桥KS3数学课程的重点之一。它们帮助我们在同一幅图中展示两组数值数据，并观察变量之间是否存在关系。在包括第221页第2题在内的许多试卷中，你都需要描点、描述相关性、画出最佳拟合线并利用它进行预测。本文将逐步复习这些核心技能。

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1. What is a scatter graph? | 什么是散点图？

A scatter graph is a diagram that plots pairs of numerical data as points on a coordinate grid. Each point represents one observation. One variable is shown on the horizontal x-axis and the other is shown on the vertical y-axis.

散点图是一种将成对数值数据作为点绘制在坐标网格上的图。每个点表示一次观测。一个变量显示在水平 x 轴上，另一个变量显示在垂直 y 轴上。

Scatter graphs are used when we want to investigate a possible relationship between two continuous variables, such as height and arm span, temperature and ice cream sales, or revision time and test score. They are especially important in Cambridge Checkpoint data-handling questions.

当我们想要研究两个连续变量之间可能的关系时，会使用散点图，例如身高与臂展、温度与冰淇淋销量，或复习时间与考试成绩。它们在剑桥 Checkpoint 数据处理题中尤其重要。

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2. Plotting points accurately | 准确描点

To plot a point, read each pair of values as (x, y). The first number tells you how far to move along the x-axis, and the second number tells you how far to move up the y-axis. Always check the scale on each axis before you start.

要描点，请按 (x, y) 读取每对数值。第一个数告诉你沿 x 轴移动多远，第二个数告诉你沿 y 轴向上移动多远。开始前务必检查每个轴上的刻度。

Use a sharp pencil and mark a small cross rather than a dot. A cross is easier to see and is accepted in Cambridge KS3 exams. Label each axis with the variable name and its unit, for example ‘Height (cm)’ or ‘Test score (%)’.

使用削尖的铅笔并画小十字而不是圆点。十字更容易看清，并且在剑桥KS3考试中被接受。给每个坐标轴标上变量名称和单位，例如 ‘身高 (cm)’ 或 ‘考试分数 (%)’。

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3. Choosing a suitable scale | 选择合适的刻度

A good scale spreads the data points across most of the grid. If the scale is too small, all the points will be crowded into a corner. If the scale is too large, it will be difficult to read values accurately.

合适的刻度能让数据点分布在网格的大部分区域。如果刻度太小，所有点都会挤在一个角落里。如果刻度太大，就很难准确读取数值。

Use equal intervals on each axis, such as 2, 5, or 10, depending on the data. Do not use uneven intervals, because this distorts the pattern and can make weak correlation look strong.

每个轴上使用相等的间隔，例如 2、5 或 10，具体取决于数据。不要使用不均匀的间隔，因为这会扭曲分布模式，并可能使弱相关看起来很强。

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4. Independent and dependent variables | 自变量与因变量

The independent variable is the one you control or change. It is usually plotted on the x-axis. The dependent variable is the one you measure or observe, and it is plotted on the y-axis.

自变量是你控制或改变的变量。它通常绘制在 x 轴上。因变量是你测量或观察的变量，它绘制在 y 轴上。

For example, if you investigate how test score depends on hours of revision, then hours of revision is the independent variable and test score is the dependent variable. In some questions the variables do not have a clear cause-and-effect relationship, but a convention still helps when drawing the graph.

例如，如果你研究考试成绩如何取决于复习小时数，那么复习小时数是自变量，考试成绩是因变量。在某些问题中，变量之间并没有明确的因果关系，但在绘制图形时，使用惯例仍然有帮助。

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5. Describing correlation | 描述相关性

Correlation describes the pattern of points on a scatter graph. If the points tend to go upwards from left to right, the correlation is positive. If they tend to go downwards, the correlation is negative. If there is no clear pattern, there is no correlation.

相关性描述散点图上点的分布模式。如果点从左到右呈上升趋势，则为正相关。如果呈下降趋势，则为负相关。如果没有明显模式，则无相关。

You should also describe the strength of the correlation. If the points are close to a straight line, the correlation is strong. If they are widely spread but still show a trend, the correlation is weak or moderate.

你还应描述相关性的强度。如果点靠近一条直线，则相关性很强。如果点分布较散但仍显示出趋势，则相关性较弱或中等。

Type of correlation	Pattern of points	Example
Positive	Rises from left to right	Temperature and cold drink sales
Negative	Falls from left to right	Age of car and value
No correlation	Random scatter	Shoe size and test score

In Cambridge KS3 answers, write a sentence such as ‘There is a strong positive correlation between temperature and ice cream sales.’ Mention both direction and strength.

在剑桥KS3答案中，写出类似“温度与冰淇淋销量之间存在强正相关”的句子。要同时提到方向和强度。

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6. Line of best fit | 最佳拟合线

When the scatter graph shows a linear correlation, we can draw a line of best fit. This line should follow the general trend of the points and pass through the middle of the data, with roughly the same number of points above and below the line.

当散点图显示线性相关时，我们可以画一条最佳拟合线。这条线应沿点的总体趋势，穿过数据中间，线上方和线下方的点数大致相同。

A line of best fit does not have to pass through the origin, and it does not have to pass through any specific point. Use a ruler and draw one straight line, not a curve. If the correlation is weak or there is no correlation, a line of best fit is not appropriate.

最佳拟合线不必经过原点，也不必经过任何特定点。使用直尺画一条直线，而不是曲线。如果相关性较弱或没有相关性，就不适合画最佳拟合线。

Think of the line as a summary of the linear trend. It helps you see the relationship more clearly and is the basis for estimating unknown values.

可以把这条线看作线性趋势的总结。它帮助你更清楚地看到关系，并且是估计未知值的基础。

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7. Making predictions using the line of best fit | 利用最佳拟合线进行预测

Once you have drawn a line of best fit, you can use it to estimate an unknown value. Find one variable on an axis, move vertically or horizontally to the line, then read the corresponding value on the other axis.

画出最佳拟合线后，你可以用它来估计未知值。在一个轴上找到一个变量，向直线作垂直线或水平线，然后在另一个轴上读取相应的值。

Predictions within the range of the plotted data are called interpolation and are usually reliable. Predictions outside the range are called extrapolation and may be less reliable because the trend could change.

在已绘数据范围内的预测称为内插，通常比较可靠。超出数据范围的预测称为外推，可能不太可靠，因为趋势可能会改变。

In exam answers, state that the estimate is only approximate and based on the trend shown by the data. For example, write ‘Using the line of best fit, a revised score of about 88 is expected for 6 hours of revision.’

在考试答案中，要说明估计值只是近似的，并且基于数据所显示的趋势。例如，写“根据最佳拟合线，复习6小时的考试分数预计约为88分”。

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8. Identifying outliers | 识别异常值

An outlier is a point that lies far away from the overall pattern of the scatter graph. It does not fit the general trend and may be caused by a measurement error or an unusual result.

异常值是远离散点图整体模式的点。它不符合总体趋势，可能是由测量误差或异常结果造成的。

If you find an outlier, you should mention it when describing the correlation. For example, you can say ‘Apart from one outlier, there is a positive correlation.’ When drawing a line of best fit, an outlier can be ignored because it would pull the line away from the main trend.

如果发现异常值，在描述相关性时应提到它。例如，你可以说“除了一个异常值之外，存在正相关”。在画最佳拟合线时，可以忽略异常值，因为它会使直线偏离主要趋势。

Do not delete an outlier from the diagram unless the question asks you to. Instead, draw attention to it and explain its effect on the result.

除非题目要求，否则不要从图中删除异常值。相反，应指出它并解释它对结果的影响。

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9. Common mistakes to avoid | 常见错误

A common mistake is drawing a line of best fit through the origin even when the points do not suggest that. Always let the data determine the position of the line.

一个常见错误是即使数据点并不表明如此，也把最佳拟合线画过原点。始终让数据决定直线的位置。

Another mistake is confusing positive and negative correlation. Check the direction of the slope: uphill is positive, downhill is negative. Also remember that a strong correlation means the points are close to the line, not that the line is steep.

另一个常见错误是混淆正相关和负相关。检查斜线的方向：向上为正，向下为负。还要记住，强相关意味着点靠近直线，而不是直线很陡。

Students sometimes forget to label the axes or use uneven scales. A scatter graph without clear labels or with a distorted scale can lead to wrong interpretations. Finally, do not connect the points with a line segment. A scatter graph uses a line of best fit, not a dot-to-dot drawing.

学生有时会忘记给坐标轴贴标签或使用不均匀的刻度。没有清晰标签或刻度失真的散点图会导致错误解读。最后，不要用线段连接各个点。散点图使用最佳拟合线，而不是逐点连线。

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10. Worked example based on page 221 question 2 | 第221页第2题式例题

Question: The table shows the number of hours five students revised and their test scores. Plot the data, describe the correlation, draw a line of best fit, and estimate the score for a student who revised for 6 hours.

题目：表格显示五名学生的复习小时数和考试分数。绘制数据，描述相关性，画出最佳拟合线，并估计复习6小时的学生的分数。

Data: (1, 30), (2, 42), (3, 55), (4, 65), (5, 78). Plotting these points shows an upward trend, so there is a strong positive correlation.

数据：(1, 30)、(2, 42)、(3, 55)、(4, 65)、(5, 78)。描出这些点后呈现上升趋势，因此存在强正相关。

Draw a line of best fit through the middle of the points. To estimate the score for 6 hours, extend the line or read the trend; a value around 88 to 90 is reasonable. Show your working by drawing a dashed line from 6 on the x-axis to the line of best fit and then across to the y-axis.

穿过这些点的中间画一条最佳拟合线。要估计6小时的分数，延长直线或读取趋势；88到90左右的值是合理的。通过从 x 轴上的 6 向最佳拟合线画一条虚线，再延伸到 y 轴，展示你的解题过程。

This type of question tests plotting, interpretation, and prediction in one task, exactly like page 221 question 2 in many Cambridge KS3 practice resources.

这类题目像许多剑桥KS3练习资源中的第221页第2题一样，在一个任务中考查描图、解读和预测。

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11. Practice questions | 练习题目

Try these quick questions to check your understanding. For each one, write a sentence describing the correlation.

尝试以下快速练习来检查你的理解。对于每一题，写一句话描述相关性。

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📚 KS3 Cambridge Mathematics: Solving Linear Equations | KS3 剑桥数学：解线性方程

Linear equations are one of the most important building blocks in KS3 mathematics. They appear in many topics, from algebra and problem-solving to graphs and real-life situations. In this article, we will explore how to solve linear equations step by step, using clear methods and examples that match the Cambridge KS3 curriculum.

线性方程是 KS3 数学中最重要的基础模块之一。它们出现在许多主题中，从代数、解题到图像和现实情境。本文将逐步探讨如何解线性方程，使用符合剑桥 KS3 课程要求的清晰方法和示例。

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1. What is a Linear Equation? | 什么是线性方程？

A linear equation is an equation where the unknown variable, usually written as x, has an exponent of 1. This means there are no terms like x² or x³. For example, 2x + 3 = 11 is a linear equation because the highest power of x is 1. The graph of a linear equation is always a straight line, which is why it is called ‘linear’.

线性方程是指未知变量（通常写作 x）的指数为 1 的方程。这意味着方程中没有 x² 或 x³ 这样的项。例如，2x + 3 = 11 是一个线性方程，因为 x 的最高次幂是 1。线性方程的图像总是一条直线，这就是它被称为 ‘线性’ 的原因。

2x + 3 = 11

x + 5 = 12

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2. Key Vocabulary | 关键术语

Before solving equations, it helps to know the key terms. The left-hand side (LHS) is the expression to the left of the equals sign, and the right-hand side (RHS) is the expression to the right. A solution is the value of x that makes the equation true. Inverse operations are opposite operations, such as addition and subtraction, or multiplication and division. A constant is a number on its own, while a coefficient is the number multiplying the variable.

在解方程之前，了解关键术语会很有帮助。左边（LHS）是等号左边的表达式，右边（RHS）是等号右边的表达式。解是使方程成立的 x 值。逆运算是相反的运算，例如加法和减法，或乘法和除法。常数是单独出现的数字，而系数是乘以变量的数字。

• In 3x + 4 = 10, the LHS is 3x + 4 and the RHS is 10.
• In 3x + 4 = 10, 3 is the coefficient of x, and 4 is the constant.
• In 3x + 4 = 10, the solution is x = 2 because 3 × 2 + 4 = 10.

在 3x + 4 = 10 中，左边是 3x + 4，右边是 10。

在 3x + 4 = 10 中，3 是 x 的系数，4 是常数。

在 3x + 4 = 10 中，解是 x = 2，因为 3 × 2 + 4 = 10。

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3. The Balance Method | 天平法

The balance method is a visual way to understand equations. Imagine an old-fashioned balance scale. The equals sign means both sides are perfectly balanced. Whatever operation you do to one side, you must do the same to the other side to keep the balance. This is the golden rule of solving equations.

天平法是一种理解方程的直观方法。想象一台老式天平。等号意味着两边完全平衡。无论你对一边进行什么运算，都必须对另一边进行同样的运算，以保持平衡。这是解方程的黄金法则。

If you add, subtract, multiply, or divide one side by a number, you must do exactly the same to the other side. This rule keeps the equation true and helps you isolate the unknown variable.

如果你对一边进行加、减、乘或除以某个数，你必须对另一边进行完全相同的操作。这条规则使方程保持成立，并帮助你分离出未知变量。

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4. Solving One-Step Equations | 解一步方程

A one-step equation requires only one inverse operation to solve. For example, to solve x + 7 = 15, subtract 7 from both sides: x + 7 − 7 = 15 − 7, so x = 8. Similarly, to solve 4x = 20, divide both sides by 4: x = 5. For x − 3 = 9, add 3 to both sides to get x = 12.

一步方程只需要一个逆运算即可求解。例如，解 x + 7 = 15，两边减去 7：x + 7 − 7 = 15 − 7，因此 x = 8。类似地，解 4x = 20，两边除以 4：x = 5。对于 x − 3 = 9，两边加上 3 得到 x = 12。

x + 7 = 15 → x = 15 − 7 → x = 8

4x = 20 → x = 20 ÷ 4 → x = 5

If the coefficient is negative, such as −x = 7, multiply both sides by −1 to find x = −7. Remember that −x means −1 × x.

如果系数为负，例如 −x = 7，两边乘以 −1 得到 x = −7。请记住 −x 表示 −1 × x。

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5. Solving Two-Step Equations | 解两步方程

Two-step equations involve two operations. For example, solve 3x − 4 = 11. First, add 4 to both sides: 3x − 4 + 4 = 11 + 4, giving 3x = 15. Then divide both sides by 3: x = 5. Always undo addition or subtraction before multiplication or division.

两步方程涉及两个运算。例如，解 3x − 4 = 11。首先，两边加上 4：3x − 4 + 4 = 11 + 4，得到 3x = 15。然后两边除以 3：x = 5。一定要先处理加减，再处理乘除。

Another example is x ÷ 2 + 6 = 13. First subtract 6 from both sides: x ÷ 2 = 7. Then multiply both sides by 2: x = 14. Work backwards through the order of operations to isolate x.

另一个例子是 x ÷ 2 + 6 = 13。首先两边减去 6：x ÷ 2 = 7。然后两边乘以 2：x = 14。按照运算顺序倒推，以分离出 x。

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6. Equations with Brackets | 含有括号的方程

If an equation has brackets, expand them first. For example, 2(x + 3) = 14. Expand the bracket: 2x + 6 = 14. Then subtract 6 from both sides: 2x = 8. Finally divide by 2: x = 4. You can check by substituting x = 4 back into the original equation.

如果方程含有括号，应先把括号展开。例如，2(x + 3) = 14。展开括号：2x + 6 = 14。然后两边减去 6：2x = 8。最后除以 2：x = 4。你可以把 x = 4 代回原方程进行检验。

For 3(x − 5) = 12, expand to get 3x − 15 = 12. Add 15 to both sides: 3x = 27. Divide by 3: x = 9. Be careful with negative signs inside the brackets.

对于 3(x − 5) = 12，展开得到 3x − 15 = 12。两边加上 15：3x = 27。除以 3：x = 9。注意括号内的负号。

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7. Equations with Unknowns on Both Sides | 未知数在两侧的方程

When x appears on both sides, collect the x terms on one side first. For example, 5x + 2 = 3x + 10. Subtract 3x from both sides: 2x + 2 = 10. Then subtract 2 from both sides: 2x = 8. Finally divide by 2: x = 4.

当 x 出现在两边时，首先把含 x 的项移到同一边。例如，5x + 2 = 3x + 10。两边减去 3x：2x + 2 = 10。然后两边减去 2：2x = 8。最后除以 2：x = 4。

If the equation is 7x − 3 = 2x + 12, subtract 2x from both sides: 5x − 3 = 12. Add 3 to both sides: 5x = 15. Divide by 5: x = 3. Choose the side with the larger coefficient to keep x positive.

如果方程是 7x − 3 = 2x + 12，两边减去 2x：5x − 3 = 12。两边加上 3：5x = 15。除以 5：x = 3。选择系数较大的一边来保持 x 为正。

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8. Checking Your Solution | 检验解

Always check your answer by substituting it into the original equation. For x = 4 in 2(x + 3) = 14, the LHS becomes 2(4 + 3) = 2 × 7 = 14, which equals the RHS. This confirms the solution is correct.

一定要把答案代回原方程进行检验。对于 2(x + 3) = 14 中的 x = 4，左边变为 2(4 + 3) = 2 × 7 = 14，与右边相等。这证明解是正确的。

Checking helps you catch small arithmetic errors. Write out the substitution clearly, showing each step, so you can identify exactly where any mistake occurred.

检验有助于发现小的算术错误。清晰地写出代入过程，展示每一步，这样你就能准确找到错误发生的位置。

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9. Common Mistakes | 常见错误

One common mistake is forgetting to perform the same operation on both sides. Another is mishandling negative numbers, for example x − 5 = 3 means x = 8, not x = −2. A third mistake is dividing only one term by a number instead of the whole side. Always treat each side as a complete expression.

一个常见错误是忘记对两边进行相同的运算。另一个错误是负数处理不当，例如 x − 5 = 3 意味着 x = 8，而不是 x = −2。第三个错误是只把一个项除以某个数，而没有对整边进行除法。应始终把每一边看作一个完整的表达式。

Mistake	Wrong	Correct
Forgetting to do the same to both sides	x + 3 = 7 → x = 7	x + 3 = 7 → x = 4
Incorrect negative signs	x − 5 = 3 → x = −2	x − 5 = 3 → x = 8
Dividing only part of a side	2x + 6 = 10 → x + 6 = 5	2x + 6 = 10 → 2x = 4 → x = 2

错误：忘记对两边进行相同操作。错误：负数符号不正确。错误：只除以一边的一部分。正确：完整地除以每一项。

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10. Practice Questions | 练习题

Try these questions to test yourself. Solve each equation and check your answer by substitution.

尝试以下题目来检验自己。解每个方程并通过代入检验你的答案。

• Solve x + 9 = 17
• Solve 7x = 42
• Solve 4x − 5 = 11
• Solve 3(x − 2) = 18
• Solve 6x + 1 = 2x + 17
• Solve 5 − 2x = x + 11

Answers: x = 8, x = 6, x = 4, x = 8, x = 4, x = −2. Work through the steps slowly and use the balance method to explain your reasoning.

答案：x = 8，x = 6，x = 4，x = 8，x = 4，x = −2。慢慢完成每个步骤，并使用天平法解释你的推理。

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11. Real-Life Applications | 实际应用

Linear equations are used to solve real problems, such as calculating costs, distances, or unknown measurements. For example, if a taxi charges a fixed fee of £3 plus £2 per mile and the total fare is £15, the equation 2m + 3 = 15 can be solved to find m = 6 miles.

线性方程可用于解决实际问题，如计算费用、距离或未知测量值。例如，如果出租车收取固定费用 3 英镑，每英里 2 英镑，总费用为 15 英镑，则方程 2m + 3 = 15 可解出 m = 6 英里。

Another example is working out how many weeks you need to save pocket money. If you start with £10 and save £5 per week, and you need £45, the equation 5w + 10 = 45 gives w = 7 weeks.

另一个例子是计算你需要存多少周的零用钱。如果你从 10 英镑开始，每周存 5 英镑，你需要 45 英镑，方程 5w + 10 = 45 得出 w = 7 周。

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12. Summary | 总结

To solve linear equations confidently, remember to use inverse operations, keep the balance, expand brackets first, collect like terms, and always check your answer. Regular practice will make these steps automatic and prepare you for more advanced algebra.

要自信地解线性方程，请记住使用逆运算、保持平衡、先展开括号、合并同类项，并始终检验答案。定期练习将使这些步骤变得自动化，并为更高级的代数做好准备。

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📚 Solving Linear Equations with Brackets: Cambridge KS3 Exercise p214 Q2 | 剑桥 KS3 数学：含括号的一次方程求解（第214页第2题）

Linear equations with brackets are a core skill in the Cambridge KS3 mathematics curriculum. In this revision guide, we work through the type of question found on page 214, question 2, where you must expand brackets, collect like terms, and solve for the unknown variable.

含括号的一次方程是剑桥 KS3 数学课程中的核心技能。在本复习指南中，我们将讲解第214页第2题这类题目，你需要先去括号、合并同类项，然后求出未知数的值。

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1. Understanding the Problem | 理解题意

A linear equation with brackets is an equation such as 3(x – 2) = 2x + 5. The aim is to find the value of x that makes the left side equal to the right side.

含括号的一次方程是指像 3(x – 2) = 2x + 5 这样的等式。我们的目标是求出使左边等于右边的 x 的值。

On page 214 question 2, you may be asked to solve, simplify, or check a solution. Always read the instruction carefully before starting your working.

在第214页第2题中，你可能会被要求求解、化简或检验答案。开始计算前，一定要仔细阅读题目要求。

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2. Key Vocabulary | 关键术语

Linear equation: an equation where the unknown appears only to the power of 1, such as 4x + 7 = 19.

一次方程：未知数的最高次数为 1 的方程，例如 4x + 7 = 19。

Brackets: symbols like ( ) that show multiplication or grouping. Expanding brackets means multiplying out the terms inside.

括号：像 ( ) 这样的符号，表示乘法或分组。去括号就是把括号内的每一项乘出来。

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3. Expanding Brackets | 去括号

When a number is written directly outside a bracket, multiply every term inside the bracket by that number.

当一个数字直接写在括号外时，要用这个数字乘以括号内的每一项。

a(b + c) = ab + ac

For example, 3(x – 2) expands to 3x – 6 because 3 × x = 3x and 3 × (-2) = -6.

例如，3(x – 2) 去括号后得到 3x – 6，因为 3 × x = 3x，而 3 × (-2) = -6。

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4. Collecting Like Terms | 合并同类项

Like terms contain exactly the same variable part. You can only add or subtract like terms.

同类项含有完全相同的字母部分。只有同类项才能相加或相减。

For example, 5x – 2x = 3x but 5x – 2 cannot be simplified further because they are not like terms.

例如，5x – 2x = 3x，但 5x – 2 不能再化简，因为它们不是同类项。

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5. Isolating the Unknown | 移项求解未知数

After expanding and collecting like terms, isolate the unknown on one side of the equation.

去括号并合并同类项后，要把未知数单独移到方程的一边。

• If a term is added, subtract it from both sides.
• 如果某一项是加上的，就在等号两边同时减去它。
• If a term is subtracted, add it to both sides.
• 如果某一项是减去的，就在等号两边同时加上它。
• If the unknown is multiplied by a number, divide both sides by that number.
• 如果未知数乘以一个数，就在等号两边同时除以这个数。

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6. Worked Example 1: 3(x – 2) = 2x + 5 | 例题一：3(x – 2) = 2x + 5

Step 1: Expand the bracket on the left side.

第一步：展开左边的括号。

3x – 6 = 2x + 5

Step 2: Subtract 2x from both sides to collect the x terms on the left.

第二步：两边同时减去 2x，把含 x 的项移到左边。

3x – 2x – 6 = 5

Step 3: Simplify and then add 6 to both sides.

第三步：化简，然后两边同时加上 6。

x = 11

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7. Worked Example 2: 5(2x – 1) = 3(x + 4) | 例题二：5(2x – 1) = 3(x + 4)

Expand both brackets first.

首先展开两边的括号。

10x – 5 = 3x + 12

Subtract 3x from both sides and add 5 to both sides.

两边同时减去 3x，并同时加上 5。

10x – 3x = 12 + 5

Simplify and solve.

化简并求解。

7x = 17

x = 17 ÷ 7 = 2 3/7

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8. Checking Your Answer | 检验答案

Substitute your value of x back into the original equation to check it works.

把求出的 x 值代回原方程，检验是否成立。

For example, if x = 11 in 3(x – 2) = 2x + 5, the left side is 3(11 – 2) = 3 × 9 = 27, and the right side is 2 × 11 + 5 = 22 + 5 = 27. Both sides are equal, so the answer is correct.

例如，对于 3(x – 2) = 2x + 5，当 x = 11 时，左边是 3(11 – 2) = 3 × 9 = 27，右边是 2 × 11 + 5 = 22 + 5 = 27。两边相等，因此答案正确。

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9. Common Mistakes | 常见错误

Mistake 1: Forgetting to multiply the second term inside the bracket by the outside number.

错误一：忘记用括号外的数去乘括号内的第二项。

Incorrect: 3(x – 2) = 3x – 2. Correct: 3(x – 2) = 3x – 6.

错误写法：3(x – 2) = 3x – 2。正确写法：3(x – 2) = 3x – 6。

Mistake 2: Moving a term to the other side without changing its sign.

错误二：把项移到另一边时没有变号。

Incorrect: x – 6 = 5 so x = 5 – 6. Correct: x – 6 = 5 so x = 5 + 6.

错误做法：x – 6 = 5，所以 x = 5 – 6。正确做法：x – 6 = 5，所以 x = 5 + 6。

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10. Practice Questions | 练习题目

Try these questions to test your understanding.

尝试以下问题来检验你的理解。

• 4(x + 3) = 2x + 18
• 2(3x – 5) = 4(x + 1)
• 7(x – 2) = 5x + 6

For each one, expand the brackets, collect like terms, solve, and check your answer by substitution.

对每一题，先去括号，再合并同类项，求解，然后用代入法检验答案。

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11. Exam Tips | 考试技巧

Show every step of your working. Even if your final answer is wrong, you can still earn method marks for correct expansion and balancing.

写出每一步计算过程。即使最终答案错误，只要去括号和移项过程正确，你仍然可以得到方法分。

Write your answer clearly, and if the answer is a fraction, simplify it fully unless the question says otherwise.

答案要写清楚；如果答案是分数，除非题目另有说明，否则要化简到最简形式。

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12. Summary | 总结

To solve a linear equation with brackets, expand the brackets first, collect like terms, isolate the unknown, and check your solution by substitution.

解含括号的一次方程时，先去掉括号，合并同类项，移项求出未知数，最后用代入法检验答案。

This method is exactly what you need for Cambridge KS3 questions like page 214 question 2, and it prepares you for more advanced algebra in later stages.

这种方法正是剑桥 KS3 第214页第2题这类题目所需要的，也为后续更高阶段的代数学习打下基础。

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📚 Cambridge KS3 Maths: Solving Angle Equations | 剑桥 KS3 数学：求解角度方程

In many Cambridge KS3 mathematics exercises, especially questions like the one on page 211, students meet problems that connect geometry with algebra. A typical example gives the angles of a triangle as expressions in x and asks you to find x by using the angle sum. This article shows a clear method for these questions, including setting up the equation, solving it, and checking your answer.

在剑桥 KS3 数学的许多练习中，尤其是类似第211页第2题的问题，学生会遇到几何与代数结合的题目。一个典型例子是给出三角形内角的 x 表达式，要求你利用内角和求出 x。本文展示这类题目的清晰解法，包括建立方程、求解和检验答案。

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1. The Core Question | 核心问题

A common question from Cambridge Checkpoint practice is: ‘The angles of a triangle are x°, (2x + 10)° and (3x – 10)°. Find the value of x and write down the size of each angle.’

剑桥 Checkpoint 练习中常见的一道题是：”三角形的三个内角分别为 x°、(2x + 10)° 和 (3x – 10)°。求 x 的值，并写出每个角的大小。”

This type of question appears again and again in KS3 progression tests because it tests two skills at once: knowing the angle sum of a triangle and solving a linear equation.

这类题目在 KS3 进阶测试中反复出现，因为它同时考查两项技能：掌握三角形内角和以及解一元一次方程。

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2. Key Angle Rule: Angles in a Triangle | 关键角度规则：三角形内角和

The most important fact is that the interior angles of a triangle always add up to 180°.

最重要的事实是：三角形的三个内角之和始终为 180°。

Angle A + Angle B + Angle C = 180°

This is true for every triangle, whether it is right-angled, isosceles or scalene. If you cut out the three corners of any triangle and place them side by side, they form a straight line, which measures 180°.

这对任何三角形都成立，无论是直角三角形、等腰三角形还是不等边三角形。如果你把任意三角形的三个角剪下来并拼在一起，它们会形成一条直线，而直线正好是 180°。

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3. Turning Angle Language into Algebra | 把角度语言转化为代数式

Each angle is already given as an algebraic expression: x, 2x + 10 and 3x – 10. The variable x stands for an unknown number of degrees.

每个角已经用代数式给出：x、2x + 10 和 3x – 10。变量 x 代表未知的度数。

Do not let the addition or subtraction worry you: 2x + 10 means ‘two times x, then add 10’. Likewise, 3x – 10 means ‘three times x, then subtract 10’.

不要被加减号困扰：2x + 10 表示 “x 的两倍再加 10″。同样，3x – 10 表示 “x 的三倍再减 10″。

You can label the expressions clearly before solving:

在解题前，你可以先清楚地标出各表达式：

• x° (the smallest angle expression / 最小角表达式)
• (2x + 10)° (the middle angle expression / 中间角表达式)
• (3x – 10)° (the largest angle expression / 最大角表达式)

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4. Setting Up the Equation | 建立方程

Add the three angle expressions and set the sum equal to 180°.

把三个角度表达式相加，并让和等于 180°。

x + (2x + 10) + (3x – 10) = 180

The brackets are not essential here, but they help you see the three separate angle terms. Writing the equation clearly is also important for earning method marks in a test.

这里的括号不是必须的，但它们能帮助你看出三个独立的角度项。清晰地写出方程对在考试中获得方法分也很重要。

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5. Solving the Equation Step by Step | 逐步解方程

First collect like terms on the left-hand side: x + 2x + 3x = 6x, and 10 – 10 = 0.

首先在左边合并同类项：x + 2x + 3x = 6x，同时 10 – 10 = 0。

6x = 180

Now divide both sides by 6 to isolate x.

现在把两边同时除以 6，以求出 x。

x = 180 ÷ 6 = 30

So the value of x is 30. Always show this division step, even if you can do it mentally.

因此 x 的值是 30。即使你能心算，也请一定展示这个除法步骤。

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6. Checking the Angles | 回代检验角度

Substitute x = 30 back into each expression:

把 x = 30 代回每个表达式：

• x° = 30°
• 2x + 10 = 2(30) + 10 = 70°
• 3x – 10 = 3(30) – 10 = 80°

You can also organise the checking work in a table:

你还可以用表格来整理检验过程：

Expression / 表达式	Value / 角度值
x	30°
2x + 10	70°
3x – 10	80°

Check the sum: 30 + 70 + 80 = 180°, so the answer is correct.

检查总和：30 + 70 + 80 = 180°，因此答案正确。

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7. Common Mistakes to Avoid | 需要避免的常见错误

Many students forget that the sum must be 180°, or they add the numbers before collecting x terms incorrectly.

许多学生忘记内角和必须是 180°，或者在合并 x 项之前错误地处理数字。

Another common error is writing 2x + 10° as 2(x + 10). These are not the same: 2(x + 10) = 2x + 20.

另一个常见错误是把 2x + 10° 写成 2(x + 10)。它们并不相同：2(x + 10) = 2x + 20。

Also, do not drop the degree symbol halfway through the working. Keep using ° with each angle until the final answer.

此外，不要在解题过程中丢掉度数符号。每个角的表达式中要保留 °，直到最终答案。

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8. Extending to Quadrilaterals and Polygons | 拓展到四边形和多边形

The same method works for quadrilaterals, but the interior angle sum is 360°.

同样的方法也适用于四边形，但四边形的内角和是 360°。

Angle A + Angle B + Angle C + Angle D = 360°

For example, if the angles of a quadrilateral are x, 2x, 3x and 4x, then:

例如，如果一个四边形的四个角分别是 x、2x、3x 和 4x，那么：

x + 2x + 3x + 4x = 360

10x = 360

x = 36

This pattern continues for higher polygons, using the general formula for the interior angle sum.

这个规律可以推广到更多边形，使用内角和的一般公式：

Sum of interior angles = (n – 2) × 180°

Here n is the number of sides

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📚 Cambridge KS3 Maths: Fractions, Decimals and Percentages | 剑桥 KS3 数学：分数、小数和百分数

Understanding how fractions, decimals and percentages are connected is a core skill in Cambridge KS3 Mathematics. This revision guide walks you through key conversions, operations and real-world applications, with plenty of examples.

理解分数、小数和百分数之间的联系是剑桥 KS3 数学的核心技能。本复习指南将带你掌握关键转换、运算和实际应用，并提供大量例题。

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1. Place Value and Decimal Basics | 位值与小数基础

Decimals are based on place value. The first digit after the decimal point is tenths, the second is hundredths, and the third is thousandths.

小数基于位值。小数点后第一位是十分位，第二位是百分位，第三位是千分位。

For example, 0.35 means 3 tenths and 5 hundredths, which is the same as 35/100.

例如，0.35 表示 3 个十分之一和 5 个百分之一，也就是 35/100。

You should be able to read and write decimals in words. 0.07 is seven hundredths, and 2.405 is two and four hundred five thousandths.

你应该能够读写小数。0.07 是百分之七，2.405 是二又千分之四百零五。

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2. Converting Fractions to Decimals | 分数转换为小数

To convert a fraction to a decimal, divide the numerator by the denominator. For simple fractions, you can use equivalent fractions with denominators of 10, 100 or 1000.

要将分数转换为小数，用分子除以分母。对于简单分数，可以利用分母为 10、100 或 1000 的等值分数。

Example: 3/4 = 75/100 = 0.75. You can also divide 3 by 4 on a calculator or using short division.

例如：3/4 = 75/100 = 0.75。你也可以用计算器或短除法计算 3 除以 4。

Some fractions give recurring decimals, such as 1/3 = 0.333… which is written as 0.3 with a dot above the 3.

有些分数会得到循环小数，例如 1/3 = 0.333… 写作 0.3，并在 3 上方加一个点。

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3. Converting Decimals to Percentages | 小数转换为百分数

To convert a decimal to a percentage, multiply the decimal by 100 and add the percent symbol. This is the same as moving the decimal point two places to the right.

要将小数转换为百分数，将小数乘以 100 并加上百分号。这相当于将小数点向右移动两位。

Example: 0.65 × 100 = 65%, so 0.65 = 65%. Another example: 0.08 = 8%.

例如：0.65 × 100 = 65%，所以 0.65 = 65%。另一个例子：0.08 = 8%。

If the decimal is greater than 1, the percentage will be greater than 100%. For instance, 1.25 = 125%.

如果小数大于 1，百分数将大于 100%。例如，1.25 = 125%。

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4. Converting Fractions to Percentages | 分数转换为百分数

To convert a fraction to a percentage, first convert the fraction to a decimal, then multiply by 100. Alternatively, find an equivalent fraction with denominator 100.

要将分数转换为百分数，先将分数转换为小数，再乘以 100。或者，找到一个分母为 100 的等值分数。

Example: 2/5 = 40/100 = 40%. Another example: 7/20 = 35/100 = 35%.

例如：2/5 = 40/100 = 40%。另一个例子：7/20 = 35/100 = 35%。

For fractions that do not easily convert to hundredths, divide the numerator by the denominator and multiply by 100. For example, 5/8 = 0.625 = 62.5%.

对于不容易转换成分母为 100 的分数，用分子除以分母再乘以 100。例如 5/8 = 0.625 = 62.5%。

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5. Ordering Fractions, Decimals and Percentages | 排序分数、小数和百分数

To compare and order fractions, decimals and percentages, convert all of them into the same form. Decimals are often the easiest to compare.

要比较和排序分数、小数和百分数，需要将它们转换成相同的形式。小数通常最容易比较。

Example: Order 3/5, 0.55 and 58% from smallest to largest. Convert: 3/5 = 0.6, 58% = 0.58. The order is 0.55, 0.58, 0.6.

例如：将 3/5、0.55 和 58% 从小到大排列。转换后：3/5 = 0.6，58% = 0.58。顺序是 0.55、0.58、0.6。

Fraction	Decimal	Percentage
1/4	0.25	25%
1/2	0.5	50%
3/4	0.75	75%

Use a number line or place value chart to help you see which value is largest or smallest.

可以使用数轴或位值表来帮助判断哪个值最大或最小。

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6. Adding and Subtracting Fractions | 分数的加法与减法

To add or subtract fractions, they must have the same denominator. If they do not, find the lowest common denominator first.

要对分数进行加法或减法，它们必须有相同的分母。如果分母不同，首先找到最小公分母。

Example: 1/3 + 1/4. The lowest common denominator is 12, so 1/3 = 4/12 and 1/4 = 3/12. Therefore 1/3 + 1/4 = 4/12 + 3/12 = 7/12.

例如：1/3 + 1/4。最小公分母是 12，因此 1/3 = 4/12，1/4 = 3/12。所以 1/3 + 1/4 = 4/12 + 3/12 = 7/12。

For mixed numbers, add or subtract the whole number parts and the fraction parts separately, borrowing if necessary.

对于带分数，分别加减整数部分和分数部分，必要时进行借位。

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7. Multiplying and Dividing Fractions | 分数的乘法与除法

To multiply fractions, multiply the numerators together and multiply the denominators together. Simplify the result if possible.

要相乘分数，将分子相乘，分母相乘。如果可能，化简结果。

Example: 2/3 × 4/5 = (2 × 4)/(3 × 5) = 8/15.

例如：2/3 × 4/5 = (2 × 4)/(3 × 5) = 8/15。

To divide by a fraction, multiply by its reciprocal. The reciprocal of a fraction is obtained by swapping the numerator and denominator.

要除以一个分数，乘以它的倒数。分数的倒数是将分子和分母互换后得到的。

Example: 3/4 ÷ 2/5 = 3/4 × 5/2 = 15/8 = 1 7/8.

例如：3/4 ÷ 2/5 = 3/4 × 5/2 = 15/8 = 1 7/8。

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8. Percentage of an Amount | 求一个数的百分数

To find a percentage of an amount, convert the percentage to a decimal and multiply by the amount. You can also find 1% first and then scale up.

要求一个数的百分数，将百分数转换为小数，再乘以这个数。你也可以先求 1%，然后再扩大。

Example: Find 15% of 240. 15% = 0.15, so 0.15 × 240 = 36.

例如：求 240 的 15%。15% = 0.15，所以 0.15 × 240 = 36。

Alternative method: 10% of 240 = 24, 5% of 240 = 12, so 15% = 24 + 12 = 36.

另一种方法：240 的 10% 是 24，240 的 5% 是 12，因此 15% = 24 + 12 = 36。

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9. Percentage Increase and Decrease | 百分数增减

To increase an amount by a percentage, find the percentage of the amount and add it to the original. To decrease, subtract it.

要将一个数增加某个百分数，先求出这个百分数对应的值，再加到原数上。减少则减去该值。

Example: Increase 80 by 25%. 25% of 80 = 20, so the new amount is 80 + 20 = 100.

例如：将 80 增加 25%。80 的 25% 是 20，所以新的数是 80 + 20 = 100。

You can also use a multiplier. For a 25% increase, the multiplier is 1.25. For a 15% decrease, the multiplier is 0.85.

也可以使用乘数。增加 25% 时，乘数是 1.25。减少 15% 时，乘数是 0.85。

New value = Original value × Multiplier

新值 = 原值 × 乘数

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10. Real-World Problems and Word Problems | 实际问题与文字题

Many KS3 exam questions ask you to apply fractions, decimals and percentages to real-life situations such as discounts, sales, interest and recipes.

许多 KS3 考试题目要求你将分数、小数和百分数应用到实际生活中，例如折扣、销售、利息和食谱等。

Example: A jacket costs £45. In a sale, the price is reduced by 20%. Find the sale price. 20% of 45 = £9, so the sale price is £45 – £9 = £36.

例如：一件夹克售价 45 英镑。促销期间价格降低 20%。求促销价。45 的 20% 是 9 英镑，因此促销价为 45 – 9 = 36 英镑。

Another example: A recipe uses 2/3 cup of sugar. You want to make 1 1/2 times the recipe. How much sugar do you need? 2/3 × 3/2 = 6/6 = 1 cup.

另一个例子：一份食谱需要 2/3 杯糖。你想做原食谱的 1 1/2 倍。需要多少糖？2/3 × 3/2 = 6/6 = 1 杯。

Always read the question carefully, identify the operation needed, and check that your answer makes sense in the context.

务必仔细审题，确定所需的运算，并检查答案在上下文中是否合理。

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📚 KS3 Cambridge Mathematics: Fractions, Decimals and Percentages | KS3 剑桥数学：分数、小数和百分数

Working confidently with fractions, decimals and percentages is one of the most important skills in the Cambridge KS3 mathematics curriculum. These three forms describe the same proportional idea, but many mixed exercises, including page 198 question 2, test your ability to convert, compare and apply them in real situations. In this article, you will review the key methods, see worked examples and practise common exam-style problems.

在剑桥 KS3 数学课程中，熟练处理分数、小数和百分数是最重要的技能之一。这三种形式描述的是同一个比例概念，但许多综合练习，包括第198页第2题，会考查你的换算、比较和实际应用能力。本文将复习核心方法，给出典型例题，并练习常见考试题型。

1. Understanding the Core Equivalences | 理解核心等值关系

A fraction, a decimal and a percentage can all represent the same part of a whole. For example, one half can be written as 1/2, 0.5 or 50%. You should memorise common equivalences because they speed up mental calculation and help you check written work.

分数、小数和百分数都可以表示同一个整体的一部分。例如，二分之一可以写成 1/2、0.5 或 50%。你应当熟记常见等值关系，因为它们能加快心算速度，也有助于检查笔算结果。

Fraction	Decimal	Percentage
1/2	0.5	50%
1/4	0.25	25%
3/4	0.75	75%
1/5	0.2	20%
1/10	0.1	10%

When you see a mixed list of fractions, decimals and percentages, the safest first step is to convert every value to the same form. This removes the risk of comparing unlike forms and makes ordering much easier.

当你看到分数、小数和百分数混合的一组数时，最稳妥的第一步是把每个值都转换成同一种形式。这可以避免比较不同形式带来的错误，并使排序更加容易。

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2. Converting Fractions to Decimals | 分数化小数

To convert a fraction to a decimal, divide the numerator by the denominator. For simple fractions such as 3/8, you can carry out the division 3 ÷ 8 = 0.375. If the denominator does not divide neatly into a power of ten, long division gives the exact or recurring decimal.

将分数化为小数，只需用分子除以分母。对于 3/8 这样的简单分数，计算 3 ÷ 8 = 0.375。如果分母不能整除 10 的幂，则用长除法得到精确小数或循环小数。

For fractions whose denominator is already 10, 100 or 1000, place the numerator behind the decimal point so that the number of decimal places matches the number of zeros. For example, 47/100 = 0.47 and 9/1000 = 0.009.

对于分母已经是 10、100 或 1000 的分数，只需把分子放在小数点后面，并使小数位数与零的个数一致。例如 47/100 = 0.47，9/1000 = 0.009。

fraction → decimal: numerator ÷ denominator

中文：分数化小数：分子除以分母。

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3. Converting Decimals to Percentages | 小数化百分数

Percent means ‘out of 100’, so a decimal is changed to a percentage by multiplying by 100. Move the decimal point two places to the right. For instance, 0.85 becomes 0.85 × 100 = 85%, while 0.035 becomes 3.5%.

百分数表示“每一百份中的多少”，因此小数化为百分数只需乘以 100，也就是把小数点向右移动两位。例如，0.85 乘以 100 得 85%，而 0.035 得 3.5%。

Remember that numbers larger than 1 give percentages greater than 100%. For example, 1.2 = 120% because 1.2 × 100 = 120. Do not assume that percentages must be less than 100; mixed numbers often produce values above 100%.

请记住，大于 1 的数对应的百分数会超过 100%。例如 1.2 = 120%，因为 1.2 × 100 = 120。不要认为百分数一定小于 100；带分数常常会产生超过 100% 的值。

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4. Converting Percentages to Fractions | 百分数化分数

To write a percentage as a fraction, put the percentage over 100 and simplify. For example, 45% = 45/100 = 9/20 after dividing the numerator and denominator by 5. If the percentage contains a decimal, multiply the numerator and denominator by 10 or 100 first.

将百分数写成分数时，先把百分数写在 100 的上方，然后约分。例如 45% = 45/100，分子分母同除以 5 后得到 9/20。如果百分数含有小数，应先给分子分母同乘 10 或 100。

For 12.5%, write 12.5/100 and multiply top and bottom by 10 to obtain 125/1000, which simplifies to 1/8. This trick removes the decimal from the numerator and makes simplification straightforward.

对于 12.5%，可写成 12.5/100，再将分子分母同乘以 10 得到 125/1000，最后约分成 1/8。这个方法可以消去分子中的小数点，使约分更直接。

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5. Ordering Mixed Numbers | 混合数排序

When a question asks you to put a set of fractions, decimals and percentages in order, change them all to decimals first. Decimals are easy to compare place by place. For example, to order 3/5, 0.58 and 62%, convert: 3/5 = 0.6, 0.58 = 0.58, 62% = 0.62. Then ascending order is 0.58, 0.6, 0.62, so 0.58, 3/5, 62%.

当题目要求把一组分数、小数和百分数按顺序排列时，先把它们都化成小数。小数按数位逐位比较非常方便。例如，要排序 3/5、0.58 和 62%，先转换：3/5 = 0.6，0.58 = 0.58，62% = 0.62。因此从小到大依次为 0.58、0.6、0.62，即 0.58、3/5、62%。

Use zero placeholders to avoid mistakes when comparing decimals. For example, write 0.6 as 0.60 so you can see that 0.58 is smaller than

Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com

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📚 Cambridge KS3 Mathematics: Solving Linear Equations | 剑桥KS3数学：一元一次方程求解

Linear equations are a central part of the Cambridge KS3 mathematics course. They help students understand the idea of an unknown value and how to find it using logical steps. This article will guide you through the balance method, inverse operations, brackets, fractions and word problems.

一元一次方程是剑桥 KS3 数学课程的核心内容。它帮助学生理解未知数的概念，以及如何通过有逻辑的步骤求出未知数。本文将带你掌握天平平衡法、逆运算、括号、分数和文字题。

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1. What Is an Equation? | 什么是方程

An equation is a mathematical statement that two expressions are equal. It contains an equals sign and often includes a letter, such as x or y, to represent an unknown number. For example, in x + 5 = 12, the letter x stands for the number that must be added to 5 to make 12.

方程是说明两个表达式相等的数学语句。它包含等号，并且常常含有一个字母，如 x 或 y，用来表示未知数。例如，在 x + 5 = 12 中，字母 x 代表与 5 相加后等于 12 的那个数。

x + 5 = 12 ⇒ x = 7

Here x = 7 because 7 + 5 = 12. This is called the solution of the equation.

这里 x = 7，因为 7 + 5 = 12。这个值就叫做方程的解。

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2. The Balance Method | 天平平衡法

Think of an equation as a balanced scale. The left side and the right side must always have the same value. If you add, subtract, multiply or divide, you must do exactly the same to both sides. This keeps the equation balanced while you isolate the unknown.

把方程想象成一架平衡的天平。左边和右边的值必须始终相等。如果你进行加、减、乘或除的运算，必须在等号两边进行完全相同的操作。这样可以在保持方程平衡的同时，把未知数单独解出来。

For example, if x + 3 = 10, you cannot simply remove the 3 from the left side. You must subtract 3 from both sides:

例如，如果 x + 3 = 10，不能只把左边的 3 去掉，而必须两边同时减去 3：

x + 3 − 3 = 10 − 3 ⇒ x = 7

This ‘same operation on both sides’ idea is the most important rule in solving equations.

这种 “两边同时进行相同运算” 的思路是解方程最重要的规则。

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3. Using Inverse Operations | 运用逆运算

Inverse operations are opposite operations that undo each other. Addition and subtraction are inverse operations, and multiplication and division are inverse operations. To solve an equation, we use inverse operations to ‘undo’ what has been done to the unknown.

逆运算是互相对立、可以相互抵消的运算。加法和减法互为逆运算，乘法和除法互为逆运算。解方程时，我们使用逆运算来 “抵消” 对未知数所做的操作。

• If x has 5 added to it, subtract 5 from both sides.
• 如果 x 被加上了 5，就两边同时减去 5。
• If x is multiplied by 4, divide both sides by 4.
• 如果 x 被乘了 4，就两边同时除以 4。

This method turns a complicated equation into a simple statement such as x = 8.

这种方法可以把一个复杂的方程变成一个简单的结果，例如 x = 8。

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4. Solving by Adding or Subtracting | 通过加减求解

When a number is added to or subtracted from the unknown, use the opposite operation to remove it. Always perform the operation on both sides.

当一个数加在未知数上或从未知数中减去时，使用相反的运算把它消掉。务必在等号两边同时进行该运算。

Example 1: Solve x − 7 = 15.

例 1：解 x − 7 = 15。

x − 7 + 7 = 15 + 7 ⇒ x = 22Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com

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