Tag: KS3

📚 Mastering Pythagoras’ Theorem for KS3 Cambridge Mathematics | 掌握剑桥 KS3 数学毕达哥拉斯定理

Welcome to this KS3 Cambridge Mathematics revision article on Pythagoras’ Theorem. This topic appears frequently in Cambridge Checkpoint tests and lays the foundation for trigonometry and vector work in later stages. Understanding how to apply the theorem to right-angled triangles will help you solve problems involving lengths, distances and real-life measurements.

欢迎阅读这篇 KS3 剑桥数学关于毕达哥拉斯定理的复习文章。该主题经常出现在剑桥 Checkpoint 考试中，并为后续的三角学和向量学习奠定基础。理解如何在直角三角形中应用该定理，将帮助你解决涉及长度、距离和实际测量的问题。

1. What is Pythagoras’ Theorem? | 什么是毕达哥拉斯定理？

Pythagoras’ Theorem states that in any right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

毕达哥拉斯定理指出：在任何一个直角三角形中，斜边长度的平方等于另外两条直角边长度的平方和。

The theorem only works for right-angled triangles, so always check that the triangle has a 90° angle before using it.

该定理只适用于直角三角形，因此在使用之前一定要确认三角形中有一个 90° 角。

a² + b² = c²

Here c represents the hypotenuse, while a and b represent the two shorter sides called the legs.

这里 c 代表斜边，a 和 b 代表两条较短的边，称为直角边。

2. Identifying the Hypotenuse | 识别斜边

The hypotenuse is the longest side of a right-angled triangle and is always opposite the right angle.

斜边是直角三角形中最长的边，并且始终位于直角（90°角）的对面。

Do not assume the hypotenuse is the sloping side in a diagram; it is defined by its position opposite the right angle.

不要假设斜边一定是图中倾斜的那条边；斜边是由它与直角相对的位置决定的。

Labelling the sides correctly as a, b and c is the first step in solving any Pythagoras problem.

正确地将三条边标记为 a、b 和 c 是解决任何毕达哥拉斯定理问题的第一步。

3. The Standard Formula | 标准公式

The formula can be written as a² + b² = c², where c is the hypotenuse and a, b are the two shorter sides.

公式可以写成 a² + b² = c²，其中 c 是斜边，a 和 b 是两条较短的直角边。

To find the hypotenuse, add the squares of the two shorter sides and then take the square root.

要求斜边，先将两条直角边的平方相加，然后再开平方。

c = √(a² + b²)

To find a shorter side, rearrange the formula to a² = c² − b² or b² = c² − a².

要求一条直角边，可将公式变形为 a² = c² − b² 或 b² = c² − a²。

This rearrangement is essential because subtracting the squares gives the missing square of the leg.

这种变形至关重要，因为减去平方后就能得到缺失直角边的平方。

4. Finding the Hypotenuse | 求斜边

Suppose a right-angled triangle has shorter sides of 6 cm and 8 cm. To find the hypotenuse, calculate 6² + 8² = 36 + 64 = 100.

假设一个直角三角形的两条直角边分别为 6 厘米和 8 厘米。要求斜边，先计算 6² + 8² = 36 + 64 = 100。

Then take the square root: √100 = 10 cm. The hypotenuse is 10 cm long.

然后开平方：√100 = 10 厘米。斜边的长度为 10 厘米。

Always include the correct unit in your final answer, as missing units can lose marks in a Cambridge Checkpoint test.

最终答案中一定要写上正确的单位，因为在剑桥 Checkpoint 考试中漏写单位会失分。

Worked example: For sides of 5 m and 12 m, c² = 5² + 12² = 25 + 144 = 169, so c = √169 = 13 m.

例题：若两条直角边分别为 5 米和 12 米，则 c² = 5² + 12² = 25 + 144 = 169，所以 c = √169 = 13 米。

5. Finding a Shorter Side | 求直角边

When the hypotenuse and one shorter side are known, use subtraction. For example, if c = 13 cm and a = 5 cm, then b² = 13² − 5² = 169 − 25 = 144.

当已知斜边和一条直角边时，使用减法。例如，若 c = 13 厘米，a = 5 厘米，则 b² = 13² − 5² = 169 − 25 = 144。

Therefore b = √144 = 12 cm. This shows that the missing shorter side is 12 cm.

因此 b = √144 = 12 厘米。这表明缺失的直角边长为 12 厘米。

Many students accidentally add the two known squares even when finding a shorter side, so always decide first whether you are looking for the hypotenuse or a leg.

许多学生在求直角边时也会错误地把两个已知数的平方相加，所以在计算前一定要先判断要求的是斜边还是直角边。

A useful check is that the hypotenuse must be the longest side. If your answer for a shorter side comes out larger than the hypotenuse, you have made an error.

一个有效的检查方法是：斜边一定是最长的边。如果求出的直角边比斜边还长，就说明计算出错了。

6. Real-Life Applications | 实际应用

Pythagoras’ Theorem is used in many practical situations, such as finding the length of a ladder needed to reach a window, calculating the diagonal of a TV screen, or measuring the distance across a park.

毕达哥拉斯定理在许多实际场景中都有应用，例如求够到窗户所需的梯子长度、计算电视屏幕的对角线，或测量公园两点之间的距离。

In ladder problems, the ladder is usually the hypotenuse, and the ground and wall form the two shorter sides of a right-angled triangle.

在梯子问题中，梯子通常作为斜边，地面和墙壁构成直角三角形的两条直角边。

For example, if a ladder is placed 3 m from a wall and reaches a height of 4 m, the length of the ladder is √(3² + 4²) = √25 = 5 m.

例如，若一把梯子底端离墙 3 米，顶端达到 4 米高，则梯子的长度为 √(3² + 4²) = √25 = 5 米。

Another common example is finding the diagonal distance across a rectangular field of length 24 m and width 7 m: √(24² + 7²) = √(576 + 49) = √625 = 25 m.

另一个常见例子是求一个长 24 米、宽 7 米的矩形场地对角线的距离：√(24² + 7²) = √(576 + 49) = √625 = 25 米。

7. Distance on a Coordinate Grid | 坐标网格上的距离

Pythagoras’ Theorem can be used to find the distance between two points on a coordinate grid. The horizontal and vertical differences form the two shorter sides.

毕达哥拉斯定理可用于求坐标网格上两点之间的距离。水平差和垂直差构成两条直角边。

If point P is (1, 2) and point Q is (4, 6), the horizontal difference is 4 − 1 = 3 and the vertical difference is 6 − 2 = 4.

如果点 P 为 (1, 2)，点 Q 为 (4, 6)，则水平差为 4 − 1 = 3，垂直差为 6 − 2 = 4。

The distance PQ is then √(3² + 4²) = √(9 + 16) = √25 = 5 units.

那么距离 PQ 为 √(3² + 4²) = √(9 + 16) = √25 = 5 个单位。

This method works for any two points because the horizontal and vertical distances always meet at a right angle.

这种方法适用于任意两点，因为水平距离和垂直距离总是以直角相交。

In coordinate problems, always subtract the x-coordinates and y-coordinates separately before squaring them.

在坐标问题中，一定要分别将 x 坐标和 y 坐标相减，然后再平方。

8. Pythagoras in 3D | 三维空间中的毕达哥拉斯定理

In three-dimensional problems

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📚 Solving Linear Equations: Cambridge KS3 Maths (p317 Exercise Focus) | 剑桥 KS3 数学：解一元一次方程（p317 习题精讲）

Linear equations are the foundation of algebra at KS3. They show a balance between two expressions, and your job is to find the unknown value that makes the equation true. This revision guide walks you through the key methods, common errors, and exam-style examples you need for the Cambridge curriculum.

线性方程是 KS3 代数的基础。它们展示两个表达式之间的平衡关系，你的任务是找出使等式成立的未知数值。本复习指南将带你掌握剑桥课程所需的关键方法、常见错误和考试题型。

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

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

线性方程是未知数（通常写作 x）只出现一次方且指数为 1 的方程。它可以写成 ax + b = c 的形式，其中 a、b、c 都是数。

2x + 3 = 11

Here, x is the unknown, 2 is the coefficient of x, 3 is the constant term, and 11 is the value the expression must equal.

这里，x 是未知数，2 是 x 的系数，3 是常数项，11 是表达式必须等于的值。

2. The Balance Method | 天平法

Think of an equation as a balance scale. Whatever you do to one side, you must do to the other side to keep it balanced. This is the golden rule of solving equations.

把方程想象成一个天平。无论你对一边做什么，都必须对另一边做同样的操作，才能保持平衡。这是解方程的黄金法则。

You can add, subtract, multiply or divide both sides by the same non-zero number without changing the solution.

你可以在两边同时加上、减去、乘以或除以同一个非零数，而不会改变方程的解。

3. Solving One-Step Equations | 解一步方程

One-step equations need only one inverse operation to isolate the unknown. Always write the operation you perform on both sides clearly.

一步方程只需要一次逆运算就能隔离未知数。始终要清楚地写出你在两边同时进行的运算。

Example 1: Solve x + 5 = 12.

例 1：解 x + 5 = 12。

x + 5 − 5 = 12 − 5 → x = 7

Example 2: Solve 3x = 18.

例 2：解 3x = 18。

3x ÷ 3 = 18 ÷ 3 → x = 6

Example 3: Solve x ÷ 4 = 5.

例 3：解 x ÷ 4 = 5。

x ÷ 4 × 4 = 5 × 4 → x = 20

4. Solving Two-Step Equations | 解两步方程

A two-step equation involves two operations attached to the unknown, such as 2x + 3 = 11. You must undo the addition or subtraction first, then undo the multiplication or division.

两步方程涉及未知数上的两步运算，例如 2x + 3 = 11。你必须先处理加法或减法，再处理乘法或除法。

Example: Solve 2x + 3 = 11.

例：解 2x + 3 = 11。

Step 1: Subtract 3 from both sides to remove the constant.

步骤 1：两边同时减 3，消去常数项。

2x + 3 − 3 = 11 − 3 → 2x = 8

Step 2: Divide both sides by 2 to isolate x.

步骤 2：两边同时除以 2，隔离 x。

2x ÷ 2 = 8 ÷ 2 → x = 4

Always work in the reverse order of operations: undo addition or subtraction before multiplication or division.

始终按照运算的逆顺序来操作：先处理加减，再处理乘除。

5. Equations with Brackets | 含括号的方程

When an equation contains brackets, expand them first using the distributive law. Then solve the resulting equation as usual.

当方程含有括号时，先用分配律展开括号，然后按常规方法解所得的方程。

Example: Solve 3(x + 2) = 21.

例：解 3(x + 2) = 21。

Step 1: Expand the bracket by multiplying each term inside by 3.

步骤 1：把括号内每一项都乘以 3，展开括号。

3x + 6 = 21

Step 2: Subtract 6 from both sides.

步骤 2：两边同时减 6。

3x + 6 − 6 = 21 − 6 → 3x = 15

Step 3: Divide both sides by 3.

步骤 3：两边同时除以 3。

x = 5

6. Equations with Unknowns on Both Sides | 未知数在方程两边的情况

If the unknown appears on both sides of the equation, collect all x terms on one side and all constant terms on the other side.

如果未知数出现在方程的两边，就把所有含 x 的项移到一边，把所有常数项移到另一边。

Example: Solve 5x − 2 = 2x + 10.

例：解 5x − 2 = 2x + 10。

Step 1: Subtract 2x from both sides so x only appears on the left.

步骤 1：两边同时减 2x，使 x 只出现在左边。

5x − 2x − 2 = 2x − 2x + 10 → 3x − 2 = 10

Step 2: Add 2 to both sides.

步骤 2：两边同时加 2。

3x − 2 + 2 = 10 + 2 → 3x = 12

Step 3: Divide both sides by 3.

步骤 3：两边同时除以 3。

x = 4

7. Equations with Fractional Coefficients | 分数系数的方程

When the unknown is divided by a number, multiply both sides by that denominator first. This clears the fraction and makes the equation easier to solve.

当未知数被一个数除时，先两边同时乘以这个分母。这样可以去分母，使方程更容易解。

Example: Solve x / 3 + 2 = 5.

例：解 x / 3 + 2 = 5。

Step 1: Subtract 2 from both sides.

步骤 1：两边同时减 2。

x / 3 + 2 − 2 = 5 − 2 → x / 3 = 3

Step 2: Multiply both sides by 3.

步骤 2：两边同时乘以 3。

x / 3 × 3 = 3 × 3 → x = 9

For equations like 2x / 5 = 6, multiply both sides by 5 first, then divide by 2.

对于像 2x / 5 = 6 这样的方程，先两边同时乘以 5，再除以 2。

2x / 5 × 5 = 6 × 5 → 2x = 30 → x = 15

8. Word Problems Leading to Linear Equations | 列方程解应用题

Word problems ask you to translate everyday language into an equation. Let the unknown be x, write down the relationship, then solve.

应用题要求你把日常语言翻译成方程。设未知数为 x，写出关系式，然后求解。

Example: Three times a number plus 5 equals 20. Find the number.

例：一个数的 3 倍加 5 等于 20。求这个数。

Let the number be x. The equation is 3x + 5 = 20.

设这个数为 x。方程为 3x + 5 = 20。

3x + 5 = 20 → 3x = 15 → x = 5

Always check the wording: ‘more than’ means addition, ‘less than’ means subtraction, ‘times’ means multiplication, and ‘divided by’ means division.

始终注意措辞：’more than’ 表示加，’less than’ 表示减，’times’ 表示乘，’divided by’ 表示除。

9. Checking Your Answers | 验证答案

Always substitute your answer back into the original equation to make sure both sides are equal. This takes only a few seconds and can catch small mistakes.

始终把你的答案代回原方程，确认两边相等。这只需几秒钟，却能发现细小的错误。

Example: For 2x + 3 = 11, we found x = 4. Substitute x = 4.

例：对于 2x + 3 = 11，我们解得 x = 4。代入 x = 4。

2 × 4 + 3 = 8 + 3 = 11 ✓

The two sides match, so the solution is correct.

两边相等，所以解是正确的。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Many marks are lost not because the method is hard, but because of small slips. Watch out for these common errors.

很多失分并不是因为方法难，而是因为小的疏忽。请注意以下常见错误。

• Forgetting to do the same thing to both sides. If you subtract 3 on the left, you must subtract 3 on the right too.
• 忘记在两边同时进行相同操作。如果你在左边减 3，右边也必须减 3。
• Only dividing part of the expression. In 2x + 3 = 11, you must subtract 3 first, not divide by 2 first.
• 只除以表达式的一部分。在 2x + 3 = 11 中，你必须先减 3，而不是先除以 2。
• Making a sign error when moving terms across the equal sign.
• 移项时出现符号错误。
• Forgetting to expand brackets completely before solving.
• 在解方程前忘记完全展开括号。

Always show each step clearly. In Cambridge KS3 exams, method marks can be awarded even if the final answer is wrong.

始终清晰地写出每一步。在剑桥 KS3 考试中，即使最终答案错误，也可能得到方法分。

11. Practice Questions | 练习题

Try these exam-style questions. Solve each equation and check your answer.

尝试以下考试题型。解每个方程并检查你的答案。

• 4x + 7 = 31
• x / 2 − 3 = 6
• 5(x − 2) = 25
• 7x + 4 = 3x + 20

Answers: x = 6, x = 18, x = 7, x = 4.

答案：x = 6，x = 18，x = 7，x = 4。

12. Summary | 总结

To solve any KS3 linear equation, keep the balance, undo operations in reverse order, expand brackets, collect like terms, and check your solution. With practice, these steps become automatic.

要解任何 KS3 线性方程，就要保持平衡、按逆顺序进行运算、展开括号、合并同类项，并检验你的解。通过练习，这些步骤会变得自然而然。

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📚 Cambridge KS3 Mathematics: Comprehensive Topic Revision for Stages 7–9 | 剑桥 KS3 数学：阶段 7–9 全面主题复习

This revision guide covers the core skills you need for Cambridge Lower Secondary Mathematics at KS3. Use it as a quick reference when working through mixed practice pages such as p301_2.pdf. Each section below pairs key ideas with worked examples to help you revise efficiently.

本复习指南涵盖剑桥初中数学 KS3 阶段所需的核心技能。你可以在完成 p301_2.pdf 等混合练习页时将其作为快速参考。以下每一节都将关键知识点与示例配对，帮助你高效复习。

1. Number Operations and Place Value | 数的运算与位值

Understanding place value, order of operations and directed numbers is essential at KS3. For example, in 3.0 × 10² the digit 3 has a value of 300, not 3, because the power of ten moves the decimal point two places to the right.

理解位值、运算顺序和有向数在 KS3 阶段至关重要。例如，在 3.0 × 10² 中，数字 3 的值是 300，而不是 3，因为 10 的幂将小数点向右移动了两位。

• BIDMAS order: Brackets, Indices, Division/Multiplication, Addition/Subtraction.
• 中文：BIDMAS 运算顺序：括号、指数、乘除、加减。
• Directed numbers: A negative sign in front of a bracket changes the sign of every term inside.
• 中文：有向数：括号前的负号会改变括号内每一项的符号。

a × (b + c) = a × b + a × c

2. Fractions, Decimals and Percentages | 分数、小数和百分比

Converting between fractions, decimals and percentages is a key KS3 skill. To compare 2/5, 0.41 and 43%, write them all as percentages. Since 2/5 = 40% and 0.41 = 41%, the order is 40% < 41% < 43%.

在分数、小数和百分比之间进行转换是 KS3 的关键技能。要比较 2/5、0.41 和 43%，把它们都写成百分比。因为 2/5 = 40%，0.41 = 41%，所以顺序是 40% < 41% < 43%。

• Fraction to decimal: Divide the numerator by the denominator.
• 中文：分数转小数：用分子除以分母。
• Decimal to percentage: Multiply by 100 and add the % sign.
• 中文：小数转百分比：乘以 100 并加上 % 符号。

2/5 = 0.4 = 40%

3. Ratio and Proportion | 比与比例

Ratios compare quantities sharing a common unit. If the ratio of boys to girls is 3 : 2, the total number of parts is 3 + 2 = 5. To divide £40 in this ratio, one share is £40 ÷ 5 = £8, so boys get £24 and girls get £16.

比用于比较具有相同单位的数量。如果男生与女生的比是 3 : 2，那么总份数是 3 + 2 = 5。将 40 英镑按这个比例分配，一份是 40 ÷ 5 = 8 英镑，所以男生得 24 英镑，女生得 16 英镑。

• Divide in a ratio: Find the value of one share, then multiply by each ratio part.
• 中文：按比例分配：先求出一份的值，再乘以比的每一部分。
• Proportion: Direct proportion means both quantities increase or decrease at the same rate.
• 中文：比例：正比例意味着两个量以相同速率增加或减少。

part = (ratio part ÷ total parts) × total amount

4. Algebraic Expressions and Simplification | 代数表达式与化简

Algebra uses letters to stand for unknown numbers. Simplify expressions by collecting like terms. For example, 3a + 2b + 5a − b can be rearranged as 3a + 5a + 2b − b, which simplifies to 8a + b.

代数用字母表示未知数。通过合并同类项来化简表达式。例如，3a + 2b + 5a − b 可以整理为 3a + 5a + 2b − b，化简后得到 8a + b。

• Like terms: Terms with exactly the same letters and powers, such as 4x and −2x.
• 中文：同类项：具有完全相同的字母和指数的项，例如 4x 和 −2x。
• Expanding brackets: Multiply the term outside by every term inside.
• 中文：去括号：用括号外的项乘以括号内的每一项。

3a + 2b + 5a − b = 8a + b

5. Solving Linear Equations | 解一元一次方程

To solve an equation, do the same operation to both sides to isolate the unknown. For example, to solve 2x + 3 = 11, first subtract 3 from both sides to get 2x = 8, then divide both sides by 2 to find x = 4.

解方程时，对方程两边进行相同的运算以分离未知数。例如，解方程 2x + 3 = 11，先两边同时减去 3，得到 2x = 8，然后两边同时除以 2，求出 x = 4。

• Check solutions: Substitute your answer back into the original equation.
• 中文：检验解：将你的答案代回原方程。
• Equations with brackets: Expand first, then simplify and solve.
• 中文：含括号的方程：先去括号，再化简并求解。

2x + 3 = 11 → 2x = 8 → x = 4

6. Sequences and the nth Term | 数列与第 n 项

A linear sequence has a constant difference between consecutive terms. The nth term formula allows you to find any term without listing all previous ones. For the sequence 5, 8, 11, 14, … the common difference is 3, so the nth term is 3n + 2.

线性数列的相邻项差恒定。第 n 项公式使你不必列出前面所有项就能求出任意项。对于数列 5、8、11、14、……，公差是 3，所以第 n 项是 3n + 2。

• Common difference d: Subtract any term from the next term.
• 中文：公差 d：用后一项减去前一项。
• nth term: Use dn + (a − d), where a is the first term.
• 中文：第 n 项：使用 dn + (a − d)，其中 a 是首项。

nth term = dn + (a − d)

7. Angles and Properties of Shapes | 角度与图形性质

Know angle facts for points, lines, triangles and quadrilaterals. Angles on a straight line add to 180°, angles around a point add to 360°, and angles in a triangle add to 180°.

掌握点、线、三角形和四边形的角度性质。直线上的角相加为 180°，一点周围的角相加为 360°，三角形内角和为 180°。

• Parallel lines: Alternate angles are equal, corresponding angles are equal.
• 中文：平行线：内错角相等，同位角相等。
• Polygon angle sum: For an n-sided polygon, the interior angle sum is (n − 2) × 180°.
• 中文：多边形内角和：n 边形的内角和为 (n − 2) × 180°。

interior angle sum = (n − 2) × 180°

8. Perimeter, Area and Volume | 周长、面积与体积

Use the correct units and formulas for perimeter, area and volume. For compound shapes, split them into simpler parts and add or subtract the relevant areas.

使用正确的单位和公式计算周长、面积和体积。对于组合图形，将其拆分为更简单的部分，然后相加或相减相应的面积。

• Trapezium area: Half the sum of the parallel sides times the height.
• 中文：梯形面积：两底之和的一半乘以高。
• Cuboid volume: Length times width times height.
• 中文：长方体体积：长乘以宽乘以高。

Area of trapezium = ½ × (a + b) × h

9. Transformations | 变换

Transformations move or change shapes. Reflection, rotation, translation and enlargement each have specific rules. A translation is described by a column vector showing horizontal and vertical movement.

变换使图形移动或改变。反射、旋转、平移和放大各有特定规则。平移用列向量表示水平和垂直移动。

• Reflection: Each point is the same perpendicular distance from the mirror line.
• 中文：反射：每个点到镜线的垂直距离相等。
• Enlargement: A scale factor k multiplies each side length by k.
• 中文：放大：比例系数 k 使每条边长度乘以 k。

scale factor k → new length = k × original length

10. Statistics and Probability | 统计与概率

Collect, represent and interpret data using bar charts, pie charts and frequency tables. Probability can be expressed as a fraction, decimal or percentage. The probabilities of all possible outcomes add to 1.

使用条形图、饼图和频数表收集、表示和解释数据。概率可以表示为分数、小数或百分比。所有可能结果的概率之和为 1。

• Mean: Sum of values divided by the number of values.
• 中文：平均数：所有数值之和除以数值的个数。
• Probability: Favourable outcomes divided by total outcomes.
• 中文：概率：有利结果数除以总结果数。

Probability = favourable outcomes ÷ total outcomes

11. Coordinates and Linear Graphs | 坐标与线性图

Plot points in all four quadrants and draw straight-line graphs. The equation y = mx + c gives the gradient and y-intercept. The gradient m measures steepness, while c is where the line crosses the y-axis.

在四个象限中描点并绘制直线图。方程 y = mx + c 给出斜率和 y 轴截距。斜率 m 衡量倾斜程度，而 c 是直线与 y 轴的交点。

• Gradient: Rise over run, or change in y divided by change in x.
• 中文：斜率：纵向变化量除以横向变化量。
• Parallel lines: Lines with the same gradient are parallel.
• 中文：平行线：斜率相同的直线互相平行。

gradient m = Δy ÷ Δx

12. Problem Solving and Reasoning | 问题解决与推理

Apply your skills to multi-step problems. Read carefully, identify knowns and unknowns, choose a strategy and check your answer. In mixed practice such as p301_2.pdf, you may need to combine several topics in one question.

将你的技能应用于多步骤问题。仔细读题，确定已知量和未知量，选择策略并检验答案。在 p301_2.pdf 等混合练习中，你可能需要在一道题中综合多个主题。

• Estimate first: Use estimation and inverse operations to check results.
• 中文：先估算：使用估算和逆运算检验结果。
• Show working: Clear steps earn method marks even if the final answer is wrong.
• 中文：展示过程：即使最终答案错误，清晰的步骤也能获得方法分。

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📚 Probability Foundations from Cambridge KS3 Maths p299_2 | 剑桥KS3数学p299_2概率基础

Probability is one of the most practical topics in the Cambridge KS3 Mathematics course. This article works through the core ideas behind a typical p299_2 style question, helping you build confidence with theoretical probability, experimental probability, sample spaces, and common exam techniques.

概率是剑桥KS3数学课程中最实用的主题之一。这篇文章围绕典型的p299_2类型题目，梳理理论概率、实验概率、样本空间和常见考试技巧，帮助你建立扎实的解题能力。

1. What is Probability? | 什么是概率？

Probability measures how likely an event is to happen. It is always written as a number between 0 and 1, where 0 means impossible and 1 means certain. You can also write probability as a fraction, decimal, or percentage.

概率用来衡量一个事件发生的可能性大小。它总是写成0到1之间的一个数，其中0表示不可能发生，1表示一定发生。概率也可以用分数、小数或百分数表示。

For example, when you flip a fair coin, the probability of getting heads is 1/2, because there are two equally likely outcomes and only one is heads.

例如，抛一枚均匀硬币时，正面朝上的概率是1/2，因为一共有两个等可能的结果，而正面只是其中一个。

2. Sample Spaces and Outcomes | 样本空间与结果

An outcome is a single possible result of an experiment. The sample space is the set of all possible outcomes. In a KS3 probability question, you must first identify all possible outcomes before calculating any probability.

一个结果是指实验中某一个可能发生的情况。样本空间是指所有可能结果的集合。在KS3概率题中，你必须先找出所有可能的结果，再计算概率。

For a fair six-sided dice, the sample space is {1, 2, 3, 4, 5, 6}. If the dice is fair, every outcome has the same chance of occurring.

对于一枚均匀的六面骰子，样本空间是{1, 2, 3, 4, 5, 6}。如果骰子是均匀的，每个结果发生的可能性都相同。

• English: List outcomes clearly before choosing the ones you need.
  中文：先清楚地列出所有结果，再从中选出你需要的结果。
• English: Check that all outcomes are equally likely only when the question says fair or random.
  中文：只有当题目说明是均匀或随机时，才能认为所有结果等可能。

3. Calculating Theoretical Probability | 计算理论概率

Theoretical probability uses reasoning to predict how likely an event is. The formula is simple: divide the number of favourable outcomes by the total number of possible outcomes.

理论概率通过推理来预测事件发生的可能性。公式很简单：用有利结果的数量除以所有可能结果的总数。

P(event) = number of favourable outcomes ÷ total number of outcomes

If a bag contains 5 red counters, 3 blue counters, and 2 green counters, then the total number of counters is 10. The probability of picking a red counter at random is 5 ÷ 10, which simplifies to 1/2.

如果一个袋子里有5个红色计数器、3个蓝色计数器和2个绿色计数器，那么计数器总数是10。随机取出一个红色计数器的概率是5 ÷ 10，化简后是1/2。

P(red) = 5 / 10 = 1/2

4. The Probability Scale | 概率标度

The probability scale is a number line from 0 to 1. Events that are impossible sit at 0, events that are certain sit at 1, and equally likely events sit exactly in the middle at 0.5. Probability can never be less than 0 or greater than 1.

概率标度是一条从0到1的数轴。不可能事件在0的位置，必然事件在1的位置，等可能事件正好在中间0.5处。概率永远不会小于0或大于1。

Many KS3 exam questions ask you to mark a probability on a scale or to classify an event as impossible, unlikely, even, likely, or certain. Always connect the fraction to a position on the number line.

很多KS3考试题会要求你在标度上标出某个概率，或者判断一个事件是不可能、不太可能、等可能、很可能还是必然。一定要把分数和数轴上的位置对应起来。

5. Complementary Events | 互补事件

Complementary events are two opposite outcomes that cover all possibilities. For example, when you roll a dice, the complement of rolling a 6 is rolling 1, 2, 3, 4, or 5. The two probabilities always add up to 1.

互补事件是两个相反且覆盖所有可能性的结果。例如，掷骰子时，掷出6的互补事件是掷出1、2、3、4或5。这两个事件的概率之和总是等于1。

P(A’) = 1 – P(A)

If the probability of raining tomorrow is 0.3, then the probability of not raining tomorrow is 1 – 0.3 = 0.7. This shortcut is very useful when finding the opposite event is easier than finding the event itself.

如果明天下雨的概率是0.3，那么明天不下雨的概率就是1 – 0.3 = 0.7。当求相反事件的概率更容易时，这个快捷方法非常有用。

6. Mutually Exclusive Events | 互斥事件

Mutually exclusive events cannot happen at the same time. For example, when you pick one counter from a bag, picking a red counter and picking a blue counter are mutually exclusive because the single counter cannot be both red and blue.

互斥事件是指不可能同时发生的事件。例如，当你从袋子里取出一个计数器时，取出红色计数器和取出蓝色计数器就是互斥事件，因为一个计数器不可能既是红色又是蓝色。

For mutually exclusive events, you can add their individual probabilities to find the probability that either one happens. For example, the probability of picking a red or blue counter from the earlier bag is 5/10 + 3/10 = 8/10, which is 4/5.

对于互斥事件，你可以把各自发生的概率相加，得到其中一个发生的概率。例如，从之前的袋子里取出红色或蓝色计数器的概率是5/10 + 3/10 = 8/10，也就是4/5。

P(A or B) = P(A) + P(B)

7. Experimental Probability and Relative Frequency | 实验概率与相对频率

Experimental probability is based on data from an actual experiment or survey. The formula is similar to theoretical probability, but the numbers come from observed outcomes rather than from reasoning.

实验概率基于真实实验或调查得到的数据。公式与理论概率相似，但数字来自观察到的结果，而不是推理。

Experimental probability = number of times the event occurs ÷ total number of trials

If you spin a spinner 50 times and it lands on blue 12 times, the experimental probability of landing on blue is 12/50, which simplifies to 6/25. This is also called the relative frequency of blue.

如果你转动一个转盘50次，其中12次停在蓝色区域，那么停在蓝色区域的实验概率是12/50，化简为6/25。这个值也叫做蓝色的相对频率。

As the number of trials increases, experimental probability usually gets closer to theoretical probability. This is an important idea in statistics and helps explain why real experiments may differ from predictions.

随着试验次数增加，实验概率通常会越来越接近理论概率。这是统计学中的一个重要概念，也解释了为什么真实实验结果可能与预测有所差异。

8. Listing Outcomes and Tree Diagrams | 列出结果与树状图

When a question involves two or more steps, listing all outcomes in a systematic way prevents missing any possibilities. A tree diagram is a useful tool for showing every combination clearly.

当题目涉及两个或更多步骤时，系统地列出所有结果可以避免遗漏。树状图是一种清晰展示每一个组合的有效工具。

For example, if you flip a coin and roll a fair four-sided dice numbered 1 to 4, the sample space has 2 × 4 = 8 equally likely outcomes. Tree diagrams show one branch for the coin and four branches from each coin outcome for the dice.

例如，如果你抛一枚硬币，同时掷一个标有1到4的均匀四面骰子，样本空间有2 × 4 = 8个等可能结果。树状图会先分出硬币的两个分支，再从每个硬币结果分出骰子的四个分支。

• English: Multiply the number of choices at each stage to find the total number of outcomes.
  中文：把每一步的选择数相乘，就得到总结果数。
• English: Check tree diagrams for repeated branches or missing combinations.
  中文：检查树状图是否有重复分支或遗漏组合。

9. Worked Example from p299_2 Style Question | p299_2 类型例题解析

Let us solve a question similar to p299_2. A bag contains 6 red sweets, 4 yellow sweets, and 2 green sweets. One sweet is taken at random. Work out the probability that the sweet is not yellow.

我们来解一道类似p299_2的题目。一个袋子里有6颗红色糖果、4颗黄色糖果和2颗绿色糖果。随机取出1颗糖果，求这颗糖果不是黄色的概率。

First, find the total number of sweets: 6 + 4 + 2 = 12. The probability of picking a yellow sweet is 4/12, which simplifies to 1/3. Therefore, the probability of not picking yellow is 1 – 1/3 = 2/3.

首先，求出糖果总数：6 + 4 + 2 = 12。取出黄色糖果的概率是4/12，化简为1/3。因此，取出不是黄色糖果的概率是1 – 1/3 = 2/3。

P(not yellow) = 1 – P(yellow) = 1 – 4/12 = 8/12 = 2/3

You could also count the non-yellow sweets directly: 6 red + 2 green = 8, so 8/12 = 2/3. Both methods give the same answer, but using the complement is often quicker.

你也可以直接数非黄色糖果的数量：6颗红色加2颗绿色共8颗，所以概率是8/12 = 2/3。这两种方法得到相同答案，但使用互补事件通常更快。

10. Common Misconceptions | 常见误区

Students often add probabilities incorrectly when events are not mutually exclusive, or they forget that probability must be between 0 and 1. Another common mistake is using the total number of favourable outcomes without simplifying the fraction.

学生经常在事件不是互斥时错误地把概率相加，或者忘记概率必须在0到1之间。另一个常见错误是写出有利结果的数量，却没有化简分数。

• English: Never write a probability as an improper fraction or a decimal greater than 1.
  中文：绝不要把概率写成假分数或大于1的小数。
• English: Make sure the denominator represents all possible outcomes, not just some of them.
  中文：确保分母代表所有可能结果，而不只是其中一部分。
• English: When using experimental probability, do not confuse the number of trials with the number of successes.
  中文：使用实验概率时，不要把试验次数与成功次数混淆。

11. Exam Tips | 考试技巧

In Cambridge KS3 assessments, probability questions usually award marks for clear working as well as the final answer. Always show the fraction before simplifying, and write a short sentence to explain your reasoning if the question asks for a worded answer.

在剑桥KS3测评中，概率题通常既给过程分也给最终答案分。一定要先写出未化简的分数，再化简；如果题目要求用文字说明，也要写出简短的理由。

When a question asks you to compare experimental and theoretical probability, mention that the difference is due to chance or a small number of trials. Use the word relative frequency when describing experimental results.

当题目要求比较实验概率和理论概率时，要提到差异是由随机性或者试验次数较少造成的。描述实验结果时使用相对频率这个术语。

12. Summary and Key Formulae | 总结与关键公式

Probability is fundamentally about comparing the number of favourable outcomes with the total possible outcomes. Once you master the basic formula, complementary events, and mutually exclusive events, most KS3 questions become straightforward.

概率的核心是比较有利结果的数量与所有可能结果的数量。一旦你掌握了基本公式、互补事件和互斥事件，大多数KS3题目都会变得很简单。

P(event) = favourable outcomes ÷ total outcomes

P(A’) = 1 – P(A)

For mutually exclusive events: P(A or B) = P(A) + P(B)

Keep practising with mixed example questions such as drawing counters, spinning spinners, rolling dice, and survey data. The more you practise, the more naturally you will choose the correct method in an exam.

坚持练习混合例题，例如取计数器、转盘、掷骰子和调查数据。练习得越多，考试时就越能自然地选择正确的方法。

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📚 Solving Linear Equations: Cambridge KS3 Mathematics Revision | 解一元一次方程：剑桥 KS3 数学复习

Linear equations are one of the first major algebraic tools you meet in Key Stage 3. They let you change a word problem into a visual balance and find an unknown value step by step. Mastery of this topic is essential for Cambridge Checkpoint because later topics such as graphs, sequences, and problem solving all rely on fluent equation solving.

一元一次方程是你在 Key Stage 3 遇到的第一个重要代数工具。它们让你能把文字题转化为一架可视的天平，并逐步求出未知数的值。掌握这一主题对剑桥 Checkpoint 考试至关重要，因为后续的图像、数列和问题解决等内容都依赖于熟练地解方程。

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 variable, usually written as x, appears only to the first power. There are no x², x³, square roots, or products of x with itself. Examples include x + 5 = 12, 3x − 4 = 11, and 2(x + 1) = 10.

一元一次方程是一种仅含有一个未知数（通常写作 x）的方程，且未知数只出现一次方。方程中没有 x²、x³、平方根或 x 与自身相乘的项。例如 x + 5 = 12、3x − 4 = 11 和 2(x + 1) = 10。

The word linear tells us that its graph is a straight line. For KS3, you will usually solve equations by rearranging rather than by drawing graphs, but knowing the connection helps you see why equations have one solution.

线性这个词告诉我们它的图像是一条直线。在 KS3 阶段，你通常通过变形来解方程，而不是画图，但了解这种联系有助于你明白为什么这类方程只有一个解。

2. The Balance Method | 平衡法

Think of an equation as a set of balance scales. The equals sign is the pivot. If you add, subtract, multiply, or divide both sides by the same non-zero amount, the scales stay balanced. This is the golden rule of equation solving.

把方程想象成一架天平。等号就是支点。如果你对方程两边同时加、减、乘或除以同一个非零的数，天平仍然保持平衡。这是解方程的金科玉律。

We can use the balance method to solve x + 7 = 15. To isolate x, subtract 7 from both sides: x + 7 − 7 = 15 − 7, which gives x = 8.

我们可以用平衡法来解 x + 7 = 15。为了单独留下 x，两边同时减去 7：x + 7 − 7 = 15 − 7，得到 x = 8。

x + 7 = 15
x = 15 − 7
x = 8

3. Inverse Operations | 逆运算

Every operation has an inverse. Addition and subtraction undo each other, and multiplication and division undo each other. When solving, use the inverse operation to remove a number from the side containing the variable.

每种运算都有逆运算。加法与减法互为逆运算，乘法与除法互为逆运算。解方程时，使用逆运算来去掉含有未知数那一边的数。

For x − 5 = 9, add 5 to both sides: x = 14. For 4x = 28, divide both sides by 4: x = 7.

例如 x − 5 = 9，两边同时加 5，得到 x = 14。对于 4x = 28，两边同时除以 4，得到 x = 7。

Keeping a short table of inverse operations nearby is useful when you start. It prevents you from using the wrong operation under time pressure.

开始时在手边放一张简短的逆运算表很有用。它可以防止你在时间紧迫时用错运算。

4. Solving Two-Step Equations | 解两步方程

A two-step equation needs two inverse operations. The standard order is to undo addition or subtraction first, then undo multiplication or division. For example, solve 2x + 5 = 17.

两步方程需要两次逆运算。标准的顺序是先去加减，再去乘除。例如，解方程 2x + 5 = 17。

First subtract 5 from both sides: 2x = 12. Then divide both sides by 2: x = 6. Always record each step clearly.

首先两边同时减去 5：2x = 12。然后两边同时除以 2：x = 6。务必清晰记录每一步。

2x + 5 = 17
2x = 17 − 5
2x = 12
x = 6

If you prefer, you can first divide every term by 2, but this usually introduces fractions: x + 2.5 = 8.5. The balance method keeps it simpler.

如果你愿意，也可以先把每一项都除以 2，但这通常会引入分数：x + 2.5 = 8.5。平衡法会让计算更简单。

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📚 Cambridge KS3 Maths: Solving Linear Equations with Perimeter Problems | 剑桥 KS3 数学：用周长问题解一元一次方程

Linear equations are one of the most useful tools in Key Stage 3 mathematics. In this lesson, we focus on a common Cambridge Lower Secondary question style: using the perimeter of a rectangle to form a linear equation. By the end, you will be able to set up, solve, and check equations from word problems.

线性方程是初中数学中最有用的工具之一。本节课我们重点讲解剑桥初中常见的题型：利用矩形的周长建立一元一次方程。学完本课后，你将能够根据文字题建立方程、求解并检验结果。

1. What Are Linear Equations? | 什么是线性方程？

A linear equation is an equation in which the highest power of the unknown is 1. For example, 2x + 5 = 17 is a linear equation because x appears only to the first power.

线性方程是指未知数的最高次数为 1 的方程。例如 2x + 5 = 17 就是一个线性方程，因为 x 只出现一次方。

The solution of an equation is the value of the unknown that makes the equation true. If we replace x with 6 in 2x + 5 = 17, we get 2 × 6 + 5 = 17, so x = 6 is the solution.

方程的解是使方程成立的未知数的值。如果把 x = 6 代入 2x + 5 = 17，就得到 2 × 6 + 5 = 17，所以 x = 6 是方程的解。

In Cambridge KS3, you will meet linear equations in many contexts, including number puzzles, angles, sequences, and perimeter problems.

在剑桥 KS3 课程中，你会在许多情境中遇到线性方程，包括数字谜题、角度、数列和周长问题。

2. Key Skills: Collecting Like Terms | 关键技能：合并同类项

Before solving equations, you need to be confident with collecting like terms. Like terms have exactly the same variable part. For example, 2x and 3x are like terms, but 2x and 3 are not.

在解方程之前，你需要熟练掌握合并同类项。同类项是指变量部分完全相同的项。例如 2x 和 3x 是同类项，但 2x 和 3 不是同类项。

To collect like terms, add or subtract the coefficients while keeping the variable unchanged. For example, 2x + 3x = 5x and 4x – x = 3x.

合并同类项时，把系数相加或相减，变量保持不变。例如 2x + 3x = 5x，4x – x = 3x。

When an expression contains brackets, expand them first. For example, 2(x + 1) becomes 2x + 2.

当表达式中含有括号时，要先去括号。例如 2(x + 1) 去括号后得到 2x + 2。

2(x + 1) = 2x + 2

3. Understanding Perimeter Expressions | 理解周长表达式

The perimeter of a shape is the total distance around its outside. For a rectangle with length l and width w, the perimeter P is given by the formula P = 2(l + w).

图形的周长是围绕它外部的总长度。对于长为 l、宽为 w 的矩形，周长 P 由公式 P = 2(l + w) 给出。

P = 2(l + w)

If the length and width are given as algebraic expressions, we can substitute them into the perimeter formula to build an equation.

如果长和宽是用代数式给出的，我们可以把它们代入周长公式，从而建立一个方程。

For example, if a rectangle has length (2x + 3) cm and width (x – 1) cm, the perimeter expression is:

例如，如果一个矩形的长为 (2x + 3) cm，宽为 (x – 1) cm，那么周长表达式为：

P = 2[(2x + 3) + (x – 1)]

4. Setting Up the Equation | 建立方程

Once you have the perimeter expression, read the word problem carefully to find the given perimeter value. In our example, the perimeter is 34 cm.

得到周长表达式后，仔细阅读文字题，找到已知的周长值。在我们的例子中，周长是 34 cm。

Set the algebraic expression equal to the known perimeter. This gives the equation:

让代数表达式等于已知

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📚 Cambridge KS3 Maths: Solving Linear Equations from Exercise 2 (p.280) | 剑桥 KS3 数学：求解一元一次方程（第280页练习2）

Welcome to this TutorHao revision lesson for Cambridge KS3 Mathematics. In this article we work through the style of questions found in Exercise 2 on page 280 of the Cambridge Lower Secondary Maths course. The focus is linear equations, balancing steps, expanding brackets and checking answers.

欢迎来到 TutorHao 的剑桥 KS3 数学复习课。本文讲解剑桥初中数学教材第280页练习2中常见的一元一次方程题型，重点包括等式平衡步骤、去括号、未知数在两边以及代入检验。

1. Understanding the Exercise 2 Context | 理解练习2的背景

Page 280 Exercise 2 in the Cambridge KS3 course usually contains a mixed set of linear equations. The questions are designed to test whether you can keep an equation balanced while isolating the unknown.

剑桥 KS3 教材第280页练习2通常包含一组混合的一元一次方程。题目旨在考查你是否能在分离未知数的同时保持等式两边平衡。

In KS3, the unknown is usually written as x, y or n, and the solution must make the original equation true. This means every step should preserve equality.

在 KS3 阶段，未知数通常写作 x、y 或 n，解必须能使原方程成立。这意味着每一步都要保持等式的相等关系。

2. Key Vocabulary and Notation | 核心词汇与符号

Before solving, make sure you know these key terms: equation, expression, variable, coefficient, constant, left-hand side (LHS), right-hand side (RHS), solution and root.

解题之前，请确保你掌握以下术语：方程、表达式、变量、系数、常数项、左边（LHS）、右边（RHS）、解和根。

For example, in 3x + 5 = 14, the coefficient is 3, the constant is 5, and x is the variable. The left-hand side is 3x + 5 and the right-hand side is 14.

例如，在 3x + 5 = 14 中，系数是 3，常数项是 5，x 是变量。左边是 3x + 5，右边是 14。

3. The Balancing Method | 平衡法

The balancing method is the golden rule: whatever you do to one side of an equation, you must do to the other side. This keeps the equation true.

平衡法是黄金法则：对方程一边进行任何运算，另一边也必须进行相同运算。这样才能保持方程成立。

Think of an equation as a set of scales. If you add 2 kg to one side, you must add 2 kg to the other side to keep it balanced. The same idea applies to subtraction, multiplication and division.

可以把方程想象成一台天平。如果在一侧加 2 kg，就必须在另一侧也加 2 kg 才能保持平衡。减法、乘法和除法也同样适用。

For 3x + 5 = 14, subtract 5 from both sides: 3x + 5 − 5 = 14 − 5, giving 3x = 9. Then divide both sides by 3: 3x ÷ 3 = 9 ÷ 3, so x = 3.

对于 3x + 5 = 14，两边同时减去 5：3x + 5 − 5 = 14 − 5，得到 3x = 9。然后两边同时除以 3：3x ÷ 3 = 9 ÷ 3，所以 x = 3。

3x + 5 = 14 → 3x = 9 → x = 3

4. Solving Two-Step Equations | 求解两步方程

A two-step equation requires exactly two inverse operations to isolate the variable. The usual order is: undo addition or subtraction first, then undo multiplication or division.

两步方程需要正好两个逆运算来分离变量。通常顺序是：先消去加法或减法，再消去乘法或除法。

Solve 5x − 7 = 18. Step 1: add 7 to both sides: 5x = 25. Step 2: divide both sides by 5: x = 5.

解 5x − 7 = 18。第一步：两边同时加 7，得 5x = 25。第二步：两边同时除以 5，得 x = 5。

Always show both steps clearly. If you skip a step, it is easy to lose track of the operations and make a sign error.

一定要清楚地写出两步。如果省略步骤，就很容易忘记运算顺序并出现符号错误。

5x − 7 = 18 → 5x = 25 → x = 5

5. Equations with Brackets | 含括号的方程

When an equation has brackets, expand them first using the distributive law. Then solve the simpler equation using the balancing method.

当方程含有括号时，先用分配律展开括号。然后用平衡法解简化后的方程。

Solve 2(x + 3) = 16. Expand: 2x + 6 = 16. Subtract 6 from both sides: 2x = 10. Divide by 2: x = 5.

解 2(x + 3) = 16。展开得 2x + 6 = 16。两边同时减 6 得 2x = 10，再除以 2 得 x = 5。

Sometimes you will see a negative outside the bracket, such as −3(x − 2) = 21. Expand carefully: −3x + 6 = 21. Then subtract 6: −3x = 15. Divide by −3: x = −5.

有时括号外是负号，例如 −3(x − 2) = 21。展开时要仔细：−3x + 6 = 21。然后减 6 得 −3x = 15，再除以 −3 得 x = −5。

6. Unknowns on Both Sides | 未知数在两边

If the unknown appears on both sides of the equation, collect the variable terms on one side and the constants on the other side.

如果未知数出现在方程两边，就先把含未知数的项移到一边，把常数项移到另一边。

Solve 7x + 4 = 3x + 20. Subtract 3x from both sides: 4x + 4 = 20. Subtract 4: 4x = 16. Divide by 4: x = 4.

解 7x + 4 = 3x + 20。两边同时减去 3x，得 4x + 4 = 20。再减 4 得 4x = 16，除以 4 得 x = 4。

If the variable terms cancel completely, the equation may have no solution or infinitely many solutions. For KS3, focus on the usual unique-solution case.

如果含未知数的项完全抵消，方程可能无解或有无穷多个解。KS3 阶段主要掌握通常有唯一解的情况。

7. Forming Equations from Word Problems | 根据文字题列方程

Many Exercise 2 questions ask you to form an equation from a written problem. Read the question carefully, define the unknown with a letter, and translate each phrase into algebraic language.

练习2中的许多题目要求你根据文字题列出方程。仔细读题，用字母表示未知数，并把每个短语转化为代数语言。

Example: “Three more than twice a number is 19.” Let the number be n. Then 2n + 3 = 19. Subtract 3: 2n = 16. Divide by 2: n = 8.

例题：”一个数的两倍加 3 等于 19。” 设这个数为 n，则 2n + 3 = 19。减 3 得 2n = 16，除以 2 得 n = 8。

Example: “The perimeter of a rectangle is 26 cm. Its length is 3 cm more than its width. Find the width.” Let width be w, length is w + 3. Perimeter: 2(w + w + 3) = 26, so 4w + 6 = 26. Subtract 6: 4w = 20. Divide by 4: w = 5 cm.

例题：”一个长方形的周长是 26 cm，长比宽多 3 cm。求宽。” 设宽为 w，则长为 w + 3。周长：2(w + w + 3) = 26，所以 4w + 6 = 26。减 6 得 4w = 20，除以 4 得 w = 5 cm。

8. Checking Your Solution | 检验你的解

Always substitute your answer back into the original equation to check. The left-hand side must equal the right-hand side.

一定要把答案代回原方程检验。左边必须等于右边。

For 5x − 7 = 18 with x = 5: LHS = 5(5) − 7 = 25 − 7 = 18 = RHS. Correct.

对于 5x − 7 = 18 且 x = 5：左边 = 5(5) − 7 = 25 − 7 = 18 = 右边。正确。

Checking also helps catch sign errors and arithmetic mistakes before you move to the next question. It is a quick way to boost your accuracy.

检验还能帮助你在做下一题之前发现符号错误和计算错误。这是提高准确率的快速方法。

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

Mistake 1: Applying an operation to only one side. Remember, every operation must be applied to both sides.

错误1：只对一边进行运算。记住，每一运算都必须同时作用于两边。

Mistake 2: Incorrect sign when expanding brackets, especially with negative multipliers.

错误2：展开括号时符号错误，尤其是括号前为负号时。

Mistake 3: Confusing the order of inverse operations. Undo addition or subtraction before multiplication or division.

错误3：逆运算顺序混淆。应先消去加减法，再消去乘除法。

Mistake 4: For

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📚 Cambridge KS3 Maths: Mixed Exercise p297_2 Revision | 剑桥 KS3 数学：p297_2 混合练习复习

This revision article covers a mixed set of Cambridge KS3 Mathematics skills similar to those found in page 297 exercise 2. You will review fractions, decimals, percentages, negative numbers, algebraic simplification, simple equations, angle facts, area and perimeter, ratio, probability, and data interpretation. Work through each section, check the worked examples, and then try similar questions on your own.

本复习文章涵盖一组剑桥 KS3 数学混合技能，题型与第 297 页第 2 题相似。你将复习分数、小数、百分比、负数、代数式化简、简单方程、角度知识、面积与周长、比、概率以及数据解读。逐节学习并检查示例，然后自己尝试类似的问题。

1. Working with Fractions | 分数运算

To add or subtract fractions, first rewrite them with a common denominator. For example, to work out 1/3 + 1/4, use denominator 12 because 12 is the lowest common multiple of 3 and 4.

进行分数加减法时，先通分为同分母。例如计算 1/3 + 1/4，使用公分母 12，因为 12 是 3 和 4 的最小公倍数。

1/3 + 1/4 = 4/12 + 3/12 = 7/12

To multiply fractions, multiply the numerators and denominators separately. Always write your final answer in its simplest form by dividing the top and bottom by any common factor.

分数相乘时，分子乘分子，分母乘分母。最后答案要化为最简形式，将分子分母同时除以公因数。

2/5 × 3/4 = 6/20 = 3/10

To divide by a fraction, keep the first fraction, change the division sign to multiplication, and flip the second fraction. Simplify if possible.

除以一个分数时，保留第一个分数，将除号改为乘号，并把第二个分数取倒数。结果能约分则约分。

3/4 ÷ 2/3 = 3/4 × 3/2 = 9/8 = 1 1/8

2. Fractions, Decimals and Percentages | 分数、小数和百分比转换

To convert a fraction to a decimal, divide the numerator by the denominator. To convert a decimal to a percentage, multiply by 100. These skills are needed for comparing quantities in mixed questions.

分数化小数的方法是分子除以分母；小数化百分比的方法是乘以 100。在混合题中比较数量时经常需要这些技能。

3/8 = 0.375 and 0.375 × 100 = 37.5%

To convert a percentage to a fraction, write the percentage over 100 and simplify. For example, 45% becomes 45/100, which simplifies to 9/20.

百分比化分数时，先写成百分之几再约分。例如 45% 写成 45/100，约分后为 9/20。

45% = 45/100 = 9/20

When comparing values, change them all to the same form. For example, 0.4, 2/5 and 40% are equal, so choosing the correct form depends on the question.

比较数值时，将所有数值化成相同形式。例如 0.4、2/5 和 40% 是相等的，因此选择哪种形式取决于题目要求。

3. Negative Numbers | 负数运算

When adding and subtracting negative numbers, a number line can help. Moving right is positive; moving left is negative. For example, start at -3 and move 5 places right to reach 2.

进行负数加减法时，数轴可以帮助理解。向右移动为正，向左移动为负。例如从 -3 出发向右移动 5 位到达 2。

-3 + 5 = 2 and -4 – 3 = -7

Multiplying or dividing two negative numbers gives a positive result. If the signs are different, the result is negative. This sign rule is often tested in mixed arithmetic questions.

两个负数相乘或相除结果为正；异号相乘或相除结果为负。混合算术题中经常考查这一符号规则。

(-6) × (-4) = 24 and 12 ÷ (-3) = -4

Use brackets to keep signs clear. For example, 5 – (-2) is the same as 5 + 2. Be careful when a negative sign appears next to a subtraction sign.

使用括号使符号清晰。例如 5 – (-2) 等于 5 + 2。当负号与减号相邻时要特别小心。

5 – (-2) = 5 + 2 = 7

4. Simplifying Algebraic Expressions | 化简代数式

Collect like terms by adding or subtracting the coefficients of terms that have the same variable. Constant terms without a variable are also like terms and should be combined separately.

合并同类项时，将含有相同变量的项的系数相加减。不含变量的常数项也属于同类项，应单独合并。

3a + 5b – 2a + 7b = a + 12b

Use the distributive law to expand brackets. Multiply each term inside the bracket by the term outside, paying attention to signs.

使用分配律展开括号。括号外的项要与括号内每一项相乘，并注意符号。

4(2x + 3) = 8x + 12

To factorise an expression, find the highest common factor of the terms and write it outside a bracket. Check your answer by expanding the bracket again.

因式分解时，找出各项的最高公因式并写在括号外。可以通过再次展开括号来检查答案。

6x + 9 = 3(2x + 3)

5. Solving Linear Equations | 解一次方程

Keep an equation balanced by doing the same operation to both sides. Your aim is to isolate the variable on one side of the equals sign.

解方程时保持等式两边平衡，对方程两边同时进行相同运算。目标是使变量单独位于等号一边。

x + 5 = 12 → x = 12 – 5 = 7

If the variable has a multiplier, divide both sides by that number. If the variable is divided by a number, multiply both sides by that number.

若变量前有系数，两边同时除以该系数；若变量除以某数，两边同时乘以该数。

5x = 35 → x = 7

For equations with x on both sides, collect variable terms on one side and constants on the other. Finish by dividing to

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

Linear equations are one of the most important topics in the Cambridge KS3 mathematics curriculum. They help learners understand algebraic balance, develop logical reasoning, and prepare for more advanced work with graphs, sequences, and simultaneous equations. In this article, we will explore what linear equations are, how to solve them step by step, and how to avoid common errors.

一元一次方程是剑桥 KS3 数学课程中最重要的主题之一。它们帮助学习者理解代数平衡，培养逻辑推理能力，并为后续的图像、数列和联立方程等更高阶内容做好准备。本文将探讨什么是一元一次方程、如何逐步求解，以及如何避免常见错误。

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

A linear equation is an algebraic statement in which two expressions are equal. The highest power of the unknown is 1, which means the variable appears without a square, cube, or higher power. In KS3 Cambridge mathematics, you will meet equations such as 2x + 3 = 11 and 5(x − 2) = 3x + 4. The goal is to find the value of the unknown, usually written as x, that makes the statement true.

一元一次方程是表示两个代数式相等的等式。未知数的最高次数为 1，也就是说变量不会出现平方、立方或更高次幂。在 KS3 剑桥数学中，你会遇到诸如 2x + 3 = 11 和 5(x − 2) = 3x + 4 这样的方程。目标是求出使等式成立的未知数（通常写作 x）的值。

A simple example is x + 5 = 12. The only value that makes this true is x = 7. A linear equation may have one solution, no solution, or infinitely many solutions, but at KS3 level the focus is usually on finding one unique solution.

一个简单的例子是 x + 5 = 12。唯一能使它成立的值为 x = 7。一元一次方程可能有一个解、没有解或有无穷多个解，但在 KS3 阶段通常重点在于求出一个唯一解。

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

Before solving equations, it helps to know the main words. The unknown is the letter whose value you are finding. A term is a single number, letter, or product such as 3x or −5. An expression is a combination of terms without an equals sign, while an equation has an equals sign. The solution is the value that makes the equation true.

在解方程之前，了解主要术语很有帮助。未知数是指你要求出其值的字母。项是一个单独的数、字母或乘积，例如 3x 或 −5。代数式是不含等号的项的组合，而方程含有等号。解是使方程成立的值。

Using clear notation is important. The equals sign shows that the left side and right side have the same value. When you write each step, keep the equals signs aligned so your working is easy to follow.

使用清晰的符号非常重要。等号表示左边和右边的值相同。书写每一步时，保持等号对齐，这样你的解题过程更易于理解。

3. The Balancing Method | 天平法

Think of an equation as a balance scale. Whatever you do to one side, you must do to the other to keep it balanced. This rule is often called the balancing method. You can add, subtract, multiply, or divide both sides by the same non-zero number.

可以把方程想象成一架天平。无论你对一边做什么，都必须对另一边做同样的事情，以保持平衡。这条规则通常被称为天平法。你可以在方程两边同时加、减、乘或除以同一个非零数。

If a = b, then a + c = b + c, a − c = b − c, a × c = b × c, a ÷ c = b ÷ c (c ≠ 0)

如果 a = b，那么 a + c = b + c，a − c = b − c，a × c = b × c，a ÷ c = b ÷ c（c ≠ 0）。

For example, with the equation x + 5 = 12, you subtract 5 from both sides so that the left side becomes x and the right side becomes 7. This gives x = 7. The balance is maintained because both sides were reduced by the same amount.

例如，对于方程 x + 5 = 12，你从两边都减去 5，这样左边变为 x，右边变为 7。于是得到 x = 7。由于两边都减少了相同的量，天平得以保持平衡。

4. Solving One-Step and Two-Step Equations | 解一步和两步方程

For one-step equations, you only need one operation to isolate x. For example, in x + 5 = 12, subtract 5 from both sides to get x = 7. In 3x = 18, divide both sides by 3 to get x = 6. The key is to perform the inverse operation.

对于一步方程，你只需一步运算就能分离出 x。例如，在 x + 5 = 12 中，两边都减去 5，得到 x = 7。在 3x = 18 中，两边都除以 3，得到 x = 6。关键是进行逆运算。

Most KS3 equations need two steps: undo addition or subtraction first, then undo multiplication or division. This is because the variable is usually combined with a constant before being multiplied. Reversing the order of operations is essential.

大多数 KS3 方程需要两步：先消去加法或减法，再消去乘法或除法。这是因为变量通常先与常数结合，再被乘以某个数。颠倒运算顺序非常关键。

Worked example: Solve 2x + 3 = 11. First subtract 3 from both sides: 2x = 8. Then divide both sides by 2: x = 4. You can check by substituting 4 back into the original equation.

示例：解 2x + 3 = 11。首先两边都减去 3：2x = 8。然后两边都除以 2：x = 4。你可以把 4 代回原方程进行检验。

2x + 3 = 11 → 2x = 8 → x = 4

2x + 3 = 11 → 2x = 8 → x = 4

5. Equations with Brackets | 含括号的方程

When an equation contains brackets, you can either expand the brackets first or divide both sides by the number outside the brackets if it is a factor. Expanding first is often clearer. For example, 3(x − 2) = 15 can be expanded to 3x − 6 = 15.

当方程含有括号时，你可以先展开括号，或者如果括号外的数是公因数，可以先在两边除以这个数。通常先展开更清晰。例如，3(x − 2) = 15 可以展开为 3x − 6 = 15。

After expanding, solve the resulting equation normally: add 6 to both sides to get 3x = 21, then divide by 3 to get x = 7. Alternatively, you could divide both sides by 3 first: x − 2 = 5, then add 2 to get x = 7. Both methods give the same answer.

展开后，正常解所得方程：两边都加 6，得到 3x = 21，然后除以 3，得到 x = 7。或者，你也可以先在两边都除以 3：x − 2 = 5，再加 2，得到 x = 7。两种方法得到相同答案。

Be careful when the bracket is preceded by a negative sign. For example, −2(x + 4) = 10 expands to −2x − 8 = 10, not −2x − 4 = 10. The sign must be distributed to every term inside the bracket.

当括号前面是负号时要小心。例如，−2(x + 4) = 10 展开后得到 −2x − 8 = 10，而不是 −2x − 4 = 10。负号必须分配给括号内的每一项。

6. Unknowns on Both Sides | 未知数在两侧

When an equation has x on both sides, collect the variable terms on one side and the constant terms on the other. You can do this by adding or subtracting the same term from both sides. This keeps the equation balanced while simplifying the form.

当方程的两边都含有 x 时，应把变量项集中到一边，常数项集中到另一边。你可以通过在两边同时加上或减去相同的项来实现。这样既保持方程平衡，又简化了形式。

For example, solve 5x − 7 = 3x + 9. Subtract 3x from both sides: 2x − 7 = 9. Then add 7 to both sides: 2x = 16. Finally divide by 2: x = 8. Always aim for a positive coefficient of x if possible.

例如，解 5x − 7 = 3x + 9。两边都减去 3x：2x − 7 = 9。然后两边都加 7：2x = 16。最后除以 2：x = 8。尽可能让 x 的系数为正数。

5x − 7 = 3x + 9 → 2x − 7 = 9 → 2x = 16 → x = 8

5x − 7 = 3x + 9 → 2x − 7 = 9 → 2x = 16 → x = 8

7. Equations Involving Fractions | 含分数的方程

If x appears in a fraction, multiply both sides by the denominator to clear it. For example, x/4 + 1 = 6. First subtract 1 from both sides: x/4 = 5. Then multiply both sides by 4: x = 20. Clearing fractions at the start can make the working simpler.

如果 x 出现在分数中，应在两边都乘以分母来消去分数。例如，x/4 + 1 = 6。首先两边都减去 1：x/4 = 5。然后两边都乘以 4：x = 20。先消去分数可以使解题过程更简单。

If there are several fractions, multiply every term by the lowest common denominator. For example, in x/2 + x/3 = 5, the lowest common denominator is 6. Multiplying gives 3x + 2x = 30, so 5x = 30 and x = 6.

如果有多个分数，应将每一项都乘以最小公分母。例如，在 x/2 + x/3 = 5 中，最小公分母是 6。乘以 6 得到 3x + 2x = 30，所以 5x = 30，x = 6。

When multiplying, remember to apply the multiplication to every term on both sides of the equation. Missing a constant term is a very common mistake in KS3 algebra.

相乘时，记得对方程两边的每一项都进行乘法。漏乘常数项是 KS3 代数中非常常见的错误。

8. Checking Your Solution | 检验你的解

Always substitute your answer back into the original equation. Replace x with your answer and check that the left side equals the right side. This catches arithmetic mistakes and confirms that the solution is correct.

一定要把你的答案代回原方程进行检验。用你的答案替换 x，检查左边是否等于右边。这能发现算术错误，并确认解是正确的。

For example, if you solved 2x + 3 = 11 and got x = 4, substitute: left side = 2(4) + 3 = 8 + 3 = 11, which equals the right side. The solution is therefore correct.

例如，如果你解 2x + 3 = 11 得到 x = 4，代入检验：左边 = 2(4) + 3 = 8 + 3 = 11，与右边相等。因此解是正确的。

Checking is especially useful in tests because it takes only a few seconds and can save marks. If the two sides do not match, retrace your steps to find where the error occurred.

检验在考试中特别有用，因为它只需几秒钟，却能保住分数。如果两边不相等，就重新检查你的步骤，找出错误发生的地方。

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

Many errors in solving linear equations come from rushing or forgetting the balancing rule. Below are some common mistakes to watch for.

解一元一次方程时的许多错误都来自急于求成或忘记天平法则。以下是一些需要注意的常见错误。

• Forgetting to balance both sides: if you add 5 to one side, you must add 5 to the other side too. 忘记两边同时操作：如果你在一边加 5，另一边也必须加 5。
• Distributing signs incorrectly, especially with negative brackets: −(x − 3) = −x + 3, not −x − 3. 符号分配错误，尤其是负数括号：−(x − 3) = −x + 3，而不是 −x − 3。
• Undoing operations in the wrong order: subtraction and addition should be undone before multiplication and division. 运算逆序错误：应先消去加法和减法，再消去乘法和除法。
• Losing a negative sign when moving terms across the equals sign. 移项时丢失负号。
• Multiplying only part of an equation when clearing fractions. 消去分数时只乘方程的一部分。

To avoid these mistakes, write every step clearly, use brackets when needed, and always check your final answer in the original equation.

为了避免这些错误，每一步都要写清楚，必要时使用括号，并始终把最终答案代回原方程检验。

10. Worked Exam-Style Example | 考试风格例题解析

Let us solve a typical KS3 exam-style equation: 2(x + 3) − 4 = 3x − 5. First expand the bracket on the left: 2x + 6 − 4 = 3x − 5. Simplify: 2x + 2 = 3x − 5.

让我们解一道典型的 KS3 考试风格方程：2(x + 3) − 4 = 3x − 5。首先展开左边的括号：2x + 6 − 4 = 3x − 5。化简：2x + 2 = 3x − 5。

Next, subtract 2x from both sides to collect the x terms on the right: 2 = x − 5. Then add 5 to both sides: 7 = x. So the solution is x = 7.

接下来，两边都减去 2x，把 x 项集中到右边：2 = x − 5。然后两边都加 5：7 = x。因此解为 x = 7。

Check by substituting: left side = 2(7 + 3) − 4 = 2(10) − 4 = 20 − 4 = 16. Right side = 3(7) − 5 = 21 − 5 = 16. Both sides equal 16, so the solution is verified.

代入检验：左边 = 2(7 + 3) − 4 = 2(10) − 4 = 20 − 4 = 16。右边 = 3(7) − 5 = 21 − 5 = 16。两边都等于 16，因此解得到验证。

11. Practice Questions | 练习题

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

尝试以下题目来检验你的理解。解出每个方程，然后通过代入检验你的答案。

• 4x − 7 = 21 答案：x = 7
• 6 + 2x = 14 答案：x = 4
• 5(x + 2) = 35 答案：x = 5
• 7x + 3 = 2x + 23 答案：x = 4
• x/3 − 2 = 4 答案：x = 18
• 3x + 5 = 2(x + 8) 答案：x = 11

If you got all of them correct, you are ready to move on to inequalities and formulae. If any were incorrect, review the relevant section and practise a few more examples.

如果你全部做对了，就可以继续学习不等式和公式。如果有题目做错了，请复习相关小节，并多练习几个例子。

12. Summary | 总结

Linear equations are solved by keeping the equation balanced while isolating the unknown. Use inverse operations in the correct order: undo addition or subtraction, then undo multiplication or division. Expand brackets carefully, clear fractions by multiplying by the denominator, and collect like terms when the unknown appears on both sides.

解一元一次方程时，要在分离未知数的同时保持方程平衡。按正确顺序使用逆运算：先消去加法或减法，再消去乘法或除法。仔细展开括号，乘以分母消去分数，当未知数出现在两边时合并同类项。

Always check your solution by substituting it into the original equation. This habit improves accuracy and builds confidence for harder algebra topics in Cambridge KS3 and beyond.

始终通过代入原方程来检验你的解。这个习惯能提高准确性，并为剑桥 KS3 及以后更难的代数主题建立信心。

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📚 Cambridge KS3 Maths: Bearings, Scale Drawings and Loci | 剑桥KS3数学：方位角、比例绘图与轨迹

Bearings, scale drawings, and loci are central to the Cambridge KS3 mathematics curriculum because they link geometry with real-world navigation and map-reading. This article builds up the rules step by step, gives worked examples, and highlights the common errors that appear in school and checkpoint-style papers.

方位角、比例绘图和轨迹是剑桥KS3数学课程的核心内容，因为它们将几何与现实世界中的导航和地图阅读联系起来。本文逐步建立相关规则，给出例题解析，并指出在学校和Checkpoint类型试卷中常见的错误。

1. What Are Bearings? | 什么是方位角？

A bearing is a precise way of stating direction. It is always measured from the north line, moving clockwise, and is written as a three-digit angle such as 045° or 130°.

方位角是一种精确表示方向的方法。它总是从北线开始按顺时针方向测量，并用三位数字书写，例如 045° 或 130°。

Bearings are used in navigation, surveying, aviation, and map reading. Instead of saying ‘roughly north-east’, a bearing gives an exact angle such as 052°.

方位角用于导航、测量、航空和地图阅读。它不是粗略地说 ‘大致东北’，而是给出精确角度，例如 052°。

If an object is directly east of you, its bearing is 090°. If it is directly south, the bearing is 180°, and if it is directly west, the bearing is 270°.

如果一个物体在你的正东方向，它的方位角是 090°。如果在正南方向，方位角是 180°，如果在正西方向，方位角是 270°。

In KS3 questions, you will usually be asked to measure a bearing on a diagram, draw a bearing from a point, or use bearings to plot a path.

在KS3题目中，通常要求你在图上测量方位角、从一个点画方位角，或使用方位角绘制路径。

2. Three Key Rules for Measuring Bearings | 测量方位角的三个关键规则

Rule 1: Always measure from north. At the starting point, draw a faint vertical north line before placing your protractor.

规则1：始终从北方测量。在起点处，先画一条淡淡的垂直北线，再放置量角器。

Rule 2: Measure clockwise. The angle must be read in the clockwise direction from the north line to the line pointing towards the destination.

规则2：按顺时针方向测量。角度必须从北线开始按顺时针方向读到指向目的地的直线。

Rule 3: Use three digits. A bearing of 60° must be written as 060°; a bearing of 8° must be written as 008°.

规则3：使用三位数字。60° 的方位角必须写作 060°；8° 的方位角必须写作 008°。

Many students lose marks because they measure from the target point instead of the starting point. The phrase ‘the bearing of B from A’ means the angle is measured at A.

许多学生因为从目标点而不是起点测量而失分。’B 从 A 的方位角’ 这个短语意味着角度是在 A 点测量的。

Always check whether your bearing is less than 360°. If you obtain an angle larger than 360°, subtract 360° or re-check the clockwise direction.

始终检查你的方位角是否小于 360°。如果得到的角度大于 360°，应减去 360° 或重新检查顺时针方向。

3. Compass Directions and Bearings | 罗盘方向与方位角

Compass directions can be converted directly into bearings. North is 000°, east is 090°, south is 180°, and west is 270°.

罗盘方向可以直接转换为方位角。北是 000°，东是 090°，南是 180°，西是 270°。

For intermediate directions, north-east is 045°, south-east is 135°, south-west is 225°, and north-west is 315°.

对于中间方向，东北是 045°，东南是 135°，西南是 225°，西北是 315°。

Practising these conversions helps you judge whether a measured bearing is reasonable. For example, a bearing of 200° should point roughly south-west, not north-east.

练习这些转换有助于你判断测得的方位角是否合理。例如，200° 的方位角应该大致指向西南，而不是东北。

In navigation, direction is often given as a compass direction in speech, but written as a bearing in calculations. Being able to switch between the two is an important KS3 skill.

在导航中，方向在口头表述时常用罗盘方向，但在计算中写作方位角。能够在两者之间切换是一项重要的KS3技能。

4. Constructing a Bearing Diagram | 绘制方位角示意图

To construct a bearing diagram, place the centre of your protractor on the starting point and align 0° with the north line.

要绘制方位角示意图，将量角器的中心放在起点上，并将 0° 与北线对齐。

Read the required angle clockwise from the protractor and mark a small point. Then draw a straight line from the starting point through that mark.

从量角器上按顺时针读出所需角度，并标记一个小点。然后从起点画一条经过该标记的直线。

If the bearing is greater than 180°, it is often easier to measure the smaller anticlockwise angle

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📚 Solving Linear Equations in Cambridge KS3 Maths | 剑桥KS3数学解一元一次方程

A linear equation is a key topic in Key Stage 3 mathematics and appears regularly in Cambridge Lower Secondary exercises. This article explains how to solve equations step by step, using the type of challenge you might meet in a textbook exercise such as page 275, question 2 from the Cambridge KS3 scheme.

一元一次方程是KS3数学中的核心主题，经常出现在剑桥初中练习中。本文逐步解释如何解方程，并使用你在剑桥KS3教材练习（例如第275页第2题）中可能遇到的题型作为示例。

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

A linear equation is an equation in which the unknown appears only to the power of 1. For example, 3x + 5 = 20 and 2(x – 3) = 8 are linear equations because x is not squared or cubed.

一元一次方程是未知数只出现一次方（即指数为1）的方程。例如 3x + 5 = 20 和 2(x – 3) = 8 都是线性方程，因为 x 没有平方项或立方项。

In a Cambridge KS3 exercise such as the one in p275_2.pdf, the question usually asks you to find the value of the unknown that makes the equation true. This value is called the solution or root of the equation.

在剑桥KS3练习（例如 p275_2.pdf 中的题目）里，通常要求找出使等式成立的未知数的值。这个值叫作方程的解或根。

2. The Balance Method | 天平法

Think of an equation as a balanced set of scales. Whatever you do to one side of the equation, you must do exactly the same to the other side to keep the equation balanced.

把方程想象成一座平衡的天平。你对等式一边做的任何运算，必须对另一边做完全相同的运算，才能保持方程平衡。

This balance method helps you isolate the unknown without changing the solution. You can add, subtract, multiply or divide both sides by the same non-zero number.

天平法帮助你在不改变解的情况下把未知数单独分离出来。你可以在两边同时加、减、乘或除以同一个非零数。

3. Solving One-Step Equations | 解一步方程

For a one-step equation such as x + 7 = 15, subtract 7 from both sides: x + 7 – 7 = 15 – 7, so x = 8.

对于一步方程如 x + 7 = 15，两边同时减去7：x + 7 – 7 = 15 – 7，因此 x = 8。

x + 7 = 15 → x = 8

If the equation is 5x = 35, divide both sides by 5: 5x ÷ 5 = 35 ÷ 5, so x = 7.

如果方程是 5x = 35，两边同时除以5：5x ÷ 5 = 35 ÷ 5，因此 x = 7。

5x = 35 → x = 7

Always use the inverse operation to undo what is being done to x. Addition and subtraction are inverse operations, and multiplication and division are inverse operations.

始终使用逆运算来消除对 x 进行的运算。加法与减法互为逆运算，乘法与除法互为逆运算。

4. Solving Two-Step Equations | 解两步方程

A two-step equation such as 4x + 3 = 23 needs two inverse operations. First subtract 3 from both sides to get 4x = 20. Then divide both sides by 4 to get x = 5.

两步方程如 4x + 3 = 23 需要两步逆运算。首先从两边减去3得到 4x = 20。然后两边除以4得到 x = 5。

4x + 3 = 23
4x = 20
x = 5

The key idea is to undo the addition or subtraction first, and then undo the multiplication or division. This order makes the working clear and reduces mistakes.

关键思路是先消去加法或减法，然后再消去乘法或除法。这个顺序让过程更清晰，减少错误。

5. Removing Brackets First | 先去括号

When an equation contains brackets, expand them first unless there is a good reason not to. For example, 2(x – 3) = 8 becomes 2x – 6 = 8 after expansion.

当方程含有括号时，通常先展开括号，除非有充分理由不这样做。例如 2(x – 3) = 8 展开后变成 2x – 6 = 8。

Then add 6 to both sides: 2x = 14. Finally divide by 2: x = 7.

然后两边加6：2x = 14。最后除以2：x = 7。

2(x – 3) = 8
2x – 6 = 8
2x = 14
x = 7

Remember that expanding a bracket must multiply every term inside the bracket by the number outside. So 3(x + 4) becomes 3x + 12, not 3x + 4.

记住展开括号必须用括号外的数乘以括号内的每一项。因此 3(x + 4) 变成 3x + 12，而不是 3x + 4。

6. Unknowns on Both Sides | 未知数在等式两边

If unknowns appear on both sides of the equation, collect the unknown terms on one side first. For 5x + 2 = 3x + 10, subtract 3x from both sides to get 2x + 2 = 10.

如果未知数出现在等式两边，先把未知数项移到同一边。例如 5x + 2 = 3x + 10，两边减去 3x 得到 2x + 2 = 10。

Then subtract 2 from both sides: 2x = 8, so x = 4.

然后两边减去2：2x = 8，因此 x = 4。

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

Always check that you have moved the correct term. Subtracting the smaller x-term from both sides keeps the coefficient positive, which is usually easier.

始终检查你是否移动了正确的项。从两边减去较小的 x 项可以让系数保持为正，通常更容易计算。

7. Equations Containing Fractions | 含分数的方程

For equations with fractions, multiply every term by the lowest common denominator to clear the fractions. For example, x/3 + 1 = 5 can be solved by multiplying every term by 3: x + 3 = 15, so x = 12.

对于含分数的方程，将每一项乘以最小公分母来消去分母。例如 x/3 + 1 = 5，可以把每一项乘以3：x + 3 = 15，因此 x = 12。

x/3 + 1 = 5
x + 3 = 15
x = 12

Always check that the original denominators are not zero. In KS3 work, the denominators are usually positive whole numbers, so this is just a quick habit to develop.

始终检查原分母是否不为零。在KS3学习中，分母通常是正整数，因此这只是一个需要养成的好习惯。

8. Forming an Equation from a Word Problem | 从文字题列方程

Word problems often hide a linear equation. Read the problem carefully, define the unknown as a letter, and translate the words into algebra.

文字题通常隐藏着一元一次方程。仔细阅读题目，用字母定义未知数，并将文字转化为代数式。

Example: ‘A number is tripled, then 5 is added, and the result is 26.’ Let the number be x. Then 3x + 5 = 26.

示例：“一个数乘以3，再加上5，结果是26。”设这个数为 x。那么 3x + 5 = 26。

Solve: 3x = 21, so x = 7.

解方程：3x = 21，因此 x = 7。

3x + 5 = 26
3x = 21
x = 7

Writing the equation correctly is often more difficult than solving it. Underline the operations described in the problem and put them in the correct order.

正确列出方程通常比解方程更难。在题目中划出描述运算的词语，并按正确顺序转化为代数式。

9. Checking Your Answer by Substitution | 代入检验答案

After finding a solution, substitute it back into the original equation to check that the left-hand side equals the right-hand side.

求出解后，把解代入原方程，检验左边是否等于右边。

For x = 5 in 4x + 3 = 23, the left-hand side is 4 × 5 + 3 = 23, which matches the right-hand side.

在 4x + 3 = 23 中令 x = 5，左边为 4 × 5 + 3 = 23，与右边相等。

This checking step is especially important in exams because it catches careless arithmetic errors before you move on to the next question.

代入检验这一步在考试中特别重要，因为它能在你做下一题之前发现粗心的算术错误。

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

A common mistake is forgetting to apply an operation to both sides of the equation. For example, changing x + 3 = 7 to x = 7 – 3 is correct, but changing x + 3 = 7 to x = 7 + 3 is wrong because the operation was not applied correctly.

一个常见错误是忘记对等式两边都执行运算。例如把 x + 3 = 7 变成 x = 7 – 3 是正确的，但把 x + 3 = 7 变成 x = 7 + 3 就错了，因为运算方向没有正确使用。

Another common error is only dividing one term by a number when the whole side should be divided. For 2x + 4 = 10, you must write 2x = 6 after subtracting 4, not divide only one term prematurely.

另一个常见错误是只将某一项除以一个数，而应该将整个边除以该数。对于 2x + 4 = 10，你必须先减去4得到 2x = 6，而不是过早地只除某一项。

Also remember that expanding a bracket must multiply every term inside, so 2(x – 3) is 2x – 6, not 2x – 3.

还要记住展开括号必须乘以括号内的每一项，所以 2(x – 3) 是 2x – 6，而不是 2x – 3。

11. Practice Questions with Worked Solutions | 练习题与解答

Try these short practice questions, then compare your working with the solutions below. Each one follows the methods described in the sections above.

尝试以下短练习，然后与下面的解答对比。每道题都遵循上述各节中描述的方法。

• Question 1: 3x + 7 = 22 | 题目1：3x + 7 = 22
• Step: 3x = 15, so x = 5 | 步骤：3x = 15，因此 x = 5
• Question 2: 2(x + 4) = 18 | 题目2：2(x + 4) = 18
• Step: 2x + 8 = 18, 2x = 10, so x = 5 | 步骤：2x + 8 = 18，2x = 10，因此 x = 5
• Question 3: 5x – 3 = 2x + 9 | 题目3：5x – 3 = 2x + 9
• Step: 3x = 12, so x = 4 | 步骤：3x = 12，因此 x = 4
• Question 4: x/4 + 2 = 6 | 题目4：x/4 + 2 = 6
• Step: x/4 = 4, so x = 16 | 步骤：x/4 = 4，因此 x = 16

If you obtained the same solutions, you are ready to tackle similar questions in your Cambridge KS3 exercise file. If not, go back to the relevant section and retry the examples.

如果你得到了相同的答案，就说明你已经准备好应对剑桥KS3练习文件中的类似题目。如果没有，请回到相关小节重新尝试例题。

12. Key Summary | 关键总结

Remember to keep the equation balanced by doing the same operation to both sides. Use inverse operations to isolate the unknown. Always check your answer by substitution.

记住通过两边执行相同运算来保持方程平衡。使用逆运算分离未知数。始终代入检验答案。

Linear equations are a foundation for many later topics such as graphs, sequences and formulae. Building strong solving skills now will help you throughout your Cambridge KS3 course and beyond.

一元一次方程是后续许多主题的基础，例如图像、数列和公式。现在打下扎实的解题基础，将帮助你在整个剑桥KS3课程及以后的学习中受益。

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📚 Bearings and Scale Drawings: Reading and Mapping for Cambridge KS3 | 方位角与比例绘图：剑桥KS3数学精讲

This revision guide supports the Cambridge KS3 exercise set p274_2 and covers bearings and scale drawings, two skills that often appear together in checkpoint-style questions.

本讲对应剑桥KS3练习 p274_2 所考查的内容，讲解方位角与比例绘图这两项常在同一题型中出现的技能。

1. Understanding Bearings | 认识方位角

A bearing is an angle measured clockwise from the north line at a given point. It tells you the direction from one place to another without using left, right, up or down.

方位角是从某一点的北线开始、沿顺时针方向测量的角。它用来表示从一个地点到另一个地点的方向，而不使用左、右、上、下等含糊说法。

Bearings are used in navigation, air traffic control, orienteering and map work. In Cambridge KS3 questions, bearings are always given as three-figure angles.

方位角用于航海、航空交通管制、定向越野和地图工作。在剑桥KS3题目中，方位角始终写成三位数角度。

Bearing of B from A = direction measured clockwise from the north line at A

2. The Three Golden Rules for Bearings | 方位角的三个黄金规则

There are three rules you must follow every time you write or measure a bearing. They are simple but examiners check them carefully.

每次书写或测量方位角时，都需要遵守三条规则。这些规则很简单，但考官会仔细检查。

• Rule 1: Always measure from the north line drawn at the starting point. 中文：务必从起点处画出的北线开始测量。
• Rule 2: Always measure clockwise from the north line. 中文：务必从北线顺时针测量。
• Rule 3: Always write the bearing as three digits, such as 045° or 005°. 中文：务必用三位数书写，例如 045° 或 005°。

A bearing of 80° is written 080°, and a bearing of 8° is written 008°. This avoids confusion in navigation and on diagrams.

80° 的方位角写作 080°，8° 的方位角写作 008°。这可以避免导航和图表中的混淆。

3. Measuring a Bearing with a Protractor | 用量角器测量方位角

To measure the bearing of B from A, first draw a clear north line at point A. Place your protractor so that 0° is on the north line and the centre is at A.

要测量 B 从 A 的方位角，先在 A 点画一条清晰的北线。将量角器放在图上，使 0° 对准北线，中心点对准 A。

Read the angle clockwise from north until you reach the line AB. If the angle is less than 100°, add a leading zero when writing it as a bearing.

从北线开始顺时针读出到达 AB 线的角度。如果角度小于 100°，书写时要在前面补零。

Always draw the north line yourself if it is not given. Use a ruler to keep it vertical and parallel to the map edge.

如果图上没有给出北线，要自己画出。用尺子保持北线竖直并与地图边缘平行。

4. Compass Directions and Common Bearings | 罗盘方向与常见方位角

Compass directions are a quick way to state a bearing. You should know the exact three-figure bearing for each main compass point.

罗盘方向是快速表示方位角的方法。你需要知道每个主要罗盘方向对应的准确三位数方位角。

5. Back Bearings and Reverse Directions | 反方位角与相反方向

A back bearing is the bearing you would use to travel in the opposite direction. If you walk from A to B on a bearing of 060°, the back bearing from B to A is not 060°.

反方位角是沿相反方向行走时使用的方位角。如果你从 A 到 B 的方位角是 060°，那么从 B 到 A 的反方位角不是 060°。

To find a back bearing, add 180° if the original bearing is less than 180°, or subtract 180° if it is 180° or greater.

求反方位角时，如果原方位角小于 180°，则加上 180°；如果原方位角大于或等于 180°，则减去 180°。

If θ < 180°, back bearing = θ + 180°; if θ ≥ 180°, back bearing = θ − 180°

Example: bearing 060° → back bearing 240°; bearing 300° → back bearing 120°.

例如：方位角 060° → 反方位角 240°；方位角 300° → 反方位角 120°。

6. Introducing Scale Drawings | 认识比例绘图

A scale drawing is a diagram that represents a real object or distance at a fixed ratio. Bearings are often combined with scale drawings to show a journey or map.

比例绘图是以固定比例表示真实物体或距离的图。方位角常与比例绘图结合，用来表示行程或地图。

A scale is written as 1 : n. This means 1 unit on the drawing represents n units in real life. Both parts of the ratio must use the same unit.

比例尺写作 1 : n。这表示图上的 1 个单位代表实际中的 n 个单位。比例的两部分必须使用相同单位。

For example, a scale of 1 : 50 000 means 1 cm on the drawing represents 50 000 cm in real life. Since 50 000 cm = 500 m = 0.5 km, 1 cm represents 0.5 km.

例如，比例尺 1 : 50 000 表示图上 1 cm 代表实际 50 000 cm。因为 50 000 cm = 500 m = 0.5 km，所以 1 cm 代表 0.5 km。

7. Using Scales: From Drawing to Real Life | 使用比例尺：从图上到实际

To convert a drawing length to a real length, multiply the drawing length by the scale factor n. To convert a real length to a drawing length, divide by n.

将图上长度换算为实际长度时，将图上长度乘以比例因子 n。将实际长度换算为图上长度时，除以 n。

Real length = drawing length × n

For example, on a 1 : 25 000 map, a road measuring 4 cm on the map has real length 4 × 25 000 = 100 000 cm = 1 km.

例如，在 1 : 25 000 地图上，一条路在图上量得 4 cm，实际长度就是 4 × 25 000 = 100 000 cm = 1 km。

If a real fence is 60 m long and the scale is 1 : 200, its drawing length is 60 ÷ 200 = 0.3 m = 30 cm.

如果实际围栏长 60 m，比例尺是 1 : 200，那么图上的长度就是 60 ÷ 200 = 0.3 m = 30 cm。

8. Constructing a Scale Drawing | 制作比例绘图

When a question asks you to construct a scale drawing, follow these steps carefully. Accuracy with ruler and protractor is essential.

当题目要求绘制比例图时，请按照以下步骤仔细操作。准确使用尺子和量角器非常重要。

• Step 1: Choose a suitable scale, such as 1 cm : 5 km. 中文：选择合适的比例尺，例如 1 cm : 5 km。
• Step 2: Convert all real distances into drawing lengths. 中文：把所有实际距离换算成图上长度。
• Step 3: Draw a north line at the starting point. 中文：在起点画出北线。
• Step 4: Use a protractor to measure the bearing and draw a faint guide line. 中文：用量角器测量方位角，并画出浅色辅助线。
• Step 5: Measure the required drawing length along this line and mark the point. 中文：沿这条线量出所需图上长度，并标出点。
• Step 6: If there is a second bearing, draw a new north line at the new point before measuring the next angle. 中文：如果有第二个方位角，先在新点画出新北线，再测量下一个角度。

9. Worked Example: Locating a Ship | 例题：确定船只位置

A ship leaves port P. It sails 15 km on a bearing of 060° to point A, then 10 km on a bearing of 150° to point B. Use a scale of 1 cm : 5 km to draw the journey, then find the distance and bearing of B from P.

一艘船离开港口 P。它沿 060° 方位角航行 15 km 到达 A 点，再沿 150° 方位角航行 10 km 到达 B 点。用 1 cm : 5 km 的比例绘图，并求 B 点相对于 P 的距离和方位角。

First convert distances: 15 km → 3 cm, and 10 km → 2 cm. Draw a north line at P, mark 060° clockwise, and measure a 3 cm line to locate A.

首先换算距离：15 km → 3 cm，10 km → 2 cm。在 P 点画北线，顺时针标出 060°，沿该方向量 3 cm，标出 A 点。

At A, draw a new north line, measure 150° clockwise, and draw a 2 cm line to locate B. Finally, measure PB with a ruler and the bearing of B from P with a protractor.

在 A 点画出新北线，顺时针测量 150°，画出 2 cm 线段标出 B 点。最后用尺子量 PB，用量角器量 B 点相对于 P 的方位角。

From the accurate drawing, PB is about 3.6 cm, so the real distance is 3.6 × 5 = 18 km. The measured bearing is approximately 094°.

根据精确绘图，PB 约为 3.6 cm，因此实际距离为 3.6 × 5 = 18 km。测得的方位角约为 094°。

10. Common Mistakes and Exam Tips | 常见错误与考试提示

One of the most common errors is measuring the bearing anticlockwise instead of clockwise. Always start at the north line and move clockwise.

最常见的错误之一是把方位角按逆时针方向测量。务必从北线开始顺时针测量。

Another common mistake is forgetting that a bearing must have three digits. Writing 60° instead of 060° can lose marks even if the angle is correct.

另一个常见错误是忘记方位角必须写成三位数。把 60° 写成 060° 以外的形式，即使角度正确也可能丢分。

When working with scale drawings, always convert units before calculating. If the scale is 1 cm : 5 km, do not mix cm and km without converting.

使用比例绘图时，计算前务必统一单位。如果比例尺是 1 cm : 5 km，不要在未换算的情况下混合使用 cm 和 km。

Finally, show your north lines and construction marks clearly. Examiners award working marks for correct method even if the final measurement is slightly off.

最后，清楚地画出北线和作图痕迹。即使最终测量略有误差，过程方法正确也能获得步骤分。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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📚 Cambridge KS3 Maths: Linear Equations Practice (p259_2) | 剑桥 KS3 数学：一元一次方程练习（p259_2）

Welcome to this Cambridge KS3 Mathematics revision article based on Practice Sheet p259_2. The focus is linear equations, one of the most important algebraic skills at Key Stage 3. We will cover balancing, two-step equations, brackets, fractions, checking, word problems and common errors. Work through each section slowly and attempt the practice questions yourself.

欢迎阅读这篇基于练习卷 p259_2 的剑桥 KS3 数学复习文章。重点是一元一次方程，这是 Key Stage 3 代数中最重要的技能之一。我们将学习天平法、两步方程、去括号、分数系数、检验解、文字题和常见错误。请慢慢完成每个部分，并自己尝试练习题。

1. Understanding Linear Equations | 理解一元一次方程

A linear equation is an equation in which the unknown, often written as x, appears only to the power of 1. The equation forms a straight line when graphed. At KS3, most equations have one unknown and one solution.

一元一次方程是未知数（通常写作 x）只出现一次幂的方程。它在图像上形成一条直线。在 KS3 阶段，大多数方程只有一个未知数和一个解。

Examples include x + 5 = 12, 3y − 2 = 16 and 4(n + 1) = 28. The goal is to find the value of the unknown that makes the equation true.

例如 x + 5 = 12、3y − 2 = 16 和 4(n + 1) = 28。目标是求出使方程成立的未知数的值。

x + 5 = 12 and 3y − 2 = 16

2. Balancing Method | 天平法

Think of an equation as a balance. Whatever you do to one side, you must do to the other side. This keeps the equation true.

把方程想像成一座天平。你对一边做的任何操作，也必须对另一边做，这样方程才能保持成立。

To solve x + 7 = 15, subtract 7 from both sides: x + 7 − 7 = 15 − 7, so x = 8.

要解 x + 7 = 15，两边同时减去 7：x + 7 − 7 = 15 − 7，因此 x = 8。

x + 7 = 15 → x = 15 − 7 → x = 8

Always use inverse operations: addition is undone by subtraction, multiplication by division, and vice versa.

总是使用逆运算：加法由减法抵消，乘法由除法抵消，反之亦然。

3. Solving Two-Step Equations | 解两步方程

A two-step equation has two operations attached to the unknown. For example, in 2x + 3 = 11, the unknown x is first multiplied by 2 and then 3 is added.

两步方程中有两个运算作用在未知数上。例如，在 2x + 3 = 11 中，未知数 x 先乘以 2，然后加 3。

Undo the addition or subtraction first, then undo the multiplication or division. This order is the reverse of the order of operations, sometimes called BIDMAS or BODMAS.

先消去加法或减法，再消去乘法或除法。这个顺序与运算顺序（有时称为 BIDMAS 或 BODMAS）相反。

2x + 3 = 11 → 2x = 8 → x = 4

4. Equations with Brackets | 带括号的方程

When an equation contains brackets, expand them first using the distributive law. Multiply the term outside by each term inside the brackets.

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

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