Year 7 Edexcel Further Maths: Summer Preview and Bridging Course | Year 7 Edexcel 进阶数学：暑期预习与衔接课程

📚 Year 7 Edexcel Further Maths: Summer Preview and Bridging Course | Year 7 Edexcel 进阶数学：暑期预习与衔接课程

Welcome to your summer preparation for Year 7 Edexcel Further Mathematics! This bridging course is designed to strengthen your numerical fluency, algebraic thinking, and logical reasoning, giving you a confident start in advanced junior mathematics. Through exploring new concepts and revisiting key ideas, you will build problem-solving skills that form the foundation for IGCSE and beyond.

欢迎参加 Year 7 Edexcel 进阶数学暑期预习课程！本衔接课程旨在巩固你的数感、代数思维和逻辑推理能力，让你在初中进阶数学学习中自信启航。通过探索新概念并重温核心思想，你将培养解决问题的能力，为 IGCSE 及更高阶段打下坚实基础。

---

1. Number and Place Value Beyond the Basics | 超越基础的数字与位值

Going deeper into place value, you will work with numbers up to billions and decimals with many places. Understanding powers of ten is essential for reading and writing large figures accurately.

更进一步学习位值，你将处理高达十亿的数字和多位小数。理解十的幂对于准确读写大数至关重要。

Negative numbers are used in real-life contexts like temperature and debt. You will practise ordering, adding and subtracting negatives using the number line, so that calculations below zero become automatic.

负数用于温度、负债等现实情境。你将练习使用数轴对负数进行排序、加减，使零下的运算变得自然而然。

Squaring and cubing numbers, together with square roots and cube roots, are introduced early. For instance, 5² = 25 and ³√27 = 3. These ideas link geometry and algebra.

平方、立方以及平方根、立方根被提早引入。例如 5² = 25，³√27 = 3。这些概念将几何与代数联系起来。

Prime factorisation helps you break numbers into their building blocks, which is a powerful tool for highest common factor (HCF) and lowest common multiple (LCM). A number like 60 can be written as a product of primes: 2² × 3 × 5.

质因数分解能将数字拆分成其构建块，这是求最大公因数（HCF）和最小公倍数（LCM）的强大工具。像60这样的数可写成质数乘积：2² × 3 × 5。

60 = 2² × 3 × 5

---

2. Algebraic Expressions and Manipulation | 代数表达式与运算

Letters are used to represent unknown numbers or variables. An expression like 3x + 2 changes its value when x changes. Learning to read and write expressions is the first step in algebra.

字母用来表示未知数或变量。如 3x + 2 这样的表达式会随 x 的变化而改变值。学会读写表达式是代数的第一步。

You will learn to collect like terms: 5a + 3b – 2a + b simplifies to 3a + 4b. This tidy-up process makes expressions easier to handle and compare.

你将学会合并同类项：5a + 3b – 2a + b 简化为 3a + 4b。这个整理过程让表达式更易处理和比较。

Expanding brackets using the distributive law is a key skill. Remember to multiply the term outside by every term inside: 2(x + 5) = 2x + 10, and a more challenging example is 3(2y – 4) = 6y – 12.

使用分配律展开括号是一项关键技能。记住要把外面的项乘以括号内的每一项：2(x + 5) = 2x + 10，更有难度的例子如 3(2y – 4) = 6y – 12。

Writing expressions for real-life situations – such as total cost = 2p + 3 – builds modelling ability and shows how algebra describes the world around you.

为现实情境写出表达式——例如 总成本 = 2p + 3——能培养建模能力，体现代数如何描述你周围的世界。

3(2y – 4) = 6y – 12

---

3. Solving Linear Equations | 解线性方程

An equation states that two expressions are equal, and solving means finding the unknown value that makes the statement true. This balances both sides like a scale.

方程说明两个表达式相等，解方程就是求出使等式成立的未知数值。这就像天平一样保持两边平衡。

Using the balance method, whatever operation you do to one side, you must do to the other. For x + 7 = 15, subtract 7 from both sides to get x = 8.

使用平衡法，你对一边做什么运算，另一边也必须做同样的运算。对于 x + 7 = 15，两边同时减去7得到 x = 8。

Two-step equations like 3x – 4 = 11 require undoing operations in reverse order: add 4, then divide by 3, giving x = 5. This builds logical sequencing skills.

两步方程如 3x – 4 = 11 需要逆向运算：先加4，再除以3，得出 x = 5。这能培养逻辑顺序感。

Equations with brackets should be expanded first, then solved just like the others. For example, solve 2(x + 3) = 10: expand to 2x + 6 = 10, then 2x = 4, so x = 2.

含有括号的方程应先展开再求解。例如，解 2(x + 3) = 10：展开得 2x + 6 = 10，再得 2x = 4，所以 x = 2。

2(x + 3) = 10 → 2x + 6 = 10 → 2x = 4 → x = 2

---

4. Geometric Reasoning and Properties | 几何推理与性质

You will classify angles as acute (less than 90°), right (90°), obtuse (between 90° and 180°) and reflex (greater than 180°). Naming angles correctly is the first step in geometric communication.

你将角度分类为锐角（小于90°）、直角（90°）、钝角（90°到180°之间）和优角（大于180°）。正确命名角是几何交流的第一步。

Vertically opposite angles are equal. Angles on a straight line sum to 180°, and angles around a point add to 360°. These facts will help you find unknown angles without measuring.

对顶角相等。直线上的角之和为180°，绕一点一周的角之和为360°。这些性质能帮你不用量角器就求得出未知角度。

When parallel lines are cut by a transversal, corresponding angles are equal, and alternate angles are equal. You can use these relationships to solve angle puzzles in complex diagrams.

当平行线被一条截线切割时，同位角相等，内错角相等。你可以利用这些关系解决复杂图形中的角度谜题。

The sum of interior angles of a triangle is always 180°. For a quadrilateral it is 360°, and this extends to any polygon with the formula (n – 2) × 180°, where n is the number of sides.

三角形的内角和总是180°。四边形的内角和为360°，进而推广到任意多边形公式 (n – 2) × 180°，其中 n 是边数。

Sum of interior angles of a pentagon = (5 – 2) × 180° = 540°

---

5. Ratio, Proportion and Rates | 比、比例与率

A ratio compares two or more quantities. Simplifying 8:12 to 2:3 by dividing by the highest common factor makes ratios neater and easier to use.

比用来比较两个或更多数量。将 8:12 除以最大公因数简化为 2:3，使比更整洁、更好用。

Sharing an amount in a given ratio uses total parts. To divide £60 in the ratio 3:2, total parts = 5, so each part is £12, giving £36 and £24. This method works for any share-out task.

按给定比例分配金额使用总份数概念。按 3:2 分配 60 英镑，总份数为5，每份12英镑，得到36英镑和24英镑。这个方法适用于任何分配任务。

Direct proportion means as one quantity doubles, the other also doubles. You can use a multiplier or a ratio table to find unknown values when quantities stay in step.

正比例意味着一个量翻倍时另一个也翻倍。当两个量保持同步时，你可以使用乘数或比率表格来求未知值。

Rates involve different units, like speed (km/h) or price per litre. Converting units and using the formula speed = distance ÷ time allows you to compare and predict movement and cost.

率涉及不同单位，如速度（公里/小时）或每升价格。通过换算单位并使用公式 速度 = 路程 ÷ 时间，你能够比较和预测运动与成本。

Speed = Distance ÷ Time

---

6. Statistics and Data Handling | 统计与数据处理

The mean is found by summing all values and dividing by the total number. The median is the middle value when data are ordered, and the mode is the most frequent value. Each gives a different view of the ‘centre’ of data.

平均数由所有值之和除以总数求得。中位数是将数据排序后的中间值，众数是出现次数最多的值。三者从不同角度展示了数据的“中心”。

Published by TutorHao | Year 7 进阶数学 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）