📚 Year 9 SQA Mathematics: Summer Preparation and Bridging Course | Year 9 SQA 数学：暑期预习与衔接课程
This summer programme is designed to help you bridge into Year 9 SQA Mathematics with confidence. We will revisit key ideas from Year 8 and introduce core topics you will meet in the first term, including algebra, geometry, number skills, statistics and probability. By working through short, focused explanations and examples, you can build a solid foundation before the new school year starts.
这个暑期课程旨在帮助你自信地衔接到 Year 9 的 SQA 数学课程。我们将重温 Year 8 的核心概念，并介绍你将在第一学期遇到的重要专题，包括代数、几何、数字技能、统计与概率。通过简明扼要的讲解和例题，你可以在新学年开始前打下扎实的基础。
1. Algebraic Expressions and Substitution | 代数表达式与代换
In algebra, letters stand for unknown numbers or variables. An expression like 3a + 2b does not have an equals sign. When you are given values for the letters, you can substitute them in to find the numerical value.
在代数中，字母代表未知数或变量。像 3a + 2b 这样的式子没有等号。当你知道字母的值时，就可以把它们代入，求出数值。
For example, if a = 4 and b = 5, then 3a + 2b = 3 × 4 + 2 × 5. First multiply: 12 + 10 = 22. Always remember the order of operations – do multiplication before addition unless brackets say otherwise.
例如，若 a = 4 且 b = 5，则 3a + 2b = 3 × 4 + 2 × 5。先算乘法：12 + 10 = 22。务必牢记运算顺序——先乘除后加减，除非有括号另有指示。
When an expression contains a power, such as x², substitute the value and then square it. For x² + 3x when x = 2, you get 2² + 3 × 2 = 4 + 6 = 10. Practise with positive and negative numbers, as negatives raised to an even power become positive.
当表达式含有乘方时，例如 x²，先代入数值再求平方。对于 x² + 3x，当 x = 2 时，得到 2² + 3 × 2 = 4 + 6 = 10。练习时要包括正数和负数，因为负数取偶次方会变成正数。
2. Solving Linear Equations | 解一元一次方程
An equation states that two expressions are equal. To solve an equation like 2x + 3 = 11, we need to find the value of x that makes the statement true. The goal is to isolate the variable on one side, using inverse operations.
方程表示两个表达式相等。要求解如 2x + 3 = 11 的方程，我们需要找出让等式成立的 x 值。目标是通过逆运算将变量单独移到一边。
Start by subtracting 3 from both sides: 2x = 8. Then divide both sides by 2: x = 4. Always check your answer by substituting back: 2(4) + 3 = 8 + 3 = 11, which matches the original equation.
首先从两边减去 3：2x = 8。然后两边同除以 2：x = 4。每次都要将答案代回检验：2(4) + 3 = 8 + 3 = 11，与原始方程一致。
When equations involve brackets, expand them first. For 3(x − 2) = 12, expand to 3x − 6 = 12, then add 6 to get 3x = 18 and divide by 3: x = 6. If letters appear on both sides, collect like terms onto one side.
当方程含有括号时，要先去括号。对于 3(x − 2) = 12，去括号得 3x − 6 = 12，然后加 6 得 3x = 18，再除以 3：x = 6。如果字母同时在两边，将同类项移到同一边。
3. Sequences and the nth Term | 序列与第 n 项
A number sequence follows a rule. In Year 9 you will work with linear sequences, where the difference between consecutive terms is constant. For example: 5, 9, 13, 17, … has a common difference of +4.
数列遵循一定的规律。在 Year 9，你将学习等差数列，即相邻两项的差是常数。例如：5, 9, 13, 17, … 的公差是 +4。
The rule for the nth term allows you to find any term without listing all the previous ones. For the sequence above, the nth term is 4n + 1. Check: when n = 1, 4(1)+1=5; n = 2, 4(2)+1=9; the rule works.
第 n 项公式让你无需列出前面所有项就能找到任意一项。对于上述数列，第 n 项是 4n + 1。检验一下：n = 1 时，4(1)+1=5；n = 2 时，4(2)+1=9；这个公式是成立的。
ntify the common difference (this becomes the coefficient of n) and then work out the zero‑term adjustment. For sequence 7, 10, 13, 16, … is 3, so start with 3n. Then 3(1) = 3, but the first term is 7, so add 4: the nth term is 3n + 4.
要找出第 n 项，先确定公差（它将成为 n 的系数），再算出零项调整值。对于数列 7, 10, 13, 16, … 公差为 3，可初步写为 3n。代入 n=1 得 3，但第一项是 7，所以加 4：第 n 项为 3n + 4。
4. Fractions, Decimals and Percentages | 分数、小数与百分数
Being fluent in converting between fractions, decimals and percentages is essential. For example, ½ = 0.5 = 50%. For quarters, ¼ = 0.25 = 25% and ¾ = 0.75 = 75%. You should memorise these key equivalences.
熟练地在分数、小数和百分数之间进行转换非常重要。例如，½ = 0.5 = 50%。四分之一的话，¼ = 0.25 = 25%，¾ = 0.75 = 75%。你需要熟记这些基本等值关系。
To express a fraction as a percentage, divide the numerator by the denominator and multiply by 100. For 3/8, divide 3 by 8 to get 0.375, then multiply by 100 → 37.5%. When comparing quantities, it is often easier to turn all values into percentages or decimals.
要把分数化成百分数，用分子除以分母再乘以 100。对于 3/8，先算 3 ÷ 8 得 0.375，再乘以 100 → 37.5%。在比较数量时，通常把所有数值都化成百分数或小数会更方便。
Percentages are used for increases and decreases. A 15% increase on £60 is £60 × 1.15 = £69. A 20% discount on £45 is £45 × 0.80 = £36. Practise finding percentages of amounts without a calculator by using 10% and 1% building blocks.
百分数也用于增长和折扣。£60 增长 15% 之后是 £60 × 1.15 = £69。£45 打八折是 £45 × 0.80 = £36。练习不使用计算器求出一个量的百分数，方法是借助 10% 和 1% 作为积木块。
5. Ratios and Proportions | 比与比例
A ratio compares two or more quantities. If a recipe uses 2 parts of rice to 5 parts of water, the ratio is 2 : 5. Ratios can be simplified just like fractions – divide both numbers by their highest common factor.
比用来比较两个或多个量。如果一个食谱用 2 份米配 5 份水，那么米和水的比是 2 : 5。比可以像分数一样化简——将两个数同时除以它们的最大公因数。
To share an amount in a given ratio, add the parts to find the total number of shares. For a ratio 3 : 2 with £50, total parts = 5. One part = £50 ÷ 5 = £10, so the amounts are 3 × £10 = £30 and 2 × £10 = £20.
要把一个总量按给定的比分配，先把份数相加，求出总份数。对于比 3 : 2 和 £50，总份数为 5。每份 = £50 ÷ 5 = £10，所以两份金额分别是 3 × £10 = £30 和 2 × £10 = £20。
Proportion problems often involve scaling. If 5 pencils cost £3, the cost of 8 pencils can be found by finding the unit rate: one pencil costs £3 ÷ 5 = £0.60, so 8 pencils cost 8 × £0.60 = £4.80. This is direct proportion.
比例问题经常涉及缩放。若 5 支铅笔花费 £3，要求 8 支铅笔的费用，可以先算出单价：一支铅笔 £3 ÷ 5 = £0.60，8 支就是 8 × £0.60 = £4.80。这属于正比例关系。
6. Angles and Polygons | 角与多边形
You need to know angle facts: angles on a straight line add to 180°, angles around a point total 360°, and vertically opposite angles are equal. When parallel lines are cut by a transversal, corresponding and alternate angles are equal.
你需要掌握角的基本性质：平角（一直线）为 180°，绕一点一周的角总和为 360°，对顶角相等。当两条平行线被一条横截线切割时，同位角与内错角分别相等。
In polygons, the sum of interior angles of an n‑sided polygon is (n − 2) × 180°. For a pentagon (5 sides), the interior angles add to (5 − 2) × 180° = 540°. If the pentagon is regular, each interior angle is 540° ÷ 5 = 108°.
在多边形中，n 边形的内角和为 (n − 2) × 180°。一个五边形 (5 条边)，内角和为 (5 − 2) × 180° = 540°。若是正五边形，每个内角为 540° ÷ 5 = 108°。
The sum of exterior angles of any convex polygon is always 360°. This can help you find missing angles when one exterior angle is unknown. For a regular polygon, each exterior angle is 360° ÷ n.
任何凸多边形的外角和总是 360°。这一性质在某个外角未知时非常有用。对于正多边形，每个外角等于 360° ÷ n。
7. Pythagoras' Theorem | 勾股定理
Pythagoras' theorem applies to right‑angled triangles. It says: a² + b² = c², where c is the hypotenuse (the longest side, opposite the right angle), and a, b are the other two sides.
勾股定理适用于直角三角形。它的内容是：a² + b² = c²，其中 c 为斜边（最长边，直角所对的边），a 和 b 是另外两条直角边。
If a triangle has legs of 6 cm and 8 cm, the hypotenuse is √6² + 8² = √36 + 64 = √100 = 10 cm. You can also use the theorem to check whether a triangle is right‑angled: if the side lengths satisfy the equation, the triangle is right‑angled.
如果一个直角三角形的两条直角边分别为 6 cm 和 8 cm，斜边长度就是 √6² + 8² = √36 + 64 = √100 = 10 cm。你也可以用该定理来检验一个三角形是否为直角三角形：如果三边长度满足等式，那么它就是直角三角形。
When the hypotenuse is known but one leg is missing, rearrange the equation. For example, if c = 13 and a = 5, then b = √13² − 5² = √169 − 25 = √144 = 12. Always remember to square root at the end.
如果斜边已知，但缺一条直角边，要对方程进行变形。例如，已知 c = 13 且 a = 5，那么 b = √13² − 5² = √169 − 25 = √144 = 12。务必记得最后要开平方根。
8. Statistics and Data Handling | 统计与数据处理
In Year 9 you will learn to calculate the mean, median, mode and range. The mean is the sum of all values divided by the number of values. The median is the middle value when the data are arranged in order.
在 Year 9，你将学习计算平均数、中位数、众数和极差。平均数等于所有数值之和除以数据的个数。中位数则是数据按大小排列后位于中间的那个数。
For a small data set like 3, 6, 8, 8, 10, 12, 14, the mean is (3+6+8+8+10+12+14) ÷ 7 = 61 ÷ 7 ≈ 8.71. The median is 8, the mode is 8 (the value that appears most often), and the range is 14 − 3 = 11.
对于一个小的数据集，如 3, 6, 8, 8, 10, 12, 14，平均数为 (3+6+8+8+10+12+14) ÷ 7 = 61 ÷ 7 ≈ 8.71。中位数为 8，众数为 8（出现最多的数），极差为 14 − 3 = 11。
Charts such as bar charts, line graphs and pie charts help visualise data. When drawing a pie chart, remember that the angle for each sector equals (frequency ÷ total) × 360°. Always label your axes clearly and give charts a title.
条形图、线形图和饼图等图表有助于直观展示数据。绘制饼图时，记住每个扇区的角度等于 (频数 ÷ 总数) × 360°。一定要清晰标注坐标轴，并给图表加上标题。
9. Introduction to Probability | 概率入门
The probability scale runs from 0 (impossible) to 1 (certain). For a fair coin, P(heads) = ½.
概率衡量一个事件发生的可能性大小。它可以用分数、小数或百分数表示。概率的数值范围从 0（不可能）到 1（必然）。对于一枚均匀的硬币，P(头像面) = ½。
The probability of an event not happening is 1 − P(event). If the probability of rain is 0.3, the probability of no rain is 1 − 0.3 = 0.7. When all outcomes are equally likely, probability = (number of favourable outcomes) ÷ (total number of outcomes).
一个事件不发生的概率为 1 − P(事件)。如果下雨的概率是 0.3，那么不下雨的概率就是 1 − 0.3 = 0.7。当所有结果等可能时，概率 = (有利结果数) ÷ (所有可能结果总数)。
In experiments, the expected number of successes is probability × number of trials. If you roll a fair six‑sided die 300 times, you would expect about 300 × 1/6 = 50 sixes. This does not guarantee exactly 50, but it is the long‑run average.
在试验中，预期的成功次数为 概率 × 试验次数。如果你投掷一枚均匀的六面骰子 300 次，你可以预期约 300 × 1/6 = 50 次出现点数 6。这并不意味着恰好是 50 次，而是长期的平均趋势。
10. Measurement and Unit Conversion | 测量与单位换算
Metric units are widely used in SQA questions. Remember the common relationships: 1 km = 1000 m, 1 m = 100 cm, 1 cm = 10 mm; for mass, 1 kg = 1000 g; for capacity, 1 litre = 1000 ml. Being confident in these conversions saves time.
公制单位在 SQA 试题中广泛使用。记住常见的换算关系：1 km = 1000 m，1 m = 100 cm，1 cm = 10 mm；质量单位中，1 kg = 1000 g；容量单位中，1 L = 1000 mL。熟练掌握这些换算可以节省时间。
Area and volume units are squared or cubed. 1 m² is 100 cm × 100 cm = 10 000 cm² match the dimension: for area use 10² = 100 per step in the scale, for volume 10³ = 1000.
面积和体积单位是平方或立方的。 1 m² 等于 100 cm × 100 cm = 10 000 cm²。换算时，要乘以或除以与维度相对应的 10 的幂：面积单位每级差 10² = 100，体积单位每级差 10³ = 1000。
Perimeter is the distance around a shape. For a rectangle, P = 2(l + w). Area of a rectangle is A = l × w, and area of a triangle is A = ½ × base × height. Always include the correct units in your final answer, such as cm, m² or mm³.
周长是围绕形状一周的长度。对于长方形，周长 P = 2(长 + 宽)。长方形的面积是 A = 长 × 宽，三角形的面积是 A = ½ × 底 × 高。最终答案务必带上正确的单位，如 cm、m² 或 mm³。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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