📚 Essential Maths 9C Homework Book: Key Topic Summaries | 《Essential Maths 9C 练习册》知识点精讲
The Essential Maths 9C Homework Book consolidates advanced KS3 mathematics, bridging the gap to GCSE. This compressed revision guide delivers clear summaries, worked examples, and key formulas for every major topic. Use it alongside your homework to reinforce understanding, spot patterns, and build confidence in problem solving.
《Essential Maths 9C 练习册》巩固 KS3 高阶数学知识，为 GCSE 搭建桥梁。这份压缩复习指南为你梳理每个重要专题的核心总结、例题与关键公式。配合课后练习使用，能加深理解、发现规律，并提升解题信心。
1. Algebraic Manipulation and Factorisation | 代数运算与因式分解
Expanding single brackets uses the distributive law: a(b + c) = ab + ac. For instance, 3(2x – 5) = 6x – 15. Multiply the term outside by every term inside.
去括号运用分配律 a(b + c) = ab + ac。例如 3(2x – 5) = 6x – 15。将括号外的项与括号内每一项相乘。
When expanding double brackets, apply the FOIL method: (x + a)(x + b) = x² + (a+b)x + ab. For example, (x + 3)(x – 2) = x² + x – 6.
展开双括号可以使用首外内尾法：(x + a)(x + b) = x² + (a+b)x + ab。如 (x + 3)(x – 2) = x² + x – 6。
Factorising extracts the highest common factor (HCF). For 4x + 12, the HCF is 4, giving 4(x + 3). Always check for a common factor before trying other methods.
因式分解先提取公因数。4x + 12 的公因数为 4，得到 4(x + 3)。尝试其它方法前务必检查公因数。
To factorise a quadratic trinomial like x² + 7x + 10, find two numbers that multiply to +10 and add to +7. The numbers 2 and 5 work, so x² + 7x + 10 = (x + 2)(x + 5).
分解二次三项式如 x² + 7x + 10，找两个乘积为 +10、和为 +7 的数。2 和 5 符合，因此 x² + 7x + 10 = (x + 2)(x + 5)。
The difference of two squares pattern is a² – b² = (a + b)(a – b). Example: x² – 9 = (x + 3)(x – 3). Recognising this structure saves time.
平方差公式为 a² – b² = (a + b)(a – b)。例如 x² – 9 = (x + 3)(x – 3)。识别这一结构能快速分解。
a² – b² = (a + b)(a – b)
2. Linear and Simultaneous Equations | 一次方程与联立方程
To solve a linear equation, perform the same operation on both sides to isolate the unknown. For 2x + 5 = 13, subtract 5 then divide by 2: x = 4.
解一次方程时，对方程两边做相同运算以分离未知数。如 2x + 5 = 13，先减 5 再除以 2，得 x = 4。
Equations with brackets require expansion first: 3(2x – 1) = 15 → 6x – 3 = 15 → 6x = 18 → x = 3.
含括号的方程需先去括号：3(2x – 1) = 15 → 6x – 3 = 15 → 6x = 18 → x = 3。
When unknowns appear on both sides, collect like terms on one side. Example: 5x + 2 = 3x + 10 → 2x = 8 → x = 4.
当未知数出现在等式两边时，将含未知数的项移到同一边。如 5x + 2 = 3x + 10 → 2x = 8 → x = 4。
Simultaneous equations can be solved by elimination. For 2x + y = 7 and x – y = 2, add the equations: 3x = 9 → x = 3, then y = 1.
解联立方程可用消元法。如 2x + y = 7 和 x – y = 2，两式相加得 3x = 9 → x = 3，代入得 y = 1。
Substitution is ideal when one variable is already the subject. If y = 2x + 1 and x + y = 7, substitute: x + (2x + 1) = 7 → 3x = 6 → x = 2, y = 5.
当已有一个未知数为对象时可用代入法。如 y = 2x + 1 且 x + y = 7，代入得 x + (2x + 1) = 7 → 3x = 6 → x = 2, y = 5。
Graphical solutions plot both lines; the intersection gives the solution. Always check your answer in both original equations.
图像法画出两条直线，交点即为解。务必将答案代回原方程验证。
3. Inequalities and Number Lines | 不等式与数轴
Inequalities use symbols: < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to). 3 ≤ x < 7 means x can be 3 up to but not including 7.
不等式符号：<（小于），>（大于），≤（小于等于），≥（大于等于）。3 ≤ x < 7 表示 x 从 3 到 7，含 3 但不含 7。
Solving linear inequalities follows the same steps as equations, but remember: multiplying or dividing by a negative number reverses the inequality sign. –2x > 8 → x < –4.
解一次不等式步骤与方程相同，但要记住：乘或除以负数时不等号方向反转。–2x > 8 → x < –4。
Represent inequalities on a number line: use an open circle for < or > and a closed circle for ≤ or ≥. Shade the region that satisfies the inequality.
在数轴上表示不等式：用空心圆表示 < 或 >，实心圆表示 ≤ 或 ≥，并将满足条件的区域涂色。
Compound inequalities like 2 < x ≤ 5 can be solved by splitting into two simple inequalities, or handled directly if the variable is isolated in the middle.
复合不等式如 2 < x ≤ 5 可拆成两个简单不等式，或当变量在中间时直接处理。
Listing integer solutions: for –1 < x ≤ 4, the integers are 0, 1, 2, 3, 4. Always check the endpoints carefully.
列出整数解：对 –1 < x ≤ 4，整数解为 0,1,2,3,4。务必仔细检查端点。
4. Sequences and the nth Term | 数列与第 n 项
A linear sequence has a constant first difference. The nth term formula is an + b, where a is the common difference and b is the zero‑th term. For 5, 8, 11, 14,… the difference is 3, so nth term = 3n + 2.
一次数列有恒定的首阶差。第 n 项公式为 an + b，其中 a 为公差，b 为零项。数列 5,8,11,14,… 的公差为 3，因此第 n 项 = 3n + 2。
To find the nth term of a linear sequence, identify the jump (a), then work backwards to find b when n=0. Check with n=1.
求一次数列第 n 项时，先确定公差 a，再反推 n=0 时的 b，并用 n=1 验证。
Simple quadratic sequences have a second difference that is constant. The nth term involves n². For 2, 5, 10, 17, 26,… the rule is n² + 1.
简单二次数列的二次差为常数，第 n 项包含 n²。数列 2,5,10,17,26,… 的通项为 n² + 1。
Linear nth term: T(n) = a + (n – 1)d
When given a position-to-term rule, generate terms by substituting n=1,2,3,… For T(n) = 2n² – 1, the first three terms are 1, 7, 17.
给定位置与项的规律时，代入 n=1,2,3,… 生成各项。T(n) = 2n² – 1 的前三项为 1,7,17。
5. Graphs of Functions | 函数图像
The equation of a straight line is y = mx + c, where m is the gradient and c is the y‑intercept. A line with gradient 2 and y‑intercept –3 is y = 2x – 3.
直线方程为 y = mx + c，m 为斜率，c 为 y 轴截距。斜率为 2、截距为 –3 的直线为 y = 2x – 3。
Parallel lines have the same gradient. Perpendicular lines have gradients that multiply to –1, though this concept deepens at GCSE.
平行线斜率相同。垂直线斜率乘积为 –1，这一概念在 GCSE 阶段会更深入。
Quadratic graphs y = x² + bx + c produce a parabola. Find the y‑intercept (c), roots by solving x² + bx + c = 0, and the vertex by completing the square or using symmetry.
二次函数 y = x² + bx + c 的图像为抛物线。通过常数项得 y 截距，解方程找与 x 轴交点，顶点可通过配方法或对称性求得。
Cubic graphs, e.g. y = x³, have an S‑shape. Reciprocal graphs y = 1/x have two branches and asymptotes. Learn to sketch these shapes quickly.
三次函数如 y = x³ 呈 S 形。反比例函数 y = 1/x 有两支分支和渐近线。学会快速勾勒这些图像。
y = mx + c
Plotting graphs requires a table of values. Choose a sensible range, calculate y for each x, then plot points smoothly. Label axes and scale.
绘制图像需要列出数值表。选取合适范围，计算各 x 对应的 y 值，平滑描点并连线。标注坐标轴和刻度。
6. Fractions, Decimals and Percentages | 分数、小数与百分比
Converting recurring decimals to fractions: for a simple repeat like 0.3̅, let x = 0.333…, then 10x = 3.333…, subtract: 9x = 3 → x = 1/3. For 0.15̅, use 100x.
循环小数化分数：如 0.3̅，设 x = 0.333…，则 10x = 3.333…，相减得 9x = 3 → x = 1/3。对 0.15̅ 则用 100x。
Percentage increase: multiply by (1 + r/100). To increase £80 by 15%, new amount = 80 × 1.15 = £92. Decrease uses (1 – r/100).
百分比增加：乘以 (1 + r/100)。£80 增加 15%，新值为 80 × 1.15 = £92。减少则用 (1 – r/100)。
Reverse percentages: if a price after a 20% decrease is £64, the original is 64 ÷ 0.8 = £80. Identify the multiplier that gives the current value.
逆向百分比：若降价 20% 后为 £64，原价 = 64 ÷ 0.8 = £80。找到对应现值的乘数。
Compound interest uses repeated multiplication: A = P(1 + r/100)ⁿ. For £500 at 3% for 4 years, A = 500 × (1.03)⁴. The exponent n is the number of compounding periods.
复利采用重复乘法：A = P(1 + r/100)ⁿ。£500 以 3% 存 4 年，A = 500 × (1.03)⁴。指数 n 为计息期数。
| Multiplier | Operation |
| 1 + r/100 | Increase by r% |
| 1 – r/100 | Decrease by r% |
7. Ratio, Proportion and Rates of Change | 比、比例与变化率
Simplify ratios by dividing all parts by their highest common factor. 24:36:12 simplifies to 2:3:1. Ratios can be written with decimals if needed, but simpler is better.
化简比：将所有部分除以它们的最大公因数。24:36:12 化为 2:3:1。必要时可用小数表示，但最简形式更优。
Dividing a quantity in a given ratio: to divide £120 in 3:5, total parts = 8. One part = £15, so shares are £45 and £75.
按给定比例分配量：将 £120 按 3:5 分配，总份数 8，一份 £15，因此两部分为 £45 和 £75。
Direct proportion: y = kx. If y = 10 when x = 4, then k = 2.5. The graph is a straight line through the origin.
正比例关系：y = kx。若 x=4 时 y=10，则 k=2.5。图像为过原点的直线。
Inverse proportion: y = k/x. The graph is a hyperbola. When distance is fixed, speed and time are inversely proportional.
反比例关系：y = k/x。图像为双曲线。路程固定时，速度与时间成反比。
Rates of change appear in distance‑time graphs: gradient = speed. A horizontal line means stopped; a steeper line means faster speed.
变化率体现在距离‑时间图中：斜率即速度。水平线表示静止；线越陡速度越快。
8. Pythagoras’ Theorem and Trigonometry | 勾股定理与三角学
Pythagoras’ theorem: in a right‑angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. c² = a² + b². For a=3, b=4, c=√(3²+4²)=5.
勾股定理：直角三角形中，斜边平方等于两直角边平方和。c² = a² + b²。a=3,b=4 时 c=√(3²+4²)=5。
To find a shorter side, rearrange: a² = c² – b². For c=13, b=5, a = √(13²–5²)=√144=12.
求短边时变形：a² = c² – b²。c=13,b=5，a = √(13²–5²)=√144=12。
Trigonometric ratios relate angles to side lengths: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. Remember SOH CAH TOA.
三角比联系角度与边长：sin θ = 对边/斜边，cos θ = 邻边/斜边，tan θ = 对边/邻边。记住 SOH CAH TOA。
sin θ = O/H, cos θ = A/H, tan θ = O/A
To find an angle, use the inverse function: θ = sin⁻¹(O/H). Always check your calculator is in degree mode.
求角度时用反函数：θ = sin⁻¹(对边/斜边)。始终检查计算器处于角度模式。
In 3D problems, apply Pythagoras twice or use trigonometric ratios in individual right‑angled triangles drawn within the solid.
解三维问题时，可两次应用勾股定理，或在立体图形内找直角三角形使用三角比。
9. Perimeter, Area and Volume | 周长、面积与体积
Circumference of a circle = 2πr, area = πr². For a circle of radius 7 cm, C ≈ 44.0 cm, A ≈ 153.9 cm² (using π≈3.14 or calculator π).
圆的周长 = 2πr，面积 = πr²。半径 7 cm 的圆，周长约 44.0 cm，面积约 153.9 cm²。
Area of a trapezium = ½(a+b)h, where a and b are the parallel sides and h is the perpendicular height. Combine with triangles and rectangles for composite shapes.
梯形面积 = ½(a+b)h，a、b 为平行边，h 为垂直高度。可结合三角形和矩形求组合图形面积。
Volume of a prism = area of cross‑section × length. For a cylinder, V = πr²h. Surface area sums the area of all faces.Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
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