Tag: KS3

  • Cambridge KS3 Maths: Solving Linear Equations (p141 Q2) | 剑桥 KS3 数学:解一元一次方程(第141页第2题)

    📚 Cambridge KS3 Maths: Solving Linear Equations (p141 Q2) | 剑桥 KS3 数学:解一元一次方程(第141页第2题)

    Welcome to this Cambridge KS3 Maths revision guide. We will work through a key algebra topic: solving linear equations in one variable. This includes one-step and two-step equations, equations with brackets, equations with fractions, and equations with unknowns on both sides. The methods shown here are directly relevant to Cambridge Checkpoint questions, including the exercise found on page 141 question 2.

    欢迎阅读本篇剑桥 KS3 数学复习指南。我们将系统学习一个核心代数主题:解一元一次方程。内容包括一步和两步方程、含括号的方程、含分数的方程以及未知数在等式两边的方程。这里展示的方法与 Cambridge Checkpoint 试题直接相关,包括第141页第2题的练习。


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

    A linear equation is an algebraic statement that says two expressions are equal, and the unknown variable appears only to the first power. For example, 3x + 5 = 20 and 4(x – 2) = 16 are linear equations. The word ‘linear’ comes from the fact that if you drew y = 3x + 5, you would get a straight line.

    一元一次方程是一个代数等式,表示两个表达式相等,而且未知数只以一次方的形式出现。例如 3x + 5 = 20 和 4(x – 2) = 16 都是一元一次方程。单词 ‘linear’ 源于这样一个事实:如果画出 y = 3x + 5 的图像,你会得到一条直线。

    3x + 5 = 20

    x = 5

    In this example, x = 5 is the solution because substituting 5 gives 3 × 5 + 5 = 20, which is true. Linear equations can have one solution, no solution, or infinitely many solutions, but at KS3 we mostly meet equations with exactly one solution.

    在这个例子中,x = 5 是解,因为代入 5 得到 3 × 5 + 5 = 20,等式成立。线性方程可能有一个解、

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

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

    Linear equations are the foundation of algebra at KS3. On page 140 of the Cambridge Checkpoint Maths series, you will meet equations such as x + 3 = 7, 4x = 20 and 2(x – 1) = x + 5. Mastering these skills allows you to solve problems in geometry, science and everyday life. This article explains the key methods step by step and gives you the confidence to tackle any Cambridge KS3 equation question.

    线性方程是 KS3 代数的基础。在剑桥 Checkpoint 数学系列的第 140 页,你会遇到像 x + 3 = 7、4x = 20 和 2(x – 1) = x + 5 这样的方程。掌握这些技能能帮助你解决几何、科学和日常生活中的问题。本文逐步讲解核心方法,帮助你自信应对任何剑桥 KS3 方程题。

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

    A linear equation is an equation in which the unknown, usually written as x, is raised only to the power of 1. This means there are no x² terms, no square roots of x, and no fractions where x appears in the denominator. For example, 3x + 5 = 14 is linear, but x² + 2 = 6 is not. The word ‘linear’ comes from the fact that the graph of y = 3x + 5 is a straight line.

    线性方程是指未知数(通常写作 x)的指数只为 1 的方程。也就是说,方程中没有 x² 项,没有 x 的平方根,也没有 x 出现在分母中的分数。例如,3x + 5 = 14 是线性方程,但 x² + 2 = 6 不是。’线性’ 一词来源于 y = 3x + 5 的图像是一条直线。

    In KS3, you solve linear equations by finding the value of x that makes the equation true. This value is called the solution or root. The equals sign acts like a balance: whatever you do to one side, you must do to the other side to keep the equation balanced.

    在 KS3 阶段,你通过求出使方程成立的 x 值来解线性方程。这个值叫做方程的解或根。等号的作用就像一个天平:你对等式的一边做了什么,就必须对另一边做同样的操作,以保持方程平衡。


    2. The Balance Method | 天平法

    The balance method is the most reliable way to solve linear equations. Imagine each side of the equation is a pan on a balance scale. If you add, subtract, multiply or divide one side by a number, you must do exactly the same to the other side. This keeps the pans level and the equation true.

    天平法是解线性方程最可靠的方法。把方程的两边想象成天平的两个托盘。如果你对一边进行加、减、乘或除以一个数,你必须对另一边做完全相同的操作。这样托盘才能保持水平,方程才成立。

    For example, to solve x + 6 = 15, subtract 6 from both sides: x + 6 – 6 = 15 – 6, which simplifies to x = 9. You can check by substituting 9 back into the original equation: 9 + 6 = 15, so the solution is correct.

    例如,解 x + 6 = 15,两边同时减去 6:x + 6 – 6 = 15 – 6,化简得 x = 9。你可以把 9 代回原方程检验:9 + 6 = 15,所以解是正确的。

    Writing the operation on both sides on the same line, or directly below, helps you show method marks in Cambridge exams. Even when you can see the answer mentally, always show the balancing step.

    把两边进行的操作写在同一行或正下方,有助于你在剑桥考试中拿到过程分。即使你能直接看出答案,也要写出平衡步骤。


    3. Solving with Addition and Subtraction | 用加法与减法解方程

    When a number is added to or subtracted from the unknown, undo that operation by using its inverse. If the equation says x – 4 = 10, add 4 to both sides: x – 4 + 4 = 10 + 4, giving x = 14. If the equation says x + 7 = 3, subtract 7 from both sides: x = 3 – 7 = -4.

    当未知数被加上或减去一个数时,用逆运算来消去它。如果方程是 x – 4 = 10,两边都加 4:x – 4 + 4 = 10 + 4,得到 x = 14。如果方程是 x + 7 = 3,两边都减 7:x = 3 – 7 = -4。

    Many KS3 students find negative solutions challenging. Remember that subtracting a larger number from a smaller one gives a negative result. Practise with examples such as x + 9 = 2 and x – 5 = -3 until you are comfortable moving between positive and negative values.

    许多 KS3 学生对负数解感到困难。记住,用较小的数减去较大的数会得到负数。多练习像 x + 9 = 2 和 x – 5 = -3 这样的例子,直到你能熟练处理正负数之间的转换。

    Always line up the equation vertically so that the added or subtracted number appears clearly on both sides. This reduces careless mistakes when the solution is negative.

    始终把方程垂直排列,使两边加或减的数字清晰可见。这样在解为负数时能减少粗心错误。


    4. Solving with Multiplication and Division | 用乘法与除法解方程

    If the unknown is multiplied by a number, divide both sides by that coefficient. For example, 5x = 35 becomes 5x ÷ 5 = 35 ÷ 5, so x = 7. If x is divided by a number, multiply both sides by that denominator. For x/4 = 9, multiply both sides by 4: x = 36.

    如果未知数乘以一个数,两边都除以这个系数。例如,5x = 35 变为 5x ÷ 5 = 35 ÷ 5,所以 x = 7。如果 x 除以一个数,两边都乘以这个分母。对于 x ÷ 4 = 9,两边都乘以 4:x = 36。

    Be careful with negative coefficients: in -3x = 12, divide both sides by -3 to get x = -4. A common sign error is to write x = 4 because the negative sign is forgotten.

    注意负系数:在 -3x = 12 中,两边都除以 -3,得到 x = -4。一个常见的符号错误是忘记负号而写成 x = 4。Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com

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  • Ratio and Proportion: Cambridge KS3 Core Skills | 比例与比例推理:剑桥初中数学核心技法

    📚 Ratio and Proportion: Cambridge KS3 Core Skills | 比例与比例推理:剑桥初中数学核心技法

    Ratios and proportions appear throughout the Cambridge KS3 mathematics course, from sharing money and mixing ingredients to reading scale drawings and solving direct proportion problems. The skills in this article match the style of page 134 question 2, where you must choose the correct part of a ratio and apply it to a real situation.

    比和比例贯穿剑桥初中数学课程的始终,从分配金钱、混合配料到读取比例尺图和解决正比例问题。本文所讲的技能与第134页第2题的题型一致,这类题目要求你正确找出比的各项,并将其应用到实际情境中。


    1. Understanding Ratio | 理解比

    A ratio compares two or more quantities in the same unit. For example, if a fruit bowl has 4 apples and 6 oranges, the ratio of apples to oranges is written as 4 : 6.

    比用来比较两个或两个以上相同单位的量。例如,一个水果碗里有 4 个苹果和 6 个橙子,苹果与橙子的比就写作 4 : 6。

    The order is always important: 4 : 6 is not the same as 6 : 4. The first number must match the first item named in the question.

    顺序始终非常重要:4 : 6 不等于 6 : 4。第一个数字必须对应题目中最先提到的项目。

    A ratio has no units after it is simplified. It only tells us the relative size of the parts, not the actual amounts.

    比在化简后没有单位。它只表示各部分之间的相对大小,并不表示实际数量。


    2. Simplifying Ratios | 化简比

    To simplify a ratio, divide every part by the highest common factor, or HCF. For 4 : 6, the HCF is 2, so 4 ÷ 2 : 6 ÷ 2 = 2 : 3.

    化简比时,要用最大公因数(HCF)去除比的每一项。例如 4 : 6 的最大公因数是 2,所以 4 ÷ 2 : 6 ÷ 2 = 2 : 3。

    If a ratio contains decimals or fractions, multiply every part by the same power of 10 or the same denominator first. This makes all parts whole numbers.

    如果比中含有小数或分数,要先给每一项乘上相同的 10 的幂或相同的分母,使所有部分都变成整数。

    • Example: 0.5 : 1.5 = 5 : 15 = 1 : 3 | 示例:0.5 : 1.5 = 5 : 15 = 1 : 3
    • Example: 1/2 : 1/4 = 2 : 1 | 示例:1/2 : 1/4 = 2 : 1

    3. Writing Ratios in the Form 1 : n | 写成 1 : n 的形式

    Cambridge questions often ask you to write a ratio in the form 1 : n. To do this, divide both sides of the ratio by the first number.

    剑桥考试常要求将比写成 1 : n 的形式。方法是用第一个数去除比的两边。

    For example, 3 : 12 becomes 1 : 4 because 3 ÷ 3 = 1 and 12 ÷ 3 = 4.

    例如,3 : 12 可以写成 1 : 4,因为 3 ÷ 3 = 1,12 ÷ 3 = 4。

    a : b → 1 : (b ÷ a)

    Use this form when you need to compare one part to the other relative to a single unit. It is especially useful for scale drawings and rates.

    当你需要以单位 1 为基准比较两个部分时,可以使用这种形式。它在比例尺图和速率问题中尤其有用。

    Original ratio | 原始比 Form 1 : n | 1 : n 形式
    5 : 20 1 : 4
    2 : 5 1 : 2.5

    4. Sharing in a Given Ratio | 按给定比例分配

    To share a quantity in the ratio a : b, first add the parts to find the total number of parts. Then divide the quantity by that total to find the value of one part.

    要按 a : b 分配一个量,先把各项相加求出总份数。然后用总量除以总份数,得到一份的大小。

    Finally multiply the value of one part by each ratio term. This gives the actual amount for each share.

    最后用一份的大小分别乘比的每一项,就能得到每一部分的实际数量。

    One part = Total ÷ (a + b)

    Example: Share £200 in the ratio 2 : 3.

    示例:按 2 : 3 分配 200 英镑。

    Total parts = 2 + 3 = 5, so one part = £200 ÷ 5 = £40. The shares are £40 × 2 = £80 and £40 × 3 = £120.

    总份数 = 2 + 3 = 5,所以一份 = 200 ÷ 5 = 40 英镑。两部分分别是 40 × 2 = 80 英镑和 40 × 3 = 120 英镑。

    Always check that the shares add back to the original total: £80 + £120 = £200.

    一定要检验各部分相加是否等于原来的总量:80 + 120 = 200 英镑。


    5. Three-Part Ratios | 三项比

    When three quantities are compared, such as red : blue : green = 3 : 5 : 7, the total number of parts is 3 + 5 + 7 = 15.

    当比较三个量时,例如红 : 蓝 : 绿 = 3 : 5 : 7,总份数为 3 + 5 + 7 = 15。

    Three-part ratio problems work exactly the same way as two-part ratio problems. Find one part first, then multiply by each term.

    三项比的解题方法与两项比完全相同。先求出一份的大小,再分别乘以每一项。

    Example: 300 ml of juice is made from syrup, water and sparkling water in the ratio 1 : 2 : 3.

    示例:300 毫升果汁由糖浆、水和气泡水按 1 : 2 : 3 的比例混合而成。

    Total parts = 1 + 2 + 3 = 6. One part = 300 ÷ 6 = 50 ml. Amounts are 50 ml, 100 ml and 150 ml.

    总份数 = 1 + 2 + 3 = 6。一份 = 300 ÷ 6 = 50 毫升。各量分别为 50 毫升、100 毫升和 150 毫升。

    Check: 50 + 100 + 150 = 300 ml and the simplified ratio is 50 : 100 : 150 = 1 : 2 : 3.

    检验:50 + 100 + 150 = 300 毫升,化简后为 50 : 100 : 150 = 1 : 2 : 3。


    6. Direct Proportion | 正比例

    Two quantities are in direct proportion if one increases at the same rate as the other. The ratio between the two quantities remains constant.

    如果两个量以相同的速率增长,它们就成正比例。两个量之间的比保持不变。

    For example, if 5 pens cost £3, then 10 pens cost £6. The cost per pen is £3 ÷ 5 = £0.60.

    例如,5 支笔花费 3 英镑,那么 10 支笔花费 6 英镑。每支笔的价格是 3 ÷ 5 =

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  • Mastering Percentages for Cambridge KS3 Maths | 掌握剑桥 KS3 数学百分数

    📚 Mastering Percentages for Cambridge KS3 Maths | 掌握剑桥 KS3 数学百分数

    A percentage is a fraction with denominator 100. The symbol % means ‘out of 100’, so 35% = 35/100 = 0.35. In KS3, percentages link fractions, decimals, ratio and real-life problems such as discounts, tax and interest.

    百分数是一个分母为 100 的分数。符号 % 表示“每一百”,因此 35% = 35/100 = 0.35。在 KS3 阶段,百分数将分数、小数、比与实际生活问题(如折扣、税费和利息)联系起来。

    1. What Is a Percentage? | 什么是百分数?

    A percentage is a way of expressing a number as a part of 100. It allows different quantities to be compared on the same scale. For example, 25% means 25 out of every 100, which can also be written as the fraction 25/100 = 1/4 or the decimal 0.25.

    百分数是一种把数字表示为每一百中的多少的方式。它使不同的数量能够在同一尺度上进行比较。例如,25% 表示每一百中有 25,也可以写成分数 25/100 = 1/4 或小数 0.25。

    Every percentage can be converted into a fraction or decimal, and these conversions are the foundation for many KS3 calculations.

    每个百分数都可以转换为分数或小数,这些转换是许多 KS3 计算的基础。


    2. Converting Between Fractions, Decimals and Percentages | 分数、小数与百分数互化

    To convert a decimal to a percentage, multiply by 100. To convert a percentage to a decimal, divide by 100. To convert a fraction to a percentage, multiply the fraction by 100. To convert a percentage to a fraction, write it over 100 and simplify.

    将小数化为百分数时乘以 100;将百分数化为小数时除以 100。将分数化为百分数时用分数乘以 100;将百分数化为分数时写成分母为 100 的分数并化简。

    • 0.75 × 100 = 75%
    • 40% ÷ 100 = 0.40
    • 3/5 × 100 = 60%
    • 65% = 65/100 = 13/20

    These conversions are useful when you need to compare a test score, a discount or a data proportion quickly.

    当需要快速比较考试成绩、折扣或数据比例时,这些转换非常有用。


    3. Finding a Percentage of a Quantity | 求一个量的百分数

    To find a percentage of an amount, use the rule: percentage of amount = (percentage / 100) × amount. For example, 15% of £80 is calculated as 0.15 × 80 = £12.

    要求一个量的百分数,使用规则:量的百分数 = (百分数 / 100) × 总量。例如,80 英镑的 15% 计算为 0.15 × 80 = 12 英镑。

    15% of 80 = (15 / 100) × 80 = £12

    This method is tested in questions about discounts, test marks, tax and commission.

    该方法常用于折扣、考试分数、税费和佣金类问题。


    4. Using Non-Calculator Methods: 10%, 5%, 1% and Multiples | 非计算器方法:10%、5%、1%及倍数

    You can build harder percentages from simple benchmarks. 10% = amount ÷ 10, 5% = 10% ÷ 2, and 1% = amount ÷ 100. For example, to find 35% of 240, first find 10% = 24, then 5% = 12, so 30% = 3 × 24 = 72 and 35% = 72 + 12 = 84.

    你可以从简单的基准构造更复杂的百分数。10% = 总量 ÷ 10,5% = 10% ÷ 2,1% = 总量 ÷ 100。例如,求 240 的 35%:先求 10% = 24,再求 5% = 12,因此 30% = 3 × 24 = 72,35% = 72 + 12 = 84。

    • 10% of 240 = 24
    • 5% of 240 = 12
    • 35% of 240 = 72 + 12 = 84

    This non-calculator approach is often required in Cambridge KS3 mental and written papers.

    这种非计算器方法在剑桥 KS3 的口算和笔试中经常要求使用。


    5. Percentage Increase and Decrease | 百分数增加与减少

    To increase an amount by r%, multiply by (1 + r / 100). To decrease an amount by r%, multiply by (1 – r / 100). Example: £60 increased by 15% = 60 × 1.15 = £69. £80 decreased by 25% = 80 × 0.75 = £60.

    将数量增加 r%,乘以 (1 + r / 100);减少 r%,乘以 (1 – r / 100)。例如,60 英镑增加 15% = 60 × 1.15 = 69 英镑;80 英镑减少 25% = 80 × 0.75 = 60 英镑。

    New value = Original × (1 ± r / 100)

    Using multipliers is faster and less prone to error than finding the change and then adding or subtracting.

    使用乘数比先求出变化量再加减更快,也更不容易出错。


    6. Finding the Original Value After a Percentage Change | 百分数变化后求原值

    If a final amount includes a known percentage change, divide by the multiplier. After a 20% increase, final value = original × 1.20, so original = final ÷ 1.20. After a 15% decrease, final value = original × 0.85, so original = final ÷ 0.85.

    如果最终数量包含已知的百分数变化,除以乘数即可。增加 20% 后,最终值 = 原值 × 1.20,因此原值 = 最终值 ÷ 1.20;减少 15% 后,最终值 = 原值 × 0.85,因此原值 = 最终值 ÷ 0.85。

    Example: A coat costs £69 after a 15% increase. Original price = 69 ÷ 1.15 = £60.

    例如:一件外套涨价 15% 后为 69 英镑。原价 = 69 ÷ 1.15 = 60 英镑。

    This reverse percentage skill is common in KS3 word problems and checkpoints.

    这种反向百分数技能在 KS3 应用题和 checkpoint 考试中很常见。


    7. Percentage Change Calculations | 百分数变化计算

    Percentage change = (new value – original value) / original value × 100%. If a price rises from £40 to £50, the change is £10, so percentage change = 10/40 × 100 = 25%. If it falls from £50 to £45, percentage change = -5/50 × 100 = -10%, a 10% decrease.

    百分数变化 = (新值 – 原值) / 原值 × 100%。如果价格从 40 英镑涨到 50 英镑,变化量为 10 英镑,因此百分数变化 = 10/40 × 100 = 25%;如果从 50 英镑降到 45 英镑,百分数变化 = -5/50 × 100 = -10%,即减少 10%。

    Percentage change = (New – Original) / Original × 100%

    Remember that the denominator is always the original value, not the new value.

    注意分母始终是原值,而不是新值。


    8. Percentages in Real-Life Contexts | 百分数在实际生活中的应用

    Percentages appear in discounts, VAT, profit and loss, simple interest, test results and survey data. For simple interest, use I = P × r × t / 100. For a discount, multiply the original price by (1 – discount% / 100).

    百分数出现在折扣、增值税、利润与亏损、单利、考试成绩和调查数据中。单利公式为 I = P × r × t / 100。计算折扣时,用原价乘以 (1 – 折扣率 / 100)。

    Context Example Calculation
    Discount 20% off £45 45 × 0.80 = £36
    Simple interest £200 at 5% for 2 years I = 200 × 5 × 2 / 100 = £20

    Always read the question carefully to decide whether you need a percentage of an amount, a percentage change or an original value.

    仔细审题,判断需要求的是量的百分数、百分数变化还是原值。


    9. Comparing Quantities Using Percentages | 用百分数比较数量

    Percentages standardise comparisons. For example, 17 out of 25 is 68%, while 19 out of 30 is about 63.3%, so the first is a better score. Always write both as percentages before comparing to avoid misleading by different totals.

    百分数使比较标准化。例如,25 题中对 17 题是 68%,30 题中对 19 题约为 63.3%,因此前者表现更好。比较前应先将两者写成百分数,以免因总数不同而产生误导。

    This technique is especially useful in data and statistics questions where different sample sizes appear.

    当数据与统计问题中出现不同的样本量时,这种技巧尤其有用。


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

    Do not confuse percentage increase with percentage of amount. Do not add or subtract percentages directly when the bases are different. For example, a 20% increase followed by a 10% decrease is not a net 10% increase.

    不要混淆“百分数增加”与“求一个量的百分数”。当基数不同时,不要直接相加或相减百分数。例如,先增加 20% 再减少 10%,净变化并不是增加 10%。

    Show clear working, use multipliers, and check that the final answer is sensible. In exams, look for wording such as ‘of’, ‘increased by’, ‘decreased by’, ‘original value’ and ‘percentage change’.

    展示清晰步骤,使用乘数,并检查最终答案是否合理。考试中注意“的”、“增加”、“减少”、“原值”和“百分数变化”等关键词。

    Practising these question types from Cambridge KS3 past papers will help you avoid common traps and gain confidence.

    练习剑桥 KS3 历年真题中的这些题型,有助于避开常见陷阱并增强信心。


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  • Solving Linear Equations for KS3 | KS3 线性方程求解技巧

    📚 Solving Linear Equations for KS3 | KS3 线性方程求解技巧

    A linear equation is one of the most important ideas in KS3 mathematics. It appears in almost every Cambridge checkpoint test, and the skill of solving linear equations confidently will help you in algebra, graphs, word problems and even later topics such as simultaneous equations. In this article, you will learn how to solve linear equations step by step using clear methods, and you will see how to avoid the common mistakes that many students make.

    线性方程是 KS3 数学中最重要的内容之一。它几乎出现在每一次 Cambridge checkpoint 测试中,熟练掌握解线性方程的技巧将帮助你在代数、图像、应用题甚至日后的联立方程等主题中更加自信。本文将通过清晰的方法逐步教你如何解线性方程,并帮助你避免许多学生容易犯的常见错误。


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

    A linear equation is an equation where the unknown value, usually written as x or y, is only raised to the power of 1. This means there are no terms like x², x³ or 1/x in a linear equation. For example, x + 3 = 7 and 2x – 5 = 11 are both linear equations. The word ‘linear’ comes from the fact that the graph of such an equation is always a straight line.

    线性方程是指未知数(通常写作 x 或 y)只以一次方出现的方程。也就是说,线性方程中不会出现 x²、x³ 或 1/x 这样的项。例如 x + 3 = 7 和 2x – 5 = 11 都是线性方程。’线性’ 一词来源于这类方程的图像总是形成一条直线。

    In KS3, you normally solve linear equations with one unknown. The goal is simple: find the value of the unknown that makes the equation true. The equation works like a balance, so whatever you do to one side, you must do to the other side. This is the key principle behind every method in this article.

    在 KS3 阶段,你通常只解一个未知数的线性方程。目标很简单:找出使等式成立的未知数的值。方程就像一个天平,所以你对等式一边做什么运算,另一边也必须做同样的运算。这是本文所有方法背后的关键原则。


    2. The Balance Method | 天平法

    The balance method is the most reliable way to solve linear equations. Imagine the equation as a balance scale. The left side and the right side are perfectly balanced. If you add, subtract, multiply or divide by the same number on both sides, the balance is not broken. This allows you to rearrange the equation step by step until the unknown is alone on one side.

    天平法是解线性方程最可靠的方法。把方程想象成一台天平。左边和右边完全平衡。如果你在两边同时加、减、乘或除以同一个数,天平不会失衡。这样你就可以一步一步地重新整理方程,直到未知数单独出现在一边。

    For example, if x + 4 = 9, you can subtract 4 from both sides to undo the ‘+ 4’. This gives x + 4 – 4 = 9 – 4, which simplifies to x = 5. The key idea is to always do the opposite operation: addition is undone by subtraction, subtraction is undone by addition, multiplication is undone by division, and division is undone by multiplication.

    例如,如果 x + 4 = 9,你可以在两边同时减去 4,以抵消 ‘+ 4’。这样得到 x + 4 – 4 = 9 – 4,化简后 x = 5。核心思想是始终进行相反的运算:加法用减法抵消,减法用加法抵消,乘法用除法抵消,除法用乘法抵消。

    Many students find it helpful to write each step on a new line. This makes your working clear and reduces careless errors. Always check that the operation is applied to the whole side, not just one term.

    许多学生发现每行只写一步会很有帮助。这样可以让解题过程更清晰,减少粗心错误。还要注意运算必须作用于整个一边,而不仅仅是其中的一项。


    3. Solving One-Step Equations | 解一步方程

    A one-step equation needs only one operation to find the solution. These are the building blocks for all harder equations. For addition and subtraction equations, you simply add or subtract the same number from both sides. For multiplication and division equations, you multiply or divide both sides by the same non-zero number.

    一步方程只需要一次运算就能求出解。它们是所有更复杂方程的基础。对于加法和减法方程,你只需在两边同时加或减同一个数。对于乘法和除法方程,你需要在两边同时乘以或除以同一个非零数。

    • Example 1: x + 7 = 12. Subtract 7 from both sides: x = 5.

      例 1:x + 7 = 12。两边同时减去 7:x = 5。

    • Example 2: x – 3 = 8. Add 3 to both sides: x = 11.

      例 2:x – 3 = 8。两边同时加上 3:x = 11。

    • Example 3: 4x = 20. Divide both sides by 4: x = 5.

      例 3:4x = 20。两边同时除以 4:x = 5。

    • Example 4: x ÷ 5 = 6. Multiply both sides by 5: x = 30.

      例 4:x ÷ 5 = 6。两边同时乘以 5:x = 30。

    Always remember that division and multiplication are opposite operations. If the unknown has been divided by a number, multiply by that number to undo it. If the unknown has been multiplied by a number, divide by that number to undo it.

    始终记住除法和乘法是相反的运算。如果未知数被一个数除,就乘以这个数来抵消它。如果未知数被一个数乘,就除以这个数来抵消它。


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

    Two-step equations require two operations to isolate the unknown. A typical example is 2x + 3 = 11. You must first undo the addition or subtraction, and then undo the multiplication or division. The order is usually the reverse of the order of operations, also called BIDMAS or BODMAS.

    两步方程需要两次运算才能将未知数分离出来。典型例子是 2x + 3 = 11。你必须先抵消加法或减法,然后再抵消乘法或除法。顺序通常与运算顺序(也叫 BIDMAS 或 BODMAS)相反。

    2x + 3 = 11

    Subtract 3 from both sides to undo the +3: 2x = 8. Then divide both sides by 2 to undo the ×2: x = 4. So the solution is x = 4.

    两边同时减去 3 以抵消 +3:2x = 8。然后两边同时除以 2 以抵消 ×2:x = 4。所以解是 x = 4。

    Another example is 5x – 7 = 18. First add 7 to both sides: 5x = 25. Then divide both sides by 5: x = 5. Notice how the addition or subtraction is always dealt with before the multiplication or division when you are trying to get x alone.

    另一个例子是 5x – 7 = 18。首先两边同时加上 7:5x = 25。然后两边同时除以 5:x = 5。注意当你尝试将 x 分离出来时,总是先处理加法或减法,再处理乘法或除法。


    5. Equations with Brackets | 含括号的方程

    When an equation contains brackets, such as 3(x + 2) = 15, the brackets mean multiplication. There are two common methods you can use. The first method is to expand the brackets using the distributive law. The second method is to divide both sides by the number in front of the bracket first.

    当方程含有括号时,例如 3(x + 2) = 15,括号表示乘法。你可以使用两种常见方法。第一种方法是使用分配律展开括号。第二种方法是先在两边同时除以括号前的数字。

    Using the expansion method, 3(x + 2) becomes 3x + 6. So the equation becomes 3x + 6 = 15. Subtract 6 from both sides: 3x = 9. Then divide both sides by 3: x = 3.

    使用展开法,3(x + 2) 变为 3x + 6。所以方程变为 3x + 6 = 15。两边同时减去 6:3x = 9。然后两边同时除以 3:x = 3。

    Using the division method, divide both sides by 3 first: x + 2 = 5. Then subtract 2 from both sides: x = 3. This method is often quicker when the number in front of the bracket divides evenly into the other side.

    使用除法方法,先两边同时除以 3:x + 2 = 5。然后两边同时减去 2:x = 3。当括号前的数字能整除等式另一边时,这种方法通常更快。

    Be careful with negative signs and subtraction inside brackets. For example, 2(x – 4) expands to 2x – 8, not 2x – 4. Always multiply every term inside the bracket by the number outside the bracket.

    要注意括号内的负号和减法。例如,2(x – 4) 展开为 2x – 8,而不是 2x – 4。始终将括号内的每一项都乘以括号外的数字。


    6. Equations with Unknowns on Both Sides | 未知数在等号两边

    Some linear equations have the unknown on both sides of the equals sign. For example, 7x – 3 = 4x + 9. The first goal is to collect all the x terms on one side and all the number terms on the other side. You can do this by adding or subtracting the smaller x term from both sides.

    有些线性方程的未知数出现在等号两边。例如,7x – 3 = 4x + 9。第一个目标是将所有含 x 的项移到一边,将所有数字项移到另一边。你可以通过在两边同时加上或减去较小的 x 项来实现。

    7x – 3 = 4x + 9

    Subtract 4x from both sides: 3x – 3 = 9. Then add 3 to both sides: 3x = 12. Finally divide both sides by 3: x = 4.

    两边同时减去 4x:3x – 3 = 9。然后两边同时加上 3:3x = 12。最后两边同时除以 3:x = 4。

    Always move the smaller x term so the coefficient of x stays positive. This reduces mistakes. You can move the larger x term too, but you will end up with a negative coefficient, which requires extra care.

    始终移动较小的 x 项,这样 x 的系数保持为正。这可以减少错误。你也可以移动较大的 x 项,但最终会得到负系数,需要额外小心。

    Once all the x terms are on one side, solve the remaining two-step equation as you did before. Check your answer by substituting it back into the original equation.

    当所有含 x 的项都集中到一边后,像之前一样解剩下的两步方程。将答案代入原方程进行检验。


    7. Equations with Fractions | 含分数的方程

    Equations with fractions can look difficult, but they become much easier if you multiply every term by the lowest common multiple (LCM) of the denominators. This removes the fractions and turns the equation into a simpler linear equation. Always multiply every term on both sides.

    含分数的方程看起来可能很难,但如果你将每一项都乘以所有分母的最小公倍数(LCM),问题就会变得简单得多。这样可以去掉分数,将方程转化为更简单的线性方程。一定要把两边每一项都乘以这个倍数。

    x/3 + 2 = 5

    Multiply every term by 3: x + 6 = 15. Then subtract 6 from both sides: x = 9.

    将每一项乘以 3:x + 6 = 15。然后两边同时减去 6:x = 9。

    For equations with more than one fraction, such as x/2 + x/3 = 5, the LCM of 2 and 3 is 6. Multiply every term by 6: 3x + 2x = 30. This simplifies to 5x = 30, so x = 6.

    对于含有多个分数的方程,例如 x/2 + x/3 = 5,2 和 3 的最小公倍数是 6。将每一项乘以 6:3x + 2x = 30。化简为 5x = 30,所以 x = 6。

    Be careful when a fraction has a negative sign in front of it. The negative sign applies to the whole fraction, not just the numerator or denominator. Use brackets if needed to avoid sign errors.

    当分数前有负号时要特别小心。负号适用于整个分数,而不仅仅是分子或分母。必要时使用括号以避免符号错误。


    8. Checking Your Solution | 检验解

    After you have found a value for x, you should always check it by substituting it back into the original equation. You can do this mentally or on paper. For example, if you solved 2x + 3 = 11 and found x = 4, check by replacing x with 4: 2 × 4 + 3 = 8 + 3 = 11. Since the left side equals the right side, the solution is correct.

    求出 x 的值后,你应该始终将它代入原方程进行检验。你可以在心里或纸上完成。例如,如果你解出 2x + 3 = 11 且得到 x = 4,用 4 替换 x 来检验:2 × 4 + 3 = 8 + 3 = 11。由于左边等于右边,解是正确的。

    Checking is especially important in tests. It only takes a few seconds and can catch small mistakes in signs, arithmetic or rearranging. If the two sides do not match, trace back through your working to find where the error occurred.

    在考试中检验尤为重要。它只需要几秒钟,就能发现符号、算术或移项中的小错误。如果两边不相等,就回头检查你的解题过程,找出错误所在。

    You can also use a simple table to organise your checking steps. Write the left side, the right side and the result after substitution. This helps you show clear working and earns you valuable method marks in checkpoint papers.

    你也可以用一个简单的表格来整理检验步骤。写出左边、右边和代入后的结果。这可以帮助你展示清晰的解题过程,并在 checkpoint 试卷中获得宝贵的方法分。

    Left side Right side Equal?
    2 × 4 + 3 = 11 11 Yes

    9. Common Mistakes to Avoid | 常见错误

    There are several mistakes that come up again and again when KS3 students solve linear equations. Knowing these mistakes will help you avoid them. The first mistake is forgetting to do the operation to both sides. If you subtract 3 from one side, you must subtract 3 from the other side as well.

    KS3 学生在解线性方程时经常会犯一些相同的错误。了解这些错误将帮助你避免它们。第一个错误是忘记对两边同时进行运算。如果你在一边减去 3,那么另一边也必须减去 3。

    • Mistake 1: Adding or subtracting only one side. Example: x + 4 = 10 becomes x = 10 – 4? No, write x = 10 – 4 by subtracting 4 from both sides.

      错误 1:只在一边加或减。例如 x + 4 = 10 变成 x = 10 – 4?错,应该是两边同时减去 4,写出 x = 10 – 4。

    • Mistake 2: Forgetting to multiply every term inside a bracket. Example: 3(x + 2) = 3x + 2 is wrong; it should be 3x + 6.

      错误 2:忘记乘以括号内的每一项。例如 3(x + 2) = 3x + 2 是错误的;应该是 3x + 6。

    • Mistake 3: Losing the negative sign. Example: -2x = 10 gives x = -5, not x = 5.

      错误 3:丢失负号。例如 -2x = 10 得出 x = -5,而不是 x = 5。

    • Mistake 4: Combining unlike terms. You cannot add 3x and 4 to get 7x.

      错误 4:合并不同类项。你不能把 3x 和 4 相加得到 7x。

    Another common error is dividing by a negative number and forgetting to change the sign. For instance, if -3x = 12, divide both sides by -3 to get x = -4. Always pay attention to the sign of the coefficient.

    另一个常见错误是除以负数时忘记变号。例如,如果 -3x = 12,两边同时除以 -3 得到 x = -4。始终注意系数的符号。

    Finally, many students try to solve equations by ‘moving’ terms without understanding the operation. It is better to write the operation you are performing on both sides at each step. This makes your working clear and helps you avoid sign errors.

    最后,许多学生试图通过 ‘移项’ 来解方程而不理解运算过程。更好的做法是每一步都写出你在两边执行的运算。这样可以让你的解题过程更清晰,并帮助你避免符号错误。


    10. Practice Questions | 练习题

    Use the methods from this article to solve the following equations. Try to show each step clearly and check your answers by substituting back into the original equation. These questions are typical of the types found in KS3 Cambridge checkpoint papers.

    使用本文中的方法解下列方程。尽量清楚地展示每一步,并通过代入原方程来检验答案。这些题目是 KS3 Cambridge checkpoint 试卷中常见的类型。

    • Solve x – 9 = 4.

      解方程 x – 9 = 4。

    • Solve 6x = 48.

      解方程 6x = 48。

    • Solve 3x + 5 = 20.

      解方程 3x + 5 = 20。

    • Solve 4(x – 2) = 24.

      解方程 4(x – 2) = 24。

    • Solve 5x – 7 = 3x + 9.

      解方程 5x – 7 = 3x + 9。

    • Solve x/4 + 3 = 7.

      解方程 x/4 + 3 = 7。

    For each answer, write a checking line similar to the table in Section 8. This will help you build confidence and accuracy. If you make a mistake, find the exact step where the error occurred before moving on to the next question.

    对于每个答案,写出类似第 8 节表格的检验过程。这可以帮助你建立信心并提高准确性。如果你犯了错误,在继续做下一题之前,找出错误发生的具体步骤。

    Once you can solve these equations confidently, you will be well prepared for harder topics such as inequalities, straight-line graphs and simultaneous equations in later years. Keep practising and always show your working clearly.

    当你能够自信地解出这些方程后,你就为日后更难的课题做好了准备,例如不等式、直线图像和联立方程。坚持练习,并始终清晰地展示你的解题过程。


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  • Fractions, Decimals and Percentages | KS3 数学:分数、小数与百分数

    📚 Fractions, Decimals and Percentages | KS3 数学:分数、小数与百分数

    Fractions, decimals and percentages are three different ways of describing the same value. They appear constantly in everyday life, from measuring ingredients in a recipe to calculating discounts in a shop.

    分数、小数和百分数是描述同一数值的三种不同方式。它们在日常生活中随处可见,从按食谱称量食材到在商店里计算折扣都离不开它们。

    In this KS3 revision guide, you will learn how to convert between the three forms, compare them and use them to solve problems confidently.

    在这份 KS3 复习指南中,你将学习如何在这三种形式之间进行转换、比较它们,并自信地运用它们解决问题。


    1. Understanding Fractions | 理解分数

    A fraction is written as a/b, where a is the numerator and b is the denominator. The denominator tells you how many equal parts the whole is divided into, and the numerator tells you how many of those parts are taken.

    分数写作 a/b,其中 a 是分子,b 是分母。分母表示整体被分成多少等份,分子表示取了多少份。

    For example, 3/5 means the whole is split into 5 equal parts and 3 of them are selected. A unit fraction has a numerator of 1, such as 1/2 or 1/8.

    例如,3/5 表示整体被分成 5 等份,并选取其中的 3 份。单位分数的分子为 1,如 1/2 或 1/8。

    Proper fractions have a numerator smaller than the denominator, improper fractions have a numerator greater than or equal to the denominator, and mixed numbers combine an integer with a proper fraction.

    真分数的分子小于分母,假分数的分子大于或等于分母,带分数由一个整数和一个真分数组成。

    Fractions are useful when a value is not a whole number, but decimals and percentages are often easier to compare in practical situations.

    当数值不是整数时,分数非常有用,但在实际情境中,小数和百分数通常更容易比较。


    2. Equivalent Fractions and Simplifying | 等值分数与化简

    Equivalent fractions have different numerators and denominators but represent the same value. You can create an equivalent fraction by multiplying or dividing both the numerator and the denominator by the same non-zero number.

    等值分数的分子和分母不同,但表示相同的值。你可以通过将分子和分母同时乘以或除以同一个非零数来得到等值分数。

    For example, 2/3 = 4/6 = 8/12 because each form is obtained by multiplying the top and bottom by the same factor.

    例如,2/3 = 4/6 = 8/12,因为每种形式都是将分子和分母同时乘以相同的因数得到的。

    To simplify a fraction, divide both terms by their highest common factor (HCF). The fraction 18/24 simplifies to 3/4 because the HCF of 18 and 24 is 6.

    化简分数时,将分子和分母同时除以它们的最大公因数 (HCF)。分数 18/24 化简为 3/4,因为 18 和 24 的最大公因数是 6。

    You can check if two fractions are equivalent by cross-multiplying. For example, 2/3 and 4/6 are equivalent because 2 × 6 = 3 × 4.

    你可以通过交叉相乘来检验两个分数是否等值。例如,2/3 和 4/6 等值,因为 2 × 6 = 3 × 4。

    Always simplify your final answer unless the question specifically asks for an unsimplified fraction.

    除非题目明确要求保留未化简的分数,否则最终答案一定要化简。


    3. Fractions to Decimals | 分数转小数

    To convert a fraction to a decimal, divide the numerator by the denominator. A fraction is simply a division statement.

    将分数转换为小数,用分子除以分母即可。分数本质上就是除法。

    For example, 3/8 means 3 ÷ 8 = 0.375, so 3/8 = 0.375. You can use short division or a calculator if allowed.

    例如,3/8 表示 3 ÷ 8 = 0.375,因此 3/8 = 0.375。你可以使用短除法,或者如果允许的话使用计算器。

    Some fractions produce terminating decimals, such as 1/4 = 0.25, while others produce recurring decimals, such as 1/3 = 0.333… which can be written as 0.3 with a dot over the 3.

    有些分数产生有限小数,如 1/4 = 0.25;另一些则产生循环小数,如 1/3 = 0.333…,可写作 0.3,在 3 上方加点。

    If the denominator is 10, 100 or 1000, conversion is immediate: 7/10 = 0.7, 23/100 = 0.23, and 9/1000 = 0.009.

    如果分母是 10、100 或 1000,转换非常直接:7/10 = 0.7,23/100 = 0.23,9/1000 = 0.009。

    Remember that the decimal point separates the whole number part from the fractional part, so 2/5 = 0.4 means 4 tenths, not 4 hundredths.

    记住小数点把整数部分和小数部分分开,所以 2/5 = 0.4 表示 4 个十分之一,而不是 4 个百分之一。


    4. Decimals to Fractions | 小数转分数

    To convert a terminating decimal to a fraction, write the decimal as the numerator over a power of 10, then simplify if possible.

    将有限小数转换为分数时,把小数作为分子,分母为 10 的幂,然后尽可能化简。

    For example, 0.65 = 65/100. Dividing the numerator and denominator by 5 gives 13/20.

    例如,0.65 = 65/100。将分子和分母同时除以 5,得到 13/20。

    The number of decimal places determines the denominator: one decimal place gives 10, two decimal places gives 100, three decimal places gives 1000, and so on.

    小数位数决定分母:一位小数对应 10,两位对应 100,三位对应 1000,依此类推。

    For mixed decimals such as 2.4, write the integer part separately: 2.4 = 2 + 4/10 = 2 2/5.

    对于像 2.4 这样的带小数,先写出整数部分:2.4 = 2 + 4/10 = 2 2/5。

    Always check if the fraction can be simplified. For example, 0.25 = 25/100 = 1/4, so 0.25, 25/100 and 1/4 all describe the same value.

    一定要检查分数是否可以化简。例如,0.25 = 25/100 = 1/4,所以 0.25、25/100 和 1/4 都表示相同的值。


    5. Percentages as Fractions and Decimals | 百分数作为分数与小数

    A percentage is a fraction with a denominator of 100. The symbol ‘%’ means ‘out of 100’, so 45% = 45/100 = 0.45.

    百分数是分母为 100 的分数。符号 ‘%’ 表示“每 100”,因此 45% = 45/100 = 0.45。

    To convert a percentage to a fraction, write it over 100 and simplify. For example, 25% = 25/100 = 1/4.

    将百分数转换为分数时,把它写在 100 之上并化简。例如,25% = 25/100 = 1/4。

    To convert a percentage to a decimal, divide by 100. This is the same as moving the decimal point two places to the left: 7% = 0.07 and 130%

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  • Solving Linear Equations: Cambridge KS3 Maths p124_2 | 解一元一次方程:剑桥 KS3 数学 p124_2

    📚 Solving Linear Equations: Cambridge KS3 Maths p124_2 | 解一元一次方程:剑桥 KS3 数学 p124_2

    Linear equations are the foundation of algebra in the Cambridge Lower Secondary programme. The exercise in p124_2.pdf focuses on equations that contain brackets, variables on both sides, and small fractional coefficients. This article explains the complete method, using the style of question 2 as a worked model.

    一元一次方程是剑桥初中数学代数的基础。p124_2.pdf 中的练习重点考查含有括号、两边含有未知数以及简单分数系数的方程。本文以第 2 题的题型为例,完整讲解解题方法。


    1. Understanding the Question | 理解题意

    The question in p124_2.pdf asks you to solve an equation for the unknown x. A typical example is 2(3x – 1) = 4(x + 2). Solving means finding every value of x that makes the left-hand side equal to the right-hand side.

    p124_2.pdf 中的题目要求解出未知数 x。一个典型例子是 2(3x – 1) = 4(x + 2)。解方程意味着找出所有使左边等于右边的 x 值。

    Read the question twice. Underline the brackets and any like terms before starting. This helps you choose the correct first step.

    请把题目读两遍。开始

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

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

    Working confidently with fractions, decimals and percentages is a core skill in the Cambridge KS3 Mathematics curriculum. This topic often appears on pages such as 123 of the learner’s book, where you are asked to convert, compare and calculate with all three forms. This revision guide explains the key ideas step by step and gives you clear examples.

    在剑桥 KS3 数学课程中,熟练处理分数、小数和百分比是一项核心技能。这类内容经常出现在教材第 123 页等位置,要求同学们在三种形式之间进行转换、比较和计算。本复习指南将逐步讲解关键概念,并给出清晰的示例。

    1. Understanding Fractions | 理解分数

    A fraction is written as a numerator over a denominator, for example 3/4. The denominator tells you how many equal parts make one whole, and the numerator tells you how many of those parts are being considered.

    分数写作分子在上、分母在下的形式,例如 3/4。分母表示一个整体被平均分成多少份,分子表示我们选取了多少份。

    A proper fraction has a numerator smaller than its denominator, such as 2/5. An improper fraction has a numerator equal to or greater than its denominator, such as 7/4, and a mixed number combines a whole number with a proper fraction, such as 1 3/4.

    真分数的分子小于分母,例如 2/5。假分数的分子大于或等于分母,例如 7/4。带分数由一个整数和一个真分数组成,例如 1 3/4。


    2. Equivalent Fractions and Simplest Form | 等值分数与最简形式

    Equivalent fractions represent the same quantity even though they look different. You can create an equivalent fraction by multiplying or dividing both the numerator and denominator by the same non-zero number.

    等值分数表示相同的量,尽管形式不同。将分子和分母同时乘以或除以同一个非零数,就可以得到等值分数。

    To write a fraction in its simplest form, divide the numerator and denominator by their highest common factor (HCF). For example, 12/16 becomes 3/4 because the HCF of 12 and 16 is 4.

    要把分数化为最简形式,用分子和分母的最大公因数 (HCF) 同时去除它们。例如,12/16 化简为 3/4,因为 12 和 16 的最大公因数是 4。


    3. Converting Between Improper Fractions and Mixed Numbers | 假分数与带分数互转

    To change a mixed number to an improper fraction, multiply the whole number by the denominator, add the numerator, and place the result over the original denominator. For example, 2 3/5 becomes (2 × 5 + 3)/5 = 13/5.

    将带分数化为假分数时,用整数乘以分母,加上分子,再把结果放在原分母上。例如,2 3/5 变为 (2 × 5 + 3)/5 = 13/5。

    To change an improper fraction to a mixed number, divide the numerator by the denominator. The quotient is the whole number part, the remainder is the new numerator, and the denominator stays the same.

    将假分数化为带分数时,用分子除以分母。商是整数部分,余数是新的分子,分母保持不变。


    4. Understanding Decimals | 理解小数

    Decimals are another way of writing fractions whose denominators are powers of 10. The first decimal place is tenths, the second is hundredths, and the third is thousandths.

    小数是分母为 10 的幂的分数的另一种写法。小数点后第一位是十分位,第二位是百分位,第三位是千分位。

    For example, 0.3 means 3/10, 0.25 means 25/100, and 0.375 means 375/1000. This place-value link makes converting between decimals and fractions easier.

    例如,0.3 表示 3/10,0.25 表示 25/100,0.375 表示 375/1000。这种位值联系使小数与分数的转换更加容易。


    5. Converting Fractions to Decimals | 分数转小数

    To convert a fraction to a decimal, divide the numerator by the denominator. You can use short division or a calculator: 3/8 = 3 ÷ 8 = 0.375.

    要将分数转换为小数,用分子除以分母。可以使用短除法或计算器:3/8 = 3 ÷ 8 = 0.375。

    Some fractions produce terminating decimals, such as 1/4 = 0.25. Others produce recurring decimals, such as 1/3 = 0.333… where the 3 repeats forever. A recurring decimal can be written with a dot above the repeating digit.

    有些分数会得到有限小数,例如 1/4 = 0.25。另一些则得到循环小数,例如 1/3 = 0.333…,其中 3 无限重复。循环小数可以在重复数字上方加点表示。


    6. Converting Decimals to Fractions | 小数转分数

    To convert a terminating decimal to a fraction, write the decimal as a fraction with denominator 10, 100, 1000 and so on, depending on the number of decimal places. Then simplify if possible.

    将有限小数转换为分数时,根据小数位数,将小数写成分母为 10、100、1000 等的分数,然后尽可能化简。

    For example, 0.45 = 45/100 = 9/20. For a recurring decimal like 0.444…, the conversion uses a special algebraic method: let x = 0.444…, then 10x = 4.444…, subtracting gives 9x = 4, so x = 4/9.

    例如,0.45 = 45/100 = 9/20。对于 0.444… 这样的循环小数,转换时使用一种特殊的代数方法:设 x = 0.444…,则 10x = 4.444…,相减得 9x = 4,所以 x = 4/9。


    7. Understanding Percentages | 理解百分比

    A percentage is a fraction with denominator 100. The symbol % means ‘out of 100’, so 35% = 35/100 = 0.35.

    百分比是分母为 100 的分数。百分号 % 表示 “每一百”,因此 35% = 35/100 = 0.35。

    Percentages are useful for comparing quantities because they convert different totals to a common scale of 100. For example, a test score of 17 out of 20 is 17/20 = 0.85 = 85%.

    百分比便于比较数量,因为它把不同的总量都换算成以 100 为统一标准。例如,20 分中获得 17 分就是 17/20 = 0.85 = 85%。


    8. Converting Between Fractions, Decimals and Percentages | 分数、小数和百分比互转

    To convert a percentage to a decimal, divide by 100. To convert a decimal to a percentage, multiply by 100. To convert a percentage to a fraction, write it over 100 and simplify.

    将百分比化为小数,除以 100;将小数化为百分比,乘以 100;将百分比化为分数,写成百分号下的数除以 100 再化简。

    The table below shows common conversions that are worth memorising for Cambridge KS3 assessments.

    下表列出了一些值得在剑桥 KS3

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  • KS3 Maths: Percentage Increase and Decrease | KS3 数学:百分数的增加与减少

    📚 KS3 Maths: Percentage Increase and Decrease | KS3 数学:百分数的增加与减少

    Percentage increase and decrease are essential skills in the Cambridge KS3 Mathematics curriculum. They appear in word problems about prices, discounts, interest, profit, tax and data changes. This article explains the core methods, common formulas and exam strategies step by step.

    百分数增加与减少是剑桥 KS3 数学课程中的基本技能。它们经常出现在关于价格、折扣、利息、利润、税收和数据变化的文字题中。本文将逐步讲解核心方法、常用公式和考试策略。

    1. Understanding Percentages | 理解百分数

    A percentage is a number expressed as a fraction of 100. The symbol % means ‘out of 100’, so 45% means 45 out of 100, or 45/100. In decimal form, 45% is 0.45.

    百分数是以 100 为分母的分数。符号 % 表示“每一百份中”,所以 45% 表示 100 份中的 45 份,即 45/100。写成小数就是 0.45。

    Percentage Decimal Fraction 中文含义
    25% 0.25 1/4 四分之一
    50% 0.5 1/2 二分之一
    75% 0.75 3/4 四分之三
    10% 0.1 1/10 十分之一
    5% 0.05 1/20 二十分之一

    Being able to switch quickly between percentages, decimals and fractions makes all percentage calculations easier.

    能够快速在百分数、小数和分数之间转换,会让所有百分数计算变得更加简单。


    2. Finding a Percentage of a Quantity | 求一个数量的百分之几

    To find p% of a quantity N, first write p% as a decimal by dividing by 100. Then multiply the decimal by N. For example, 15% of 240 is 0.15 × 240 = 36.

    求某个量 N 的 p%,先把 p% 除以 100 写成小数。然后用这个小数乘以 N。例如,240 的 15% 是 0.15 × 240 = 36。

    p% of N = (p ÷ 100) × N

    Always check that your answer makes sense. A useful benchmark is 10%. For 240, 10% is 24, so 15% should be 24 + 12 = 36.

    始终检查答案是否合理。10% 是一个有用的基准。对于 240,10% 是 24,因此 15% 应该是 24 + 12 = 36。


    3. Percentage Increase | 百分数增加

    Percentage increase describes how much a value grows compared with its original amount. The increase is always measured against the original value, not the new value.

    百分数增加描述一个值相对于原始值增长了多少。增加量始终是相对于原始值来衡量的,而不是相对于新值。

    Percentage increase = (increase ÷ original) × 100%

    To find a new value directly, add the increase to the original. If a quantity increases by r%, its new value can also be written as original × (1 + r/100).

    要直接求新值,将增加量加到原始值上。如果一个量增加了 r%,它的新值也可以写成 原始值 × (1 + r/100)。

    Example: a school club had

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  • KS3 Cambridge Maths: Fractions, Decimals and Percentages (p120 Q2) | 剑桥中学数学:分数、小数与百分数(p120 第2题)

    📚 KS3 Cambridge Maths: Fractions, Decimals and Percentages (p120 Q2) | 剑桥中学数学:分数、小数与百分数(p120 第2题)

    In this article we review a core KS3 Cambridge maths topic: fractions, decimals and percentages. This topic appears frequently in Checkpoint tests and in homework pages such as page 120, question 2. You will see how the three forms are linked, how to convert between them, and how to apply them to real problems.

    本文复习剑桥初中数学的核心专题:分数、小数与百分数。该专题经常出现在 Checkpoint 测试以及类似第 120 页第 2 题的作业中。你将看到这三种形式之间的联系、如何互相转换,以及如何将它们应用于实际问题。

    1. Understanding the Big Picture | 理解整体图景

    Fractions, decimals and percentages are three different ways of describing the same idea: a part of a whole. A fraction uses a numerator and denominator, a decimal uses place value, and a percentage is always out of 100.

    分数、小数和百分数是描述同一个概念的三种不同方式:整体的一部分。分数用分子和分母表示,小数使用位值,百分数则总是以 100 为基准。

    For example, one half can be written as ½, 0.5 or 50%. The ability to move quickly between these forms is a key skill in the Cambridge KS3 curriculum.

    例如,一半可以写成 ½、0.5 或 50%。在这些形式之间快速转换是剑桥 KS3 课程中的关键技能。

    Before practising conversions, make sure you understand that the three representations have equal value even though they look different.

    在练习转换之前,请确保你理解这三种表示虽然看起来不同,但数值相等。


    2. Equivalent Fractions | 等值分数

    An equivalent fraction is created by multiplying or dividing both the numerator and denominator by the same non-zero number. This does not change the value of the fraction.

    等值分数是通过将分子和分母同时乘以或除以同一个非零数而得到的分数。这样做不会改变分数的值。

    For instance, 1/2 = 2/4 = 4/8 because both parts of the fraction have been scaled by the same factor.

    例如,1/2 = 2/4 = 4/8,因为分数的分子和分母都按相同的倍数进行了缩放。

    In a typical exercise such as page 120 question 2, you may be asked to write a fraction in its simplest form. To simplify, divide the numerator and denominator by their highest common factor.

    在类似第 120 页第 2 题的典型练习中,你可能需要把分数化为最简形式。化简分数的方法是分子和分母同时除以它们的最大公因数。

    For example, 12/16 simplifies to 3/4 because the highest common factor of 12 and 16 is 4.

    例如,12/16 化简为 3/4,因为 12 和 16 的最大公因数是 4。


    3. Converting Fractions to Decimals | 分数转小数

    To convert a fraction to a decimal, divide the numerator by the denominator. You can use short division or a calculator if allowed.

    将分数转换为小数的方法是分子除以分母。可以使用短除法,或者在允许的情况下使用计算器。

    fraction → decimal: numerator ÷ denominator

    A fraction like 3/8 becomes 0.375 because 3 ÷ 8 = 0.375.

    像 3/8 这样的分数转换为 0.375,因为 3 ÷ 8 = 0.375。

    Some fractions produce terminating decimals, such as 1/4 = 0.25, while others produce recurring decimals, such as 1/3 = 0.333… In KS3 you need to recognise both forms.

    有些分数可以化为有限小数,例如 1/4 = 0.25;有些则化为循环小数,例如 1/3 = 0.333…… 在 KS3 中你需要识别这两种形式。

    If you see a fraction like 7/20, multiply the denominator to 100 first: 7/20 = 35/100 = 0.35. This method is often quicker when the denominator is a factor of 100.

    如果遇到 7/20 这样的分数,可以先把分母变成 100:7/20 = 35/100 = 0.35。当分母是 100 的因数时,这种方法通常更快。


    4. Converting Decimals to Percentages | 小数转百分数

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

    将小数转换为百分数的方法是小数乘以 100 并加上百分号。这相当于把小数点向右移动两位。

    decimal → percentage: × 100%

    For example, 0.35 becomes 35%, and 0.075 becomes 7.5%.

    例如,0.35 变成 35%,0.075 变成 7.5%。

    If the decimal is greater than 1, the percentage is greater than 100%. For instance, 1.25 = 125%. This is useful for increase problems.

    如果小数大于 1,百分数就大于 100%。例如,1.25 = 125%。这在增长问题中很有用。

    A common mistake is to write 0.5 as 0.50% instead of 50%. Remember that 0.5 is half of one, and half of 100 is 50.

    一个常见错误是把 0.5 写成 0.50% 而不是 50%。请记住,0.5 是一的一半,而 100 的一半是 50。


    5. Converting Percentages to Fractions | 百分数转分数

    To convert a percentage to a fraction, write the percentage over 100 and then simplify if possible.

    将百分数转换为分数的方法是先把百分数写成以 100 为分母的分数,然后在可能的情况下进行化简。

    percentage → fraction: percentage ÷ 100, then simplify

    For example, 25% = 25/100 = 1/4, and 62.5% = 62.5/100 = 625/1000 = 5/8.

    例如,25% = 25/100 = 1/4;62.5% = 62.5/100 = 625/1000 = 5/8。

    If the percentage is a mixed number such as 33⅓%, write it as an improper fraction first: 33⅓% = (100/3)/100 = 100/300 = 1/3.

    如果百分数是带分数,例如 33⅓%,先写成假分数:33⅓% = (100/3)/100 = 100/300 = 1/3。

    KS3 questions often ask you to match fractions, decimals and percentages, so memorising common conversions such as 1/4 = 25% = 0.25 will save time.

    KS3 题目经常要求匹配分数、小数和百分数,因此记住常见转换如 1/4 = 25% = 0.25 可以节省时间。


    6. Comparing and Ordering | 比较与排序

    To compare fractions, decimals and percentages, convert all values into the same form. Decimals are usually the easiest for ordering.

    要比较分数、小数和百分数,先将所有数值转换成同一种形式。小数通常最容易排序。

    <

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  • Sharing in a Ratio: Cambridge KS3 Maths | 按比例分配:剑桥 KS3 数学

    📚 Sharing in a Ratio: Cambridge KS3 Maths | 按比例分配:剑桥 KS3 数学

    Sharing quantities in a given ratio is a core skill in the Cambridge KS3 mathematics curriculum. It appears regularly in written exercises, including the practice set on page 119 question 2, and is essential for later work on proportion, percentages and real-life financial problems.

    按给定比例分配数量是剑桥 KS3 数学课程的核心技能。它经常出现在书面练习中,包括第119页第2题的练习,并且对后续的比例、百分数和实际财务问题都至关重要。


    1. What is a Ratio? | 什么是比?

    A ratio compares two or more quantities in the same unit. We write ratios using a colon, for example 3 : 2. This means for every 3 parts of the first quantity, there are 2 parts of the second quantity. The total number of equal parts is found by adding the numbers in the ratio.

    比用来比较两个或两个以上相同单位的量。我们用冒号写比,例如 3 : 2。这表示第一种量每占 3 份,第二种量就占 2 份。总份数等于比中各项数字之和。

    For the ratio 3 : 5, there are 3 + 5 = 8 equal parts in total. Understanding this idea is the foundation for all sharing problems.

    对于比 3 : 5,总共有 3 + 5 = 8 个相等的份。理解这个概念是解决所有分配问题的基础。

    • 3 : 5 means total parts = 8; 3 : 5 表示总份数 = 8
    • A ratio has no units; 比没有单位
    • The order of the ratio matters; 比的顺序很重要

    2. Reading a Ratio Question | 阅读比例题

    When a question says “Share £120 between A and B in the ratio 3 : 5”, it is asking you to split the whole amount into equal parts according to the ratio. Identify the total amount and the ratio terms carefully before you start.

    当题目说“按 3 : 5 的比例在 A 和 B 之间分配 120 英镑”时,是要求你按照该比例将总量分成相等的份。开始之前请仔细识别总量和比中的各项。

    Useful steps for reading the question:

    阅读题目的有用步骤:

    • Find the total amount to be shared; 找到要分配的总量
    • Find the ratio and check its order; 找到比并检查其顺序
    • Add the ratio terms to get total parts; 将比中各项相加得到总份数

    3. The Bar Model Method | 条形模型法

    A bar model is a visual way to represent sharing in a ratio. Draw one bar, divide it into equal sections matching the total parts, then label the known total. Each section represents one part.

    条形模型是一种按比例分配的可视化方法。画一个条形,把它分成与总份数相等的若干段,然后标出已知总量。每一段代表一份。

    For £120 shared in the ratio 3 : 5, draw a bar with 8 equal sections. The first 3 sections represent A, and the remaining 5 sections represent B. The whole bar equals £120.

    对于按 3 : 5 分配 120 英镑,画一个分成 8 个等份的条形。前 3 段代表 A,后 5 段代表 B。整个条形等于 120 英镑。

    Total parts = 3 + 5 = 8 | One part shown by one section | 总份数 = 3 + 5 = 8 | 一份用一段表示


    4. The Unit-Method | 单位法

    The unit-method uses three steps: find the total number of parts, divide the given amount by the total parts to find the value of one part, then multiply by each ratio term.

    单位法使用三个步骤:求出总份数,用已知数量除以总份数得到一份的值,再分别乘以比中的各项。

    This method works for money, lengths, masses, volumes and any other quantity that can be divided into equal parts.

    该方法适用于金钱、长度、质量、体积以及任何可以分成相等份的量。

    One part = Total amount ÷ Total parts | 一份 = 总量 ÷ 总份数

    • Step 1: Find total parts; 步骤1:求总份数
    • Step 2: Divide total amount by total parts; 步骤2:总量除以总份数
    • Step 3: Multiply one part by each ratio term; 步骤3:一份乘以比中各项

    5. Worked Example: Sharing Money | 示例:分钱问题

    Work through this example: Share £120 between A and B in the ratio 3 : 5.

    完成这个示例:按 3 : 5 的比例在 A 和 B 之间分配 120 英镑。

    • Total parts = 3 + 5 = 8; 总份数 = 3 + 5 = 8
    • One part = £120 ÷ 8 = £15; 一份 = 120 ÷ 8 = 15 英镑
    • A = 3 × £15 = £45; A = 3 × 15 = 45 英镑
    • B = 5 × £15 = £75; B = 5 × 15 = 75 英镑
    • Check: £45 + £75 = £120; 检查:45 + 75 = 120 英镑

    The answer is £45 for A and £75 for B. Always check that the amounts add back to the original total.

    答案是 A 得 45 英镑,B 得 75 英镑。一定要检查各数量相加是否等于原总量。


    6. Using Algebra to Share in a Ratio | 用代数按比例分配

    You can also set up an equation. Let one part be x. Then the amounts are 3x and 5x, and the sum equals the total.

    你也可以设方程。设一份为 x。那么数量就是 3x 和 5x,它们的和等于总量。

    3x + 5x = 120 → 8x = 120 → x = 15

    Then A = 3 × 15 = 45 and B = 5 × 15 = 75. This algebraic approach links ratio sharing to solving linear equations.

    那么 A = 3 × 15 = 45,B = 5 × 15 = 75。这种代数方法把按比例分配与解一元一次方程联系起来。


    7. Ratio and Fractions | 比与分数

    A ratio can be converted into fractions of the whole. In the ratio 3 : 5, the first part receives 3/8 of the total and the second receives 5/8 of the total.

    比可以转化为整体的分数。在 3 : 5 中,第一部分得到总量的 3/8,第二部分得到总量的 5/8。

    A = (3/8) × £120 = £45 | B = (5/8) × £120 = £75

    This is useful when a question asks for one share as a fraction of the whole, or when comparing two related ratio problems.

    当题目要求求某一份占整体的几分之几,或者比较两个相关的比例问题时,这种方法很有用。


    8. Common Mistakes | 常见错误

    Students often forget to add the ratio terms before dividing. Another mistake is using the ratio term as the actual amount. Always calculate the value of one part first.

    学生常常在除法之前忘记将比中各项相加。另一个错误是把比中的项直接当作实际数量。一定要先计算一份的值。

    • Mistake: dividing by 3 instead of by total parts 8; 错误:除以 3 而不是总份数 8
    • Mistake: leaving the answer as 3 : 5 without finding amounts; 错误:只写出 3 : 5 而没有求出实际数量
    • Mistake: mixing units, such as pounds and pence; 错误:单位混用,例如英镑和便士混用
    • Mistake: swapping the order of shares; 错误:调换份额顺序

    Careful reading and a clear layout will help you avoid these errors.

    认真读题和清晰的书写布局能帮助你避免这些错误。


    9. Exam-Style Practice | 考试型练习

    Try these quick questions to check your understanding:

    尝试以下快速练习来检查理解:

    • Share £200 in the ratio 3 : 7. 按 3 : 7 分配 200 英镑。
    • Divide 360 ml in the ratio 1 : 5. 按 1 : 5 分配 360 毫升。
    • The ratio of boys to girls in a class is 4 : 5. If there are 27 students in total, how many boys are there? 班级男女比例为 4 : 5。如果共有 27 名学生,男生有多少人?

    For the last question, total parts = 4 + 5 = 9, one part = 27 ÷ 9 = 3, so boys = 4 × 3 = 12.

    最后一题中,总份数 = 4 + 5 = 9,一份 = 27 ÷ 9 = 3,所以男生 = 4 × 3 = 12。


    10. Key Takeaways | 核心要点

    To share in a ratio, always find the total number of parts, divide to find one part, then multiply. Use a bar model if you need a visual check. Verify that the amounts add back to the original total.

    按比例分配时,总是先求总份数,除以总量得到一份,再乘以各项。如果需要直观检查,可以用条形图。验证各数量相加等于原总量。

    This skill is tested regularly at KS3 and forms the basis for harder proportion questions later in Cambridge IGCSE mathematics.

    这门技能在 KS3 阶段经常考查,也为剑桥 IGCSE 数学后续更复杂的比例问题打下基础。

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  • KS3 Cambridge Maths: Page 111 Q2 – Percentages to Fractions | KS3 剑桥数学:第111页第2题——百分数化分数

    📚 KS3 Cambridge Maths: Page 111 Q2 – Percentages to Fractions | KS3 剑桥数学:第111页第2题——百分数化分数

    In this KS3 Cambridge Mathematics revision article, we focus on a typical Page 111 Question 2 task: converting percentages into fractions in their simplest form. This topic is at the heart of the Cambridge Key Stage 3 curriculum and builds essential number skills for ratio, proportion, probability and everyday problem solving.

    在这篇 KS3 剑桥数学复习文章中,我们聚焦第 111 页第 2 题这一典型任务:把百分数化为最简分数。该主题是剑桥 Key Stage 3 课程的核心,为比、比例、概率和日常问题解决打下重要的数字技能基础。

    1. Understanding the Page 111 Q2 Task | 理解第111页第2题任务

    Question 2 usually presents a list of percentages such as 30%, 45%, 12.5% and 150%, and asks you to write each one as a fraction in its lowest terms.

    第 2 题通常会列出一组百分数,如 30%、45%、12.5% 和 150%,要求你把每个百分数写成最简分数。

    The instruction “in its simplest form” means you must reduce the fraction until the numerator and denominator have no common factor except 1.

    “最简形式”意味着你必须约分,直到分子和分母除了 1 之外没有公因数。

    You should show clear working, as Cambridge examiners award method marks for the conversion and simplification steps.

    你应当展示清晰的步骤,因为剑桥考官会为转换和约分步骤给予方法分。


    2. What Is a Percentage? | 什么是百分数?

    The word “percentage” comes from the Latin “per centum”, meaning “out of 100”. Therefore, x% simply means x out of 100, or x/100.

    “百分数(percentage)”一词来自拉丁语 “per centum”,意思是“每一百”。因此,x% 就是每一百中的 x,或者 x/100。

    This definition gives us a direct rule: to change a percentage to a fraction, write the percentage number over 100 and then simplify.

    这个定义给了我们一条直接规则:要把百分数化为分数,先把百分数的数值写在 100 的上方,然后进行约分。

    For example, 25% = 25/100, which simplifies to 1/4.

    例如,25% = 25/100,约分后得到 1/4。


    3. Step-by-Step Conversion Method | 分步转换方法

    Follow these three steps for every percentage in Question 2: first, write the percentage as a fraction with denominator 100; second, find the greatest common factor of the numerator and denominator; third, divide both by that factor.

    对于第 2 题中的每个百分数,按以下三步操作:第一,把百分数写成分母为 100 的分数;第二,找出分子和分母的最大公因数;第三,用该公因数同时除分子和分母。

    If the percentage has a decimal part, such as 12.5%, first eliminate the decimal by multiplying numerator and denominator by 10, 100 or 1000 as needed.

    如果百分数含有小数部分,例如 12.5%,

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  • Cambridge KS3 Mathematics: Fractions, Decimals and Percentages | 剑桥初中数学:分数、小数与百分比

    📚 Cambridge KS3 Mathematics: Fractions, Decimals and Percentages | 剑桥初中数学:分数、小数与百分比

    In the Cambridge Lower Secondary Mathematics course, fractions, decimals and percentages are treated as connected ways of describing parts of a whole. The exercises in resource p108_2.pdf require you to move flexibly between these three forms, compare their sizes, and use them in calculations. This guide explains each skill with clear examples so that you can answer similar questions accurately.

    在剑桥初中数学课程中,分数、小数和百分比被视为描述整体的一部分的三种相互关联的方式。资源 p108_2.pdf 中的练习要求你在三种形式之间灵活转换、比较大小并用于计算。本指南通过清晰的示例逐一讲解每项技能,帮助你准确解答类似问题。


    1. The Meaning of Fractions | 分数的意义

    A fraction is written as a numerator over a denominator. The denominator tells you how many equal parts the whole is divided into, and the numerator tells you how many of those parts are being considered.

    分数写作分子在上、分母在下。分母表示整体被平均分成多少份,分子表示你正在考虑其中的几份。

    For example, the fraction 3/5 means 3 parts out of 5 equal parts. It can also be read as the division 3 ÷ 5, which gives the decimal 0.6.

    例如,分数 3/5 表示 5 等份中的 3 份。它也可以读作除法 3 ÷ 5,结果是小数 0.6。

    A proper fraction has a numerator smaller than its denominator, such as 2/7. An improper fraction has a numerator equal to or greater than its denominator, such as 9/4, and can be written as a mixed number 2¼.

    真分数的分子小于分母,如 2/7。假分数的分子大于或等于分母,如 9/4,它可以写成带分数 2¼。


    2. Equivalent Fractions and Simplest Form | 等值分数与最简形式

    Equivalent fractions look different but represent exactly the same value. You

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  • KS3 Cambridge: Solving Linear Equations with Brackets | 解含括号的一元一次方程

    📚 KS3 Cambridge: Solving Linear Equations with Brackets | 解含括号的一元一次方程

    Linear equations with brackets are a key skill in KS3 Cambridge Mathematics. In this article, you will learn how to expand brackets, collect like terms, balance both sides, and check your solution step by step.

    含括号的一元一次方程是 KS3 剑桥数学中的核心技能。本文将带你一步步学习如何展开括号、合并同类项、平衡等式两边,并检验你的答案。

    This topic builds on your earlier work with expressions and equations. By the end, you will be able to solve equations such as 3(x – 2) = 12 and 5x + 4 = 2x + 19 confidently.

    本主题建立在你之前学习代数式和方程的基础上。学完本文后,你将能够自信地解出像 3(x – 2) = 12 和 5x + 4 = 2x + 19 这样的方程。


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

    A linear equation is an equation in which the highest power of the unknown is 1. It can be written in the general form ax + b = c, where a, b and c are numbers and a is not zero.

    一元一次方程是指未知数的最高次数为 1 的方程。它的一般形式为 ax + b = c,其中 a、b、c 是数,且 a 不等于 0。

    For example, 2x + 6 = 14 is linear because the variable x has an exponent of 1. Quadratic equations, such as x² + 3x = 10, are not covered in this topic.

    例如,2x + 6 = 14 是一元一次方程,因为变量 x 的指数是 1。二次方程(如 x² + 3x = 10)不在本主题范围内。

    Linear equations graph as straight lines, which is why they are called ‘linear’. Solving them means finding the value of x that makes the statement true.

    一元一次方程的图像是一条直线,因此被称为“线性”方程。解这类方程意味着求出使等式成立的 x 的值。


    2. The Distributive Law and Brackets | 分配律与括号

    Before you can solve an equation with brackets, you must understand the distributive law. This law states that multiplying a number by a sum is the same as multiplying each addend separately and then adding the results.

    在解含括号的方程之前,你必须先理解分配律。分配律指出,用一个数乘以一个和,等于分别乘以每个加数后再相加。

    a(b + c) = ab + ac

    In algebra, this means that 2(x + 3) is the same as 2 × x + 2 × 3, which simplifies to 2x + 6. You must multiply every term inside the bracket by the number outside.

    在代数中,这意味着 2(x + 3) 等同于 2 × x + 2 × 3,化简后得到 2x + 6。你必须用括号外的数乘以括号内的每一项。

    This step is essential because the equation cannot be solved correctly while the unknown is still trapped inside the brackets.

    这一步至关重要,因为当未知数还在括号内时,方程无法被正确求解。


    3. Expanding Brackets with Positive Numbers | 正数乘以括号的展开

    When the number outside the bracket is positive, the signs of the terms inside the bracket stay the same after expansion. For example, 4(x + 5) becomes 4x + 20.

    当括号外的数是正数时,展开后括号内各项的符号保持不变。例如,4(x + 5) 展开为 4x + 20。

    Consider the equation 3(x + 4) = 27. First, expand the left side to get 3x + 12 = 27. Then subtract 12 from both sides to get 3x = 15, and divide by 3 to find x = 5.

    以方程 3(x + 4) = 27 为例。首先展开左边,得到 3x + 12 = 27。然后两边同时减去 12,得到 3x = 15,再除以 3,求得 x = 5。

    Always show the expansion as a separate line in your working. This makes it easier to spot arithmetic mistakes and helps the examiner follow your method.

    解题时一定要把展开写成单独的一行。这样可以更容易发现计算错误,也有助于阅卷人理解你的解题方法。


    4. Expanding Brackets with Negative Numbers | 负数乘以括号的展开

    A negative number outside the bracket changes the sign of every term inside. For instance, -3(x – 2) becomes -3x + 6 because -3 × -2 = +6.

    括号外是负数时,会改变括号内每一项的符号。例如,-3(x – 2) 展开为 -3x + 6,因为 -3 × -2 = +6。

    Take the equation -2(x + 5) = 14. Expanding gives -2x – 10 = 14. Add 10 to both sides: -2x = 24. Finally, divide both sides by -2 to obtain x = -12.

    以方程 -2(x + 5) = 14 为例。展开得到 -2x – 10 = 14。两边同时加 10,得 -2x = 24。最后两边同时除以 -2,得到 x = -12。

    Be especially careful with double negatives. If you see -5(x – 6), the correct expansion is -5x + 30, not -5x – 30.

    要特别注意双重负号。如果你看到 -5(x – 6),正确的展开是 -5x + 30,而不是 -5x – 30。


    5. Collecting Like Terms | 合并同类项

    After expanding brackets, the equation may have several x terms and several number terms. Collect the like terms on each side before you begin balancing the equation.

    展开括号后,方程中可能有多个含 x 的项和多个数字项。在开始平衡等式之前,要先把两边的同类项分别合并。

    For example, 4x + 5 + 3x – 2 simplifies to 7x + 3. Like terms have exactly the same variable part, so 4x and 3x can be combined, but 4x and 4 cannot.

    例如,4x + 5 + 3x – 2 可以简化为 7x + 3。同类项必须含有完全相同的字母部分,所以 4x 和 3x 可以合并,但 4x 和 4 不能合并。

    Suppose you have 2(x + 3) + 5x = 35. Expanding gives 2x + 6 + 5x = 35. Collect like terms to get 7x + 6 = 35, then solve to find x = 29/7, which is approximately 4.14.

    假设你有 2(x + 3) + 5x = 35。展开得到 2x + 6 + 5x = 35。合并同类项得到 7x + 6 = 35,然后解方程,得到 x = 29/7,约等于 4.14。


    6. Using Inverse Operations | 使用逆运算

    Solving a linear equation involves using inverse operations to isolate x. Addition and subtraction are inverse operations, and so are multiplication and division.

    解一元一次方程需要用逆运算来分离出 x。加法和减法是互逆运算,乘法和除法也是互逆运算。

    For the equation 2x + 6 = 14, the order of operations is: multiply x by 2, then add 6. To undo this, subtract 6 first, then divide by 2.

    对于方程 2x + 6 = 14,运算顺序是:先把 x 乘以 2,再加 6。要逆推这个过程,就要先减 6,再除以 2。

    2x + 6 = 14 → 2x = 8 → x = 4

    Always apply the inverse operation to both sides of the equation. This keeps the equation balanced, just like keeping both sides of a scale level.

    一定要对等式两边同时进行逆运算。这能保持等式平衡,就像让天平的两端保持水平一样。


    7. Unknown on Both Sides | 未知数在等号两边

    When the unknown appears on both sides of the equation, the first goal is to gather all x terms on one side and all number terms on the other side. Choose to move the smaller x term to avoid negative coefficients.

    当未知数出现在等号两边时,第一个目标是把所有含 x 的项移到一边,把所有数字项移到另一边。最好移动较小的 x 项,以避免出现负系数。

    For 5x – 3 = 2x + 9, subtract 2x from both sides to get 3x – 3 = 9. Then add 3 to both sides to get 3x = 12. Dividing by 3 gives x = 4.

    对于 5x – 3 = 2x + 9,两边同时减去 2x,得到 3x – 3 = 9。然后两边同时加 3,得到 3x = 12。除以 3,得到 x = 4。

    If you accidentally move the larger x term instead, you will still get the same answer, but your working may include more negative signs.

    如果你不小心移动了较大的 x 项,最终答案仍然相同,但计算过程可能会出现更多负号。


    8. Equations with Fractions | 含分数的方程

    Fractions in equations can be cleared by multiplying

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  • Mastering Algebraic Expressions and Linear Equations: KS3 Cambridge p104_2 Skills | 掌握代数式与线性方程:KS3 剑桥 p104_2 核心技能

    📚 Mastering Algebraic Expressions and Linear Equations: KS3 Cambridge p104_2 Skills | 掌握代数式与线性方程:KS3 剑桥 p104_2 核心技能

    In many KS3 Cambridge mathematics resources, page 104 question 2 tests your ability to simplify algebraic expressions, expand brackets and solve linear equations. These skills are central to algebra and appear throughout the Cambridge Lower Secondary progression. This revision guide consolidates the key methods, worked examples and common mistakes so you can approach p104_2-style questions with confidence.

    在许多 KS3 剑桥数学教材中,第 104 页第 2 题考查的是化简代数式、展开括号以及解一元一次方程的能力。这些技能是代数的核心,贯穿剑桥初中数学的整个学习过程。本复习指南总结了关键方法、典型例题和常见错误,帮助你自信应对 p104_2 类型的题目。

    1. Understanding Variables, Terms and Coefficients | 理解变量、项与系数

    In algebra, letters such as x, y or n stand for unknown numbers or quantities that can change. These letters are called variables. An algebraic expression combines numbers, variables and operations. In the expression 5x + 3y – 7, the terms are 5x, 3y and -7; the coefficients are 5 and 3; and -7 is the constant term.

    在代数中,x、y 或 n 等字母表示未知数或可以变化的量,这些字母称为变量。代数式由数字、变量和运算组合而成。在表达式 5x + 3y – 7 中,项分别是 5x、3y 和 -7;系数是 5 和 3;-7 是常数项。

    A term is a single number, a variable, or a product of numbers and variables. For example, 4x, -2y, 9 and x are all terms. Identifying terms correctly helps you decide which parts can be combined when simplifying an expression.

    项是一个单独的数字、变量或数字与变量的乘积。例如 4x、-2y、9 和 x 都是项。正确识别项有助于你在化简表达式时判断哪些部分可以合并。

    • Term: a single number, variable or product (项:单个数字、变量或乘积)
    • Coefficient: the number factor in a term (系数:项中的数字因数)
    • Constant: a term with no variable (常数项:不含变量的项)

    2. Collecting Like Terms | 合并同类项

    Like terms contain exactly the same variable part, such as 3x and 7x, or 2y and -5y. You can add or subtract like terms by combining their coefficients. Unlike terms, such as 3x and 4y, cannot be combined into a single term.

    同类项含有完全相同的变量部分,例如 3x 和 7x,或 2y 和 -5y。你可以通过合并系数来加减同类项。不同类项,例如 3x 和 4y,不能合并为一个项。

    For example, simplify 3x + 5x – 2x by adding the coefficients: 3 + 5 – 2 = 6, so the result is 6x. In a longer expression, group like terms first, then simplify each group.

    例如,化简 3x + 5x – 2x 时,先把系数相加:3 + 5 – 2 = 6,因此结果是 6x。在较长的表达式中,先分组同类项,再分别化简每一组。

    4a + 3b – 2a + 5b = (4a – 2a) + (3b + 5b) = 2a + 8b

    Always keep the sign in front of each term with the term itself. This prevents mistakes when rearranging terms.

    始终把每一项前面的符号与该连在一起。这样在重新排列各项时可以避免错误。


    3. Expanding Brackets: The Distributive Law | 展开括号:分配律

    Expanding brackets means multiplying the term outside the bracket by every term inside the bracket. This uses the distributive law: a(b + c) = ab + ac. For example, 3(x + 4) expands to 3x + 12 because 3 x x = 3x and 3 x 4 = 12.

    展开括号就是用括号外面的项乘以括号内的每一项。这使用了分配律:a(b + c) = ab + ac。例如,3(x + 4) 展开后得到 3x + 12,因为 3 x x = 3x,3 x 4 = 12。

    When the bracket is multiplied by a negative term, be careful with signs: -2(x – 5) = -2x + 10 because -2 x x = -2x and -2 x -5 = +10.

    当括号前是负数时,需要特别注意符号:-2(x – 5) = -2x + 10,因为 -2 x x = -2x,而 -2 x -5 = +10。

    2(3y – 4) + 5 = 6y – 8 + 5 = 6y – 3

    After expanding, collect any like terms to simplify the expression fully. This is often required in KS3 Cambridge p104-style questions.

    展开后,合并所有同类项以完全化简表达式。这通常是 KS3 剑桥 p104 类型题目中要求的步骤。


    4. Factorising Expressions | 因式分解

    Factorising is the reverse of expanding. You rewrite an expression as a product of a common factor and a bracket. Find the highest common factor of the terms, write it outside the bracket, and divide each term by that factor inside the bracket.

    因式分解是展开的逆运算。你将一个表达式改写为一个公因式与括号的乘积。先找出各项的最高公因式,把它写在括号外,再将括号内每一项除以该公因式。

    For example, factorise 6x + 9. The highest common factor of 6 and 9 is 3, so 6x + 9 = 3(2x + 3). You can check by expanding: 3(2x + 3) = 6x + 9.

    例如,对 6x + 9 进行因式分解。6 和 9 的最高公因数是 3,所以 6x + 9 = 3(2x + 3)。你可以通过展开来检查:3(2x + 3) = 6x + 9。

    8x² + 4x = 4x(2x + 1)

    Here both terms share 4x as the highest common factor because 8x² ÷ 4x = 2x and 4x ÷ 4x = 1. Factorising is important for solving more complex equations later.

    这里两项的最高公因式是 4x,因为 8x² ÷ 4x = 2x,4x ÷ 4x = 1。因式分解对以后解决更复杂的方程非常重要。


    5. Solving One-Step Equations | 解一步方程

    A linear equation contains a variable raised only to the power of 1. To solve a one-step equation, perform the inverse operation on both sides of the equation to isolate the variable. This keeps the equation balanced.

    一元一次方程中的变量指数仅为 1。解一步方程时,在方程两边执行逆运算以分离变量。这可以保持方程平衡。

    For example, solve x + 7 = 15. Subtract 7 from both sides: x + 7 – 7 = 15 – 7, so x = 8. For the equation 3x = 21, divide both sides by 3: x = 21 ÷ 3 = 7.

    例如,解 x + 7 = 15。两边同时减去 7:x + 7 – 7 = 15 – 7,所以 x = 8。对于方程 3x = 21,两边同时除以 3:x = 21 ÷ 3 = 7。

    x – 4 = 10 → x = 10 + 4 = 14

    Always write each step clearly and keep the equals signs aligned. This helps you avoid arithmetic slips and makes your method easy to check in an exam.

    始终清晰地书写每一步,并保持等号对齐。这有助于避免计算错误,也便于考试时检查解题过程。


    6. Solving Two-Step Equations | 解两步方程

    Two-step equations involve two operations on the variable, such as multiplication and addition. Undo the operations in reverse order: first solve any addition or subtraction, then solve any multiplication or division.

    两步方程中的变量涉及两种运算,例如乘法和加法。按相反顺序逐步解除运算:先处理加法或减法,再处理乘法或除法。

    Solve 2x + 5 = 17. Step 1: subtract 5 from both sides to get 2x = 12. Step 2: divide both sides by 2 to get x = 6. Always check by substituting the value back into the original equation.

    解方程 2x + 5 = 17。第一步:两边减去 5,得到 2x = 12。第二步:两边除以 2,得到 x = 6。始终将值代回原方程进行检查。

    x/3 – 2 = 4 → x/3 = 6 → x = 18

    For equations with a negative variable, multiply or divide by -1 as the final step. For example, if -x = 7, then x = -7.

    对于变量为负数的方程,最后一步乘以或除以 -1。例如,如果 -x = 7,那么 x = -7。


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

    When an equation contains brackets, expand them first, then simplify both sides before isolating the variable. This often appears in p104_2-style questions where you must combine expanding and solving skills.

    当方程含有括号时,首先展开括号,然后化简两边,再分离变量。这类题目经常出现在 p104_2 类型的练习中,需要综合运用展开和求解技能。

    Solve 3(x + 4) = 27. Expand the left side: 3x + 12 = 27. Subtract 12 from both sides: 3x = 15. Divide by 3: x = 5.

    解方程 3(x + 4) = 27。展开左边:3x + 12 = 27。两边减去 12:3x = 15。除以 3:x = 5。

    2(x – 5) = x + 3 → 2x – 10 = x + 3 → x = 13

    After expanding, collect like terms on each side. If the variable appears on both sides, move the smaller variable term first to keep the coefficient positive.

    展开后,先合并各自边的同类项。如果两边都含有变量,先移走较小的变量项,以使系数保持为正。


    8. Writing Equations from Word Problems | 根据应用题建立方程

    Word problems require you to translate a real-life situation into an algebraic equation. Start by defining the unknown with a variable, then build an equation using the given relationships and operations.

    应用题要求你将实际情境转化为代数方程。先设未知数为变量,再利用题目给出的关系和运算建立方程。

    Example: Sam has x sweets. Emma gives him 6 more, and he now has 20. Write and solve the equation. Translation: x + 6 = 20, so x = 14.

    例如:Sam 有 x 颗糖。Emma 又给了他 6 颗,现在他有 20 颗。写出并求解方程。转化:x + 6 = 20,因此 x = 14。

    For perimeter problems, use the formula and substitute known values. If a rectangle has length 8 cm and width w cm, and perimeter 26 cm, then 2(8 + w) = 26. Solving gives w = 5 cm.

    对于周长问题,使用公式并代入已知值。如果一个矩形的长为 8 cm,宽为 w cm,周长为 26 cm,则 2(8 + w) = 26。解得 w = 5 cm。

    Always define the variable clearly and answer the original question. Include units if the problem involves measurements.

    始终明确定义变量并回答原题所问。如果题目涉及测量,答案中要包含单位。


    9. Common Errors and How to Avoid Them | 常见错误与如何避免

    One frequent mistake is forgetting to multiply all terms inside a bracket. For example, 2(x + 3) is sometimes written incorrectly as 2x + 3 instead of 2x + 6. Check that every term inside the bracket has been multiplied by the outside term.

    一个常见错误是忘记乘以括号内的所有项。例如,2(x + 3) 有时被错误地写成 2x + 3,而正确答案是 2x + 6。请检查括号内每一项是否都乘以外面的项。

    Another error is losing the negative sign when collecting like terms. The expression 5x – 2x – 3x should be simplified as (5 – 2 – 3)x = 0x = 0, not 5x – 5x = 0 incorrectly skipping the signs. Keep every sign attached to its term.

    另一个错误是合并同类项时丢失负号。表达式 5x – 2x – 3x 应化简为 (5 – 2 – 3)x = 0x = 0,而不是错误地忽略符号。要把每个符号与对应项连在一起。

    In equations, avoid applying operations to only one side. Whatever you do to one side, you must do to the other. For example, if you subtract 4 from the left, subtract 4 from the right as well.

    在方程中,避免只对一边进行运算。对一边做什么,对另一边也必须做同样的运算。例如,如果左边减去 4,右边也要减去 4。

    Common error Correct method
    3(y – 2) = 3y – 2 3(y – 2) = 3y – 6
    -2(x + 4) = -2x + 8 -2(x + 4) = -2x – 8
    x + 5 = 12 → x = 17 x + 5 = 12 → x = 7

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

    Try these three questions that mirror the structure of KS3 Cambridge p104_2 tasks. Complete all steps, then compare your answers with the solutions below.

    尝试以下三道与 KS3 剑桥 p104_2 题型结构相似的练习题。写出完整步骤,然后与下面的答案对照。

    Question 1: Simplify 7x – 3y + 2x + 5y – 4x.

    问题 1:化简 7x – 3y + 2x + 5y – 4x。

    Question 2: Solve 5(n – 3) + 2 = 27.

    问题 2:解方程 5(n – 3) + 2 = 27。

    Question 3: A rectangle has length (x + 6) cm and width 4 cm. Its perimeter is 32 cm. Form an equation and find x.

    问题 3:一个矩形的长为 (x + 6) cm,宽为 4 cm,周长为 32 cm。建立方程并求 x。

    Solutions:

    答案:

    1. 7x – 3y + 2x + 5y – 4x = (7x + 2x – 4x) + (-3y + 5y) = 5x + 2y

    2. 5(n – 3) + 2 = 27 → 5n – 15 + 2 = 27 → 5n – 13 = 27 → 5n = 40 → n = 8

    3. 2((x + 6) + 4) = 32 → 2(x + 10) = 32 → x + 10 = 16 → x = 6

    Regular practice of expanding, factorising and solving equations will build speed and accuracy for Cambridge KS3 assessments. Review each step and always check your final answer by substitution.

    经常练习展开、因式分解和解方程,有助于提高剑桥 KS3 考试的解题速度和准确率。复习每一步,并始终通过代入检验最终答案。


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  • KS3 Algebra Essentials: Simplifying Expressions and Solving Linear Equations | KS3 代数核心:化简表达式与解一元一次方程

    📚 KS3 Algebra Essentials: Simplifying Expressions and Solving Linear Equations | KS3 代数核心:化简表达式与解一元一次方程

    Algebra is one of the most important building blocks in KS3 Mathematics. In the Cambridge Lower Secondary programme, students are expected to move from simple number patterns to working confidently with letters, symbols, equations and formulae. This article focuses on simplifying algebraic expressions and solving linear equations. These skills appear throughout the KS3 curriculum and are essential for later topics such as graphs, sequences, and real-life problem solving.

    代数是 KS3 数学中最重要的基础模块之一。在剑桥初中课程中,学生需要从简单的数字规律过渡到熟练运用字母、符号、方程和公式。本文重点讲解代数表达式的化简以及一元一次方程的解法。这些技能贯穿 KS3 课程,也是后续学习图像、数列和实际问题解决的重要基础。


    1. What is an algebraic expression? | 什么是代数表达式?

    An algebraic expression is a combination of numbers, variables, and operation symbols. It does not contain an equals sign. For example, 3x + 5, 2a – 7b, and 4y² + 1 are all algebraic expressions. A variable is a letter that stands for an unknown value, and a coefficient is the number in front of a variable. In the term 3x, the coefficient is 3 and the variable is x.

    代数表达式是由数字、变量和运算符号组成的式子,它不包含等号。例如 3x + 5、2a – 7b 和 4y² + 1 都是代数表达式。变量是表示未知值的字母,系数是变量前面的数字。在项 3x 中,系数是 3,变量是 x。

    Expressions can have one term, two terms, or many terms. A term is a single number, a single variable, or a product of numbers and variables. For example, 5, x, and 2xy are each one term. When terms are separated by plus or minus signs, we can see the structure of an expression clearly.

    表达式可以有一项、两项或多项。项是单个数字、单个变量,或者数字与变量的乘积。例如 5、x 和 2xy 各是一项。当项由加号或减号分隔时,我们可以清楚地看到表达式的结构。


    2. Collecting like terms | 合并同类项

    Like terms are terms that have exactly the same variable or variables raised to the same power. For example, 3x and 5x are like terms because they both contain x to the power of 1. However, 3x and 3x² are not like terms because the powers of x are different. Numbers without variables, such as 4 and -7, are also like terms with each other.

    同类项是指含有完全相同变量且变量次数相同的项。例如 3x 和 5x 是同类项,因为它们都含有一次方的 x。但是 3x 和 3x² 不是同类项,因为 x 的次数不同。没有变量的数字,如 4 和 -7,彼此也是同类项。

    To collect like terms, add or subtract the coefficients while keeping the variable part unchanged. For example, 3x + 2x = 5x and 8y – 3y = 5y. The expression 4a + 3b + 2a – b can be simplified to 6a + 2b because 4a + 2a = 6a and 3b – b = 2b.

    合并同类项时,将系数相加或相减,同时保持变量部分不变。例如 3x + 2x = 5x,8y – 3y = 5y。表达式 4a + 3b + 2a – b 可以化简为 6a + 2b,因为 4a + 2a = 6a,3b – b = 2b。

    5x + 4y – 2x + 3y = 3x + 7y

    Always remember that you cannot combine unlike terms. For instance, 2x + 3y cannot be simplified further because x and y are different variables. Writing 2x + 3y = 5xy is a serious error and must be avoided.

    始终记住不能合并不同类项。例如 2x + 3y 无法进一步化简,因为 x 和 y 是不同的变量。把 2x + 3y 写成 5xy 是错误的,必须避免。


    3. Expanding brackets | 去括号展开

    Expanding brackets means removing the brackets by multiplying each term inside the bracket by the term outside. This uses the distributive law: a(b + c) = ab + ac. For example, 3(x + 4) expands to 3x + 12 because 3 × x = 3x and 3 × 4 = 12.

    去括号展开是指用括号外的项乘以括号内的每一项,从而去掉括号。这运用了分配律:a(b + c) = ab + ac。例如 3(x + 4) 展开后得到 3x + 12,因为 3 × x = 3x,3 × 4 = 12。

    If there is a negative sign outside the bracket, it must be multiplied by every term inside. For example, -2(x – 5) = -2x + 10 because -2 × x = -2x and -2 × (-5) = +10. A common mistake is to write -2x – 10, but the second sign should be positive because a negative times a negative gives a positive.

    如果括号外有负号,则必须乘以括号内的每一项。例如 -2(x – 5) = -2x + 10,因为 -2 × x = -2x,-2 × (-5) = +10。常见错误是写成 -2x – 10,但第二项的符号应为正,因为负数乘以负数得正数。

    4(2x – 3) = 8x – 12

    When two brackets are multiplied together, such as (x + 2)(x + 3), expand one bracket at a time. First multiply x by both terms in the second bracket, then multiply 2 by both terms: x × x = x², x × 3 = 3x, 2 × x = 2x, and 2 × 3 = 6. Combining like terms gives x² + 5x + 6.

    当两个括号相乘时,例如 (x + 2)(x + 3),每次展开一个括号。先用 x 乘以第二个括号中的两项,再用 2 乘以这两项:x × x = x²,x × 3 = 3x,2 × x = 2x,2 × 3 = 6。合并同类项得到 x² + 5x + 6。


    4. Factorising simple expressions | 因式分解简单表达式

    Factorising is the reverse of expanding brackets. It means writing an expression as a product of factors. To factorise an expression, look for the highest common factor of all the terms. For example, in the expression 6x + 9, the highest common factor of 6 and 9 is 3. Therefore, 6x + 9 = 3(2x + 3).

    因式分解是去括号展开的逆运算。它意味着把一个表达式写成因式的乘积。要对表达式进行因式分解,需要找出所有项的最高公因式。例如在表达式 6x + 9 中,6 和 9 的最高公因数是 3。因此 6x + 9 = 3(2x + 3)。

    For an expression like 8xy + 4x, the common factor is 4x because both terms contain x and the largest number that divides 8 and 4 is 4. So 8xy + 4x = 4x(2y + 1). Always check your answer by expanding the brackets again.

    对于 8xy + 4x 这样的表达式,公因式是 4x,因为两项都含有 x,且能整除 8 和 4 的最大数是 4。所以 8xy + 4x = 4x(2y + 1)。始终通过再次展开括号来检查答案。

    3x² + 6x = 3x(x + 2)

    Factorising is useful in solving equations, simplifying fractions, and understanding the structure of expressions. In KS3, you are usually expected to factorise by taking out a single common factor. More complex quadratic factorising is covered in later stages.

    因式分解在解方程、化简分式以及理解表达式结构时非常有用。在 KS3 阶段,通常只要求提取单个公因式。更复杂的二次因式分解会在后续阶段学习。


    5. What is a linear equation? | 什么是一元一次方程?

    An equation is a mathematical statement that two expressions are equal. It contains an equals sign. A linear equation is an equation where the highest power of the variable is 1. For example, x + 5 = 12, 3x – 2 = 10, and 4(x – 1) = 8 are linear equations. The word ‘linear’ comes from the fact that the graph of such an equation is a straight line.

    方程是表示两个表达式相等的数学语句,它包含等号。一元一次方程是指未知数的最高次数为 1 的方程。例如 x + 5 = 12、3x – 2 = 10 和 4(x – 1) = 8 都是一元一次方程。“一次”一词源于这类方程的图像是一条直线。

    Solving an equation means finding the value of the variable that makes the equation true. Think of an equation as a balance scale: the left side and the right side must remain equal. Whatever operation you do to one side, you must do the same to the other side. This is called the balance method.

    解方程意味着求出使方程成立的未知数的值。可以把方程看作一架天平:左边和右边必须保持相等。无论对一边进行什么运算,都必须对另一边进行相同的运算。这叫做平衡法。

    x + 5 = 12  ⇒  x = 7


    6. Solving equations with one operation | 解一步方程

    Some equations can be solved in one step. If the equation is x + 3 = 10, subtract 3 from both sides to get x = 7. If the equation is x – 4 = 9, add 4 to both sides to get x = 13. The goal is to isolate the variable on one side of the equation.

    有些方程只需一步就能解出。如果方程是 x + 3 = 10,两边同时减去 3,得到 x = 7。如果方程是 x – 4 = 9,两边同时加上 4,得到 x = 13。目标是使未知数单独出现在方程的一边。

    For multiplication, if 5x = 30, divide both sides by 5 to get x = 6. For division, if x / 4 = 7, multiply both sides by 4 to get x = 28. Always write each step clearly and keep the equals signs aligned.

    对于乘法,如果 5x = 30,两边同时除以 5,得到 x = 6。对于除法,如果 x / 4 = 7,两边同时乘以 4,得到 x = 28。始终清晰地写出每一步,并保持等号对齐。

    5x = 30  ⇒  x = 6

    A quick check can be done by substituting your answer back into the original equation. For example, 5 × 6 = 30, which is correct. This habit helps catch careless mistakes in tests and homework.

    可以通过将答案代回原方程来快速检验。例如 5 × 6 = 30,这是正确的。这个习惯有助于在考试和作业中发现粗心错误。


    7. Solving equations with two or more operations | 解两步及多步方程

    Many linear equations require more than one operation to solve. For example, to solve 2x + 5 = 13, first subtract 5 from both sides: 2x = 8. Then divide both sides by 2: x = 4. Always reverse the order of operations: deal with addition or subtraction before multiplication or division.

    许多一元一次方程需要不止一步运算才能解出。例如解 2x + 5 = 13,首先两边同时减去 5:2x = 8。然后两边同时除以 2:x = 4。总是按照与运算顺序相反的方向操作:先处理加减,再处理乘除。

    If the equation is x/3 – 2 = 7, first add 2 to both sides to get x/3 = 9. Then multiply both sides by 3 to get x = 27. This systematic approach makes even longer equations manageable.

    如果方程是 x/3 – 2 = 7,首先两边同时加上 2,得到 x/3 = 9。然后两边同时乘以 3,得到 x = 27。这种系统化的方法能使更长的方程也变得容易处理。

    2x + 5 = 13  ⇒  2x = 8  ⇒  x = 4

    When writing your solution, show each step on a new line. This makes it easier for you and your teacher to follow your reasoning. In Cambridge KS3 tests, method marks are often awarded even if the final answer is wrong, as long as the steps are logical.

    写解题过程时,每一步另起一行。这能让你和老师更容易理解你的推理。在剑桥 KS3 测试中,即使最终答案错误,只要步骤合乎逻辑,通常也能得到过程分。


    8. Equations with unknowns on both sides | 未知数在方程两边

    Some equations have the variable on both sides of the equals sign, for example 5x – 3 = 2x + 9. The first step is to collect all variable terms on one side and all constant terms on the other. Subtract 2x from both sides: 3x – 3 = 9. Then add 3 to both sides: 3x = 12. Finally divide both sides by 3: x = 4.

    有些方程的等号两边都含有未知数,例如 5x – 3 = 2x + 9。第一步是让所有含未知数的项在一边,所有常数项在另一边。两边同时减去 2x:3x – 3 = 9。然后两边同时加上 3:3x = 12。最后两边同时除以 3:x = 4。

    If the variable term has a negative coefficient, such as 7 – x = 2x + 1, add x to both sides to make the coefficient positive: 7 = 3x + 1. Then subtract 1 from both sides: 6 = 3x. Divide both sides by 3: x = 2.

    如果变量项的系数为负,例如 7 – x = 2x + 1,两边同时加上 x,使系数变为正数:7 = 3x + 1。然后两边同时减去 1:6 = 3x。两边同时除以 3:x = 2。

    5x – 3 = 2x + 9  ⇒  3x = 12  ⇒  x = 4

    Always check by substituting the value back into both sides. For x = 4, the left side is 5 × 4 – 3 = 17, and the right side is 2 × 4 + 9 = 17. Both sides are equal, so the solution is correct.

    始终通过将数值代回两边来检验。当 x = 4 时,左边为 5 × 4 – 3 = 17,右边为 2 × 4 + 9 = 17。两边相等,所以解是正确的。


    9. Equations involving brackets | 含括号的方程

    When an equation includes brackets, the first step is usually to expand the brackets using the distributive law. For example, to solve 3(x + 2) = 18, expand to get 3x + 6 = 18. Then subtract 6 from both sides: 3x = 12. Finally divide both sides by 3: x = 4.

    当方程中含有括号时,第一步通常是运用分配律展开括号。例如解 3(x + 2) = 18,展开得到 3x + 6 = 18。然后两边同时减去 6:3x = 12。最后两边同时除以 3:x = 4。

    If the equation is 2(3x – 1) = 4x + 8, expand the left side first: 6x – 2 = 4x + 8. Then subtract 4x from both sides: 2x – 2 = 8. Add 2 to both sides: 2x = 10. Divide both sides by 2: x = 5. Check: left side is 2(3 × 5 – 1) = 28, right side is 4 × 5 + 8 = 28.

    如果方程是 2(3x – 1) = 4x + 8,先展开左边:6x – 2 = 4x + 8。然后两边同时减去 4x:2x – 2 = 8。两边同时加上 2:2x = 10。两边同时除以 2:x = 5。检验:左边为 2(3 × 5 – 1) = 28,右边为 4 × 5 + 8 = 28。

    3(x + 2) = 18  ⇒  3x + 6 = 18  ⇒  x = 4


    10. Using equations in real-life problems | 用方程解决实际问题

    Linear equations often appear in word problems. To solve a word problem, follow these steps: read the question carefully, define a variable for the unknown, write an equation, solve the equation, and check that the answer makes sense in the context.

    一元一次方程经常出现在文字题中。解文字题的步骤如下:仔细读题,为未知数定义一个变量,列出方程,解方程,并检查答案在情境中是否合理。

    For example, a rectangle has a length that is 5 cm more than its width. The perimeter is 34 cm. Let the width be w cm. Then the length is (w + 5) cm. The perimeter equation is 2w + 2(w + 5) = 34. Expand: 2w + 2w + 10 = 34, so 4w + 10 = 34. Subtract 10: 4w = 24. Divide by 4: w = 6. Thus the width is 6 cm and the length is 11 cm.

    例如一个矩形的长比宽多 5 厘米。周长为 34 厘米。设宽为 w 厘米,则长为 (w + 5) 厘米。周长方程为 2w + 2(w + 5) = 34。展开:2w + 2w + 10 = 34,所以 4w + 10 = 34。减去 10:4w = 24。除以 4:w = 6。因此宽为 6 厘米,长为 11 厘米。

    Another common type is age or money problems. For example, if Ali has three times as much money as Ben, and together they have £48, then let Ben have £x. Ali has £3x. The equation is x + 3x = 48, so 4x = 48 and x = 12. Ben has £12 and Ali has £36.

    另一种常见类型是年龄或金钱问题。例如 Ali 的钱是 Ben 的三倍,他们一共有 48 英镑,设 Ben 有 £x,Ali 有 £3x。方程为 x + 3x = 48,所以 4x = 48,x = 12。Ben 有 12 英镑,Ali 有

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  • Cambridge KS3 Algebra: Solving Linear Equations with Brackets | 剑桥 KS3 代数:解含括号的一元一次方程

    📚 Cambridge KS3 Algebra: Solving Linear Equations with Brackets | 剑桥 KS3 代数:解含括号的一元一次方程

    Linear equations with brackets are a core algebra skill in the Cambridge Lower Secondary Mathematics course. You must be able to expand, simplify and solve equations such as 3(x + 2) = 21 or 2(3x − 4) = x + 5. This article guides you through the key rules, worked examples and common pitfalls.

    含括号的一元一次方程是剑桥初中数学(Cambridge Lower Secondary Mathematics)中的核心代数技能。你需要熟练掌握展开、化简并求解如 3(x + 2) = 21 或 2(3x − 4) = x + 5 这类方程。本文将带你梳理关键规则、典型例题和常见易错点。


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

    A linear equation is an algebraic statement where the highest power of the unknown is 1. It can include brackets, fractions or variables on both sides, but after simplifying it must take the form ax + b = c.

    线性方程是指未知数的最高次数为 1 的代数等式。它可以带有括号、分数或等号两边都有未知数,但化简后必须能写成 ax + b = c 的形式。

    ax + b = c, 其中 a ≠ 0


    2. Balancing the Equation | 方程平衡原则

    Equations are like a balance scale: whatever you do to one side, you must do to the other. If you add, subtract, multiply or divide one side by a number, you must do exactly the same to the opposite side to keep equality true.

    方程就像一架天平:你对等号一边做的任何操作,都必须对另一边做同样的操作。无论是加、减、乘还是除以一个数,都必须在两边同时进行,等式才能保持成立。

    Solve x + 5 = 12. Subtract 5 from both sides: x + 5 − 5 = 12 − 5, so x = 7.

    例如解 x + 5 = 12。两边同时减去 5:x + 5 − 5 = 12 − 5,因此 x = 7。


    3. Removing Brackets with the Distributive Law | 运用分配律去括号

    To remove a bracket, multiply the term outside by every term inside the bracket. This is the distributive law: a(b + c) = ab + ac. Be careful to multiply the sign as well as the number.

    去括号时,要用括号外的项分别乘以括号内的每一项。这就是分配律:a(b + c) = ab + ac。注意计算时既要乘数字,也要保留符号。

    3(x + 2) = 3 × x + 3 × 2 = 3x + 6

    For a negative multiplier, write the negative sign with the multiplier: −2(x − 4) = −2x + 8, because −2 × (−4) = +8.

    如果括号外的系数是负数,要带着负号一起去乘:−2(x − 4) = −2x + 8,因为 −2 × (−4) = +8。


    4. Collecting Like Terms First | 先合并同类项

    Before solving, simplify both sides as much as possible. Like terms are terms that have exactly the same variable part, such as 5x and −3x. Numbers without variables are also like terms with each other.

    求解之前,要先把方程左右两边尽量化简。同类项是指含有相同变量部分的项,例如 5x 和 −3x;不含变量的常数项之间也是同类项。

    Simplify 4x + 3 + 2x − 5. Combine 4x + 2x = 6x and 3 − 5 = −2, giving 6x − 2.

    例如化简 4x + 3 + 2x − 5。先合并 4x + 2x = 6x,再合并 3 − 5 = −2,得到 6x − 2。


    5. Unknowns on Both Sides | 未知数在等号两边

    If the variable appears on both sides of the equation, collect all variable terms on one side and constants on the other. Choose the side where the variable term will have a positive coefficient to reduce sign mistakes.

    如果未知数同时出现在等号两边,需要把含未知数的项移到一边,把常数项移到另一边。尽量选择让未知数系数为正的那一边,这样可以减少符号错误。

    Solve 5x + 2 = 2x + 11. Subtract 2x from both sides: 3x + 2 = 11. Subtract 2: 3x = 9. Divide by 3: x = 3.

    例如解 5x + 2 = 2x + 11。两边同时减去 2x:3x + 2 = 11。两边再减去 2:3x = 9。两边除以 3:x = 3。


    6. Using Inverse Operations | 使用逆运算

    To isolate x, undo each operation in reverse order. The inverse of addition is subtraction, the inverse of subtraction is addition, the inverse of multiplication is division, and the inverse of division is multiplication.

    要单独解出 x,需要按逆序逐步还原运算。加法的逆运算是减法,减法的逆运算是加法,乘法的逆运算是除法,除法的逆运算是乘法。

    For 3x + 6 = 21, first subtract 6, then divide by 3: 3x = 15, x = 5.

    对于 3x + 6 = 21,先减 6,再除以 3:3x = 15,x = 5。


    7. Common Sign Errors | 常见符号错误

    A very common mistake is forgetting to distribute the negative sign across every term inside the bracket. Another is changing the sign of only one term when moving it to the other side. Always rewrite each line carefully and check signs before continuing.

    一个非常常见的错误是去括号时忘记把负号分配给括号内的每一项。另一个错误是把项移到等号另一边时只改变一个项的符号。每一步都要仔细重写,并在继续之前检查符号。

    −2(3x − 5) = −6x + 10, not −6x − 10. Both the 3x and the −5 must be multiplied by −2.

    例如 −2(3x − 5) = −6x + 10,而不是 −6x − 10。3x 和 −5 都要乘以 −2。


    8. Checking Your Solution | 检验答案

    After finding x, substitute it back into the original equation. If the left side equals the right side, your solution is correct. This check is quick and can help you catch sign or arithmetic errors.

    求出 x 后,要把它代回原方程进行检验。如果左边等于右边,说明答案正确。这种检验很快,还能帮助发现符号或计算错误。

    For 3(x + 2) = 21 with x = 5: left = 3(5 + 2) = 21, right = 21, so correct.

    例如 3(x + 2) = 21,当 x = 5 时:左边 = 3(5 + 2) = 21,右边 = 21,说明答案正确。


    9. Worked Example: Two Brackets | 例题:含两个括号

    Solve 2(x + 3) = 3(x − 1). First expand both brackets: 2x + 6 = 3x − 3. Then collect variables on the left: 2x − 3x = −3 − 6, giving −x = −9, so x = 9.

    解方程 2(x + 3) = 3(x − 1)。首先展开两边括号:2x + 6 = 3x − 3。然后把含未知数的项移到左边:2x − 3x = −3 − 6,得到 −x = −9,所以 x = 9。

    Check: left = 2(9 + 3) = 24, right = 3(9 − 1) = 24, so the solution is valid.

    检验:左边 = 2(9 + 3) = 24,右边 = 3(9 − 1) = 24,所以答案成立。


    10. Practice Tips for Cambridge KS3 | 剑桥 KS3 备考提示

    For Cambridge Lower Secondary Mathematics, practise writing every step clearly and avoid doing too many operations in your head. Show the expansion, the rearrangement and the inverse operations. In exams, marks are often awarded for method even if the final answer slips.

    在剑桥初中数学中,要练习把每一个步骤写清楚,不要过多依赖心算。要展示去括号、移项和逆运算的过程。考试中即使最后答案出错,只要方法正确,也常常能拿到步骤分。

    Step | 步骤 Action | 操作
    1 Expand brackets | 去括号
    2 Collect like terms | 合并同类项
    3 Move variable terms to one side | 将含未知数项移到一边
    4 Use inverse operations to solve | 使用逆运算求解
    5 Substitute to check | 代入检验

    Remember to keep the equation balanced at all times. If you write each step on a separate line, revision and checking become much easier.

    记住要始终保持方程两边平衡。如果把每一步单独写在下一行,复习和检查都会容易得多。


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  • Cambridge KS3 Mathematics: Solving Linear Equations with Brackets and Unknowns on Both Sides | 剑桥KS3数学:解含括号及两边含未知数的一元一次方程

    📚 Cambridge KS3 Mathematics: Solving Linear Equations with Brackets and Unknowns on Both Sides | 剑桥KS3数学:解含括号及两边含未知数的一元一次方程

    Linear equations are the foundation of algebra at Key Stage 3. In this article, you will learn how to solve equations that contain brackets, fractions, and unknown terms on both sides, using the balancing method.

    一元一次方程是 KS3 代数的基础。本文将介绍如何运用天平法解含有括号、分数以及两边都含未知数的方程。


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

    A linear equation has only one unknown, and the highest power of the unknown is 1. For example, 3x + 2 = 14 and 2(x − 5) = 3x + 1 are linear equations.

    一元一次方程只有一个未知数,且未知数的最高次数为 1。例如,3x + 2 = 14 和 2(x − 5) = 3x + 1 都是一元一次方程。

    The solution is the value of x that makes the equation true.

    方程的解就是使等式成立的 x 值。


    2. The Balancing Method | 天平法

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

    把方程想象成天平。你对一边做的运算,另一边也必须做同样的运算。

    Key operations:

    关键运算:

    • Add or subtract the same amount from both sides.
    • Multiply or divide both sides by the same non-zero number.

    两边同时加上或减去同一个数;两边同时乘或除以同一个非零数。

    Example: Solve x + 5 = 12.

    例题:解 x + 5 = 12。

    x = 12 − 5 = 7

    2x = 14, so x = 7

    Use inverse operations in reverse order to isolate x.

    按相反顺序使用逆运算,把 x 单独留在等号一边。


    3. Expanding Brackets First | 先去括号

    If an equation contains brackets, expand them before collecting like terms. Use the distributive law: a(b + c) = ab + ac.

    如果方程含有括号,先展开再合并同类项。使用分配律:a(b + c) = ab + ac。

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

    例题:解 3(x + 2) = 21。

    3x + 6 = 21

    3x = 15

    x = 5

    Be careful with negative signs: −2(x − 3) = −2x + 6.

    注意负号:−2(x − 3) = −2x + 6。


    4. Unknowns on Both Sides | 两边含未知数

    When x appears on both sides, collect the x terms on one side and the numbers on the other.

    当 x 出现在等式两边时,把含 x 的项移到一边,数字移到另一边。

    Example: Solve 5x − 2 = 3x + 8.

    例题:解 5x − 2 = 3x + 8。

    5x − 3x = 8 + 2

    2x = 10

    x = 5

    Moving a term across the equals sign changes its sign.

    项移过等号时要变号。


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

    Multiply every term by the lowest common denominator to clear fractions.

    将每一项乘以最小公分母,消去分母。

    Example: Solve (x + 4) ÷ 3 = 6.

    例题:解 (x + 4) ÷ 3 = 6。

    x + 4 = 18

    x = 14

    If the equation is x/2 + 3 = 7, multiply all terms by 2 first: x + 6 = 14, so x = 8.

    如果方程是 x/2 + 3 = 7,先每一项乘 2:x + 6 = 14,所以 x = 8。


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

    Always substitute your answer back into the original equation to check.

    一定要把答案代回原方程检验。

    For 5x − 2 = 3x + 8, if x = 5, left side = 5 × 5 − 2 = 23 and right side = 3 × 5 + 8 = 23, so both sides match.

    对于 5x − 2 = 3x + 8,若 x = 5,左边 = 5 × 5 − 2 = 23,右边 = 3 × 5 + 8 = 23,两边相等。


    7. Common Mistakes to Avoid | 常见错误及避免方法

    • Forgetting to multiply every term when clearing fractions.
    • Expanding brackets with a negative multiplier incorrectly.
    • Changing the sign incorrectly when moving terms across the equals sign.

    常见错误包括:去分母时漏乘某项;括号外是负数时展开出错;移项时忘记变号。

    Take one step at a time and write each stage clearly.

    一步一步来,每一步都写清楚。


    8. Word Problems Leading to Equations | 由文字题建立方程

    Translate the words into an equation by letting x represent the unknown quantity.

    设未知量为 x,把文字翻译成方程。

    Example: Three times a number, increased by 4, equals 19. Let the number be x.

    例题:一个数的 3 倍再加 4 等于 19。设这个数为 x。

    3x + 4 = 19

    3x = 15

    x = 5

    Check: 3 × 5 + 4 = 19.

    检验:3 × 5 + 4 = 19。


    9. Practice Questions | 练习题

    Try these, then check your answers.

    尝试以下题目,然后核对答案。

    1) 4(x − 2) = 16 Answer: x = 6
    2) 7x + 3 = 4x + 15 Answer: x = 4
    3) (x + 5) ÷ 2 = 9 Answer: x = 13
    4) 2(x − 3) = x + 5 Answer: x = 11

    10. Summary and Key Takeaways | 小结与要点

    Linear equations are solved by keeping the equation balanced, expanding brackets, clearing fractions, and collecting like terms.

    解一元一次方程的关键是保持等式平衡、去括号、去分母并合并同类项。

    • Use inverse operations in order.
    • Always check your solution in the original equation.

    按顺序使用逆运算;一定要把解代回原方程检验。


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

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

    Solving linear equations is one of the most important skills in the KS3 Cambridge mathematics curriculum. It builds on number operations and prepares students for more advanced algebra, graphs and problem-solving. This article reviews key methods, common question styles and exam tips, closely matched to tasks such as those on p92_2.pdf.

    解一元一次方程是 KS3 剑桥数学课程中最重要的技能之一。它以数的运算为基础,为学生进一步学习代数、图像和问题解决打下基础。本文回顾核心方法、常见题型和考试技巧,紧扣 p92_2.pdf 等典型练习。


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

    A linear equation is an algebraic statement in which two expressions are equal, and the unknown variable appears only to the power of 1. For example, 3x + 2 = 11 is a linear equation because the variable x is not squared or cubed.

    线性方程是一个代数等式,其中两个表达式相等,未知数的次数仅为 1。例如 3x + 2 = 11 是线性方程,因为变量 x 没有被平方或立方。

    In KS3, the main goal is to find the value of the unknown that makes the equation true. This value is called the solution or root.

    在 KS3 阶段,主要目标是求出使等式成立的未知数的值。这个值称为方程的解或根。


    2. The Balance Method | 天平法(等式平衡法)

    The balance method treats an equation like a set of scales. Whatever you do to one side, you must do to the other side to keep the equation balanced.

    天平法将方程视为一架天平。对等式一边施加任何运算,必须对另一边做同样运算,才能保持等式平衡。

    For example, if you add 3 to the left side, you must also add 3 to the right side. If you divide the left side by 5, you must divide the right side by 5.

    例如,如果在左边加 3,右边也必须加 3;如果将左边除以 5,右边也必须除以 5。

    x + 3 = 10 → x + 3 – 3 = 10 – 3 → x = 7


    3. Solving One-Step Equations | 解一步方程

    A one-step equation needs only one inverse operation to find the solution. The four basic types are addition, subtraction, multiplication and division.

    一步方程只需一步逆运算即可求出解。四种基本类型是加法、减法、乘法和除法。

    Type Example Solution
    Addition x + 4 = 9 x = 5
    Subtraction x – 6 = 3 x = 9
    Multiplication 5x = 20 x = 4
    Division x ÷ 3 = 7 x = 21

    Remember to use the inverse operation: addition undoes subtraction, multiplication undoes division, and vice versa.

    记住使用逆运算:加法抵消减法,乘法抵消除法,反之亦然。


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

    A two-step equation involves two operations. The standard method is to undo the addition or subtraction first, then undo the multiplication or division.

    两步方程包含两种运算。标准方法是先消去加或减,再消去乘或除。

    2x + 5 = 17 → 2x = 12 → x = 6

    Sometimes the order is reversed if the variable is divided before a term is added: x/3 – 4 = 8 → x/3 = 12 → x = 36.

    有时顺序相反,例如变量先除以 3 再减去 4:x/3 – 4 = 8 → x/3 = 12 → x = 36。


    5. Equations with Brackets | 含括号的方程

    When an equation contains brackets, expand them first using the distributive law, then simplify and solve as usual.

    当方程含有括号时,先用分配律展开括号,然后化简并照常求解。

    3(x + 2) = 21 → 3x + 6 = 21 → 3x = 15 → x = 5

    Always multiply every term inside the bracket by the number outside. A common sign error is to forget the negative sign, as in -2(x – 3) = -2x + 6.

    必须用括号外的数乘以括号内的每一项。常见符号错误是忘记负号,例如 -2(x – 3) = -2x + 6。


    6. Equations with Variables on Both Sides | 含两边变量的方程

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

    如果未知数出现在等式两边,先将含变量的项集中到一边,将常数项集中到另一边。

    5x + 3 = 2x + 12 → 3x = 9 → x = 3

    Aim to keep the variable coefficient positive to reduce sign errors. In the above case, subtracting 2x from both sides is easier than subtracting 5x.

    尽量保持变量系数为正,以减少符号错误。在上例中,两边减去 2x 比减去 5x 更容易。


    7. Equations with Fractions | 含分数的方程

    To solve equations involving fractions, multiply every term on both sides by the least common denominator (LCD) to clear the fractions.

    解含分数的方程时,先将等式两边每一项都乘最小公分母(LCD),以消去分母。

    x/4 + 1 = 3 → x/4 = 2 → x = 8

    A more complex example: (2x)/3 = 4 → multiply both sides by 3: 2x = 12 → x = 6.

    更复杂的例子:(2x)/3 = 4 → 两边乘以 3:2x = 12 → x = 6。


    8. Translating Word Problems into Equations | 将文字题转化为方程

    Many KS3 problems ask you to form and solve an equation from a real-life situation. Use a letter for the unknown and build the equation step by step.

    许多 KS3 题目要求根据实际情境建立并解方程。用字母表示未知数,并逐步建立方程。

    Example: I think of a number, multiply it by 4, then subtract 7. The answer is 25. Let n be the number: 4n – 7 = 25, so n = 8.

    例如:我想一个数,将它乘以 4,再减去 7,结果是 25。设这个数为 n:4n – 7 = 25,因此 n = 8。


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

    The most common errors include doing an operation to only one side, forgetting to change signs when expanding brackets, and mixing up the order of inverse operations.

    最常见的错误包括只对等式一边进行运算、展开括号时忘记变号、混淆逆运算的顺序。

    Always use a written check by substituting your solution back into the original equation. This takes a few seconds and can catch many mistakes.

    始终通过代入原方程进行书面检验。这只需几秒钟,却能发现许多错误。


    10. Practice Questions and Checkpoint Tips | 练习题与 Checkpoint 应考提示

    Try these questions to consolidate your skills: 7x – 2 = 26, 4(x – 3) = 32, 3x + 5 = 2x + 17, and x/5 + 2 = 7.

    尝试以下题目以巩固技能:7x – 2 = 26,4(x – 3) = 32,3x + 5 = 2x + 17,x/5 + 2 = 7。

    In the Cambridge Checkpoint test, show every step clearly. Even if your final answer is wrong, method marks can be awarded for correct balance and inverse operations.

    在剑桥 Checkpoint 测试中,要清楚地写出每一步。即使最终答案错误,正确的等式平衡和逆运算步骤也能获得方法分。


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  • Fractions, Decimals and Percentages for KS3 Cambridge Maths | KS3剑桥数学:分数、小数与百分比

    📚 Fractions, Decimals and Percentages for KS3 Cambridge Maths | KS3剑桥数学:分数、小数与百分比

    In the Cambridge KS3 Mathematics curriculum, fractions, decimals and percentages are treated as three connected ways of describing the same idea: parts of a whole. This topic builds fluency in converting between forms, comparing values, performing calculations and applying these skills to real-world problems. A strong grasp of these number forms is essential for later work in ratio, proportion, algebra and probability.

    在剑桥 KS3 数学课程中,分数、小数和百分比被视为同一概念的三种相互关联的表达方式:整体的部分。本专题帮助学生熟练地在三种形式之间转换、比较数值、进行计算,并将这些技能应用于实际问题。扎实掌握这些数的形式对于后续学习比、比例、代数和概率至关重要。


    1. Understanding Fractions | 理解分数

    A fraction is written as a numerator over a denominator. The denominator tells you how many equal parts the whole is divided into, while the numerator tells you how many of those parts are being considered. For example, in the fraction 3/4, the whole is divided into 4 equal parts and 3 parts are selected.

    分数写作分子在上、分母在下。分母表示整体被分成多少等份,分子表示取其中的几份。例如,在分数 3/4 中,整体被分成 4 等份,取其中的 3 份。

    Fractions can represent parts of a shape, parts of a set, or a point on a number line. A proper fraction has a numerator smaller than the denominator, such as 5/8. An improper fraction has a numerator greater than or equal to the denominator, such as 9/4. A mixed number combines a whole number with a proper fraction, such as 2 1/3.

    分数可以表示图形的一部分、集合的一部分或数轴上的一个点。真分数的分子小于分母,例如 5/8。假分数的分子大于或等于分母,例如 9/4。带分数由一个整数和一个真分数组成,例如 2 1/3。

    Proper: 3/5    Improper: 7/4    Mixed: 1 3/4


    2. Equivalent Fractions and Simplifying | 等值分数与化简

    Equivalent fractions look different but have the same value. You can create an equivalent fraction by multiplying or dividing both the numerator and denominator by the same non-zero whole number. For example, 1/2 = 2/4 = 3/6 = 4/8.

    等值分数看似不同但数值相同。你可以将分子和分母同时乘以或除以同一个非零整数来得到等值分数。例如,1/2 = 2/4 = 3/6 = 4/8。

    Simplifying a fraction means dividing the numerator and denominator by their highest common factor until no common factor remains except 1. The fraction 12/18 simplifies to 2/3 because the highest common factor of 12 and 18 is 6.

    化简分数是指用分子和分母的最大公因数同时除以它们,直到除 1 以外没有其他公因数为止。分数 12/18 化简为 2/3,因为 12 和 18 的最大公因数是 6。

    12/18 = (12 ÷ 6)/(18 ÷ 6) = 2/3


    3. Decimals and Place Value | 小数与位值

    A decimal is another way to write a fraction whose denominator is a power of 10. The digits after the decimal point represent tenths, hundredths, thousandths and so on. For example, 0.35 means 3 tenths plus 5 hundredths, which is equal to 35/100.

    小数是分母为 10 的幂的分数的另一种写法。小数点后的数字分别表示十分位、百分位、千分位等。例如,0.35 表示 3 个十分之一加 5 个百分之一,等于 35/100。

    Place value is important when comparing and rounding decimals. The decimal 0.7 is greater than 0.699 because 7 tenths is larger than 6 tenths plus 9 hundredths plus 9 thousandths. Rounding 3.276 to two decimal places gives 3.28 because the digit in the thousandths place is 6, so the hundredths digit is rounded up.

    位值在比较和四舍五入小数时非常重要。小数 0.7 大于 0.699,因为 7 个十分之一大于 6 个十分之一加 9 个百分之一加 9 个千分之一。将 3.276 四舍五入到两位小数得到 3.28,因为千分位上的数字是 6,所以百分位数字向上进位。

    Tenths: 0.1 = 1/10 Hundredths: 0.01 = 1/100 Thousandths: 0.001 = 1/1000

    4. Converting Between Fractions and Decimals | 分数与小数互化

    To convert a fraction to a decimal, divide the numerator by the denominator. For example, 3/8 = 3 ÷ 8 = 0.375. Some fractions produce terminating decimals, while others produce recurring decimals such as 1/3 = 0.333… which can be written with a dot above the repeating digit.

    要将分数转换为小数,用分子除以分母。例如,3/8 = 3 ÷ 8 = 0.375。一些分数产生有限小数,而另一些产生循环小数,例如 1/3 = 0.333…,可以在重复数字上方加点表示。

    To convert a terminating decimal to a fraction, write the decimal as a fraction with denominator 10, 100 or 1000 depending on the number of decimal places, then simplify. For example, 0.45 = 45/100 = 9/20.

    要将有限小数转换为分数,根据小数位数将小数写成分母为 10、100 或 1000 的分数,然后化简。例如,0.45 = 45/100 = 9/20。

    3/8 = 0.375    and    0.45 = 9/20


    5. Percentages and Their Meaning | 百分比及其意义

    A percentage is a fraction with denominator 100. The symbol % means ‘out of 100’. For example, 47% means 47 out of 100, which is equal to 47/100 or 0.47. Percentages are useful for comparing proportions because they give a common scale.

    百分比是分母为 100 的分数。符号 % 表示“每 100 份中的份数”。例如,47% 表示 100 份中的 47 份,等于 47/100 或 0.47。百分比对于比较比例非常有用,因为它们提供了统一的尺度。

    Percentages can be greater than 100 when a quantity increases beyond its original value. For example, 150% means 150 out of 100, or 1.5 times the original amount. Percentages less than 1 are also possible, such as 0.5% which equals 0.005.

    当某个量超过原值时,百分比可以大于 100。例如,150% 表示 100 份中的 150 份,即原值的 1.5 倍。小于 1 的百分比也存在,例如 0.5% 等于 0.005。


    6. Converting Between Fractions, Decimals and Percentages | 分数、小数和百分比的互化

    The three forms are interchangeable. To convert a decimal to a percentage, multiply by 100 and add the % symbol. For example, 0.85 = 85%. To convert a percentage to a decimal, divide by 100. For example, 32% = 0.32.

    这三种形式可以相互转换。要将小数转换为百分比,乘以 100 并加上 % 符号。例如,0.85 = 85%。要将百分比转换为小数,除以 100。例如,32% = 0.32。

    To convert a fraction to a percentage, first divide the numerator by the denominator to obtain a decimal, then multiply by 100. For example, 3/5 = 0.6 = 60%. To convert a percentage to a fraction, write it as a fraction over 100 and simplify. For example, 65% = 65/100 = 13/20.

    要将分数转换为百分比,先用分子除以分母得到小数,再乘以 100。例如,3/5 = 0.6 = 60%。要将百分比转换为分数,将其写成分母为 100 的分数并化简。例如,65% = 65/100 = 13/20。

    Fraction: 1/4 Decimal: 0.25 Percentage: 25%
    Fraction: 3/10 Decimal: 0.3 Percentage: 30%
    Fraction: 7/20 Decimal: 0.35 Percentage: 35%

    7. Comparing and Ordering | 比较和排序

    To compare fractions, decimals and percentages, convert all values to the same form. Often decimals are the easiest to compare because you can compare place values directly. For example, to order 2/5, 0.44 and 39%, convert them all to decimals: 0.4, 0.44 and 0.39. In ascending order they are 0.39, 0.4, 0.44, which is 39%, 2/5, 0.44.

    要比较分数、小数和百分比,将所有数值转换为同一种形式。通常小数最容易比较,因为可以直接比较位值。例如,要对 2/5、0.44 和 39% 排序,将它们都转换为小数:0.4、0.44 和 0.39。按升序排列为 0.39、0.4、0.44,即 39%、2/5、0.44。

    When comparing fractions with different denominators, find a common denominator, convert each fraction and compare the numerators. For example, to compare 2/3 and 5/8, use 24 as the common denominator: 2/3 = 16/24 and 5/8 = 15/24, so 2/3 is greater than 5/8.

    比较不同分母的分数时,先找到公分母,转换每个分数,然后比较分子。例如,比较 2/3 和 5/8,用 24 作为公分母:2/3 = 16/24,5/8 = 15/24,所以 2/3 大于 5/8。


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

    To add or subtract fractions with the same denominator, simply add or subtract the numerators and keep the denominator. For example, 2/9 + 4/9 = 6/9 = 2/3, and 7/10 – 3/10 = 4/10 = 2/5.

    要将同分母分数相加或相减,只需将分子相加或相减,分母保持不变。例如,2/9 + 4/9 = 6/9 = 2/3,7/10 – 3/10 = 4/10 = 2/5。

    If the denominators are different, find the lowest common denominator first, rewrite each fraction as an equivalent fraction, and then add or subtract the numerators. For example, 1/4 + 2/5 = 5/20 + 8/20 = 13/20. For mixed numbers, add or subtract the whole number parts and fraction parts separately, borrowing if necessary.

    如果分母不同,先找出最小公分母,将每个分数改写为等值分数,然后再进行分子加减。例如,1/4 + 2/5 = 5/20 + 8/20 = 13/20。对于带分数,分别加减整数部分和分数部分,必要时进行借位。

    1/4 + 2/5 = 5/20 + 8/20 = 13/20


    9. Multiplying and Dividing Fractions | 分数的乘除法

    To multiply fractions, multiply the numerators together and multiply the denominators together, then simplify if possible. For example, 3/4 × 2/5 = 6/20 = 3/10. Mixed numbers should first be converted to improper fractions before multiplying.

    要将分数相乘,将分子相乘、分母相乘,然后尽可能化简。例如,3/4 × 2/5 = 6/20 = 3/10。带分数在相乘前应先转换为假分数。

    To divide by a fraction, multiply by its reciprocal. The reciprocal of a fraction is obtained by swapping its numerator and denominator. For example, 5/6 ÷ 2/3 = 5/6 × 3/2 = 15/12 = 5/4 = 1 1/4.

    要除以一个分数,乘以它的倒数。分数的倒数通过交换其分子和分母得到。例如,5/6 ÷ 2/3 = 5/6 × 3/2 = 15/12 = 5/4 = 1 1/4。

    5/6 ÷ 2/3 = 5/6 × 3/2 = 5/4 = 1 1/4


    10. Percentage of Amounts and Problems | 百分数应用与问题解决

    To find a percentage of an amount, convert the percentage to a decimal and multiply by the amount. For example, to find 15% of 240, calculate 0.15 × 240 = 36. Alternatively, find 10% first and then adjust: 10% of 240 is 24, 5% is 12, so 15% is 24 + 12 = 36.

    要求一个数的百分之几,将百分比转换为小数再乘以这个数。例如,求 240 的 15%,计算 0.15 × 240 = 36。也可以先求 10% 再调整:240 的 10% 是 24,5% 是 12,所以 15% 是 24 + 12 = 36。

    Percentage increase and decrease problems are common. To increase 80 by 25%, find 25% of 80 which is 20, then add it to the original amount: 80 + 20 = 100. Alternatively, multiply by 1.25. To decrease 60 by 30%, find 30% of 60 which is 18, then subtract: 60 – 18 = 42, or multiply by 0.70.

    百分比增减问题很常见。将 80 增加 25%,先求 80 的 25% 是 20,然后加到原数上:80 + 20 = 100。也可以乘以 1.25。将 60 减少 30%,先求 60 的 30% 是 18,然后相减:60 – 18 = 42,或者乘以 0.70。

    • Find 20% of 150: 0.20 × 150 = 30
    • Increase 40 by 15%: 40 × 1.15 = 46
    • Decrease 72 by 25%: 72 × 0.75 = 54
    • 求 150 的 20%:0.20 × 150 = 30
    • 将 40 增加 15%:40 × 1.15 = 46
    • 将 72 减少 25%:72 × 0.75 = 54

    In word problems, identify the original amount, the percentage change and whether the change is an increase or decrease. Then choose the correct multiplication or division operation. Drawing a bar model can help visualise the relationship between the part and the whole.

    在应用题中,要确定原量、百分比变化以及变化是增加还是减少。然后选择正确的乘法或除法运算。画条形模型有助于直观表示部分与整体的关系。


    Published by TutorHao | Mathematics Revision Series | aleveler.com

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