Tag: KS3

📚 Mastering Bearings: A Complete Guide for KS3 Mathematics | 精通方位角：KS3数学完整指南

Bearings are an essential topic in KS3 mathematics, providing a precise way to describe direction. Whether navigating a ship at sea, plotting a flight path, or simply reading a map, understanding bearings allows you to communicate angles accurately from one point to another. This guide will walk you through everything you need to know about bearings, from definitions to problem-solving, aligned with the Cambridge KS3 syllabus.

方位角是KS3数学中的一个重要主题，它提供了一种精确描述方向的方法。无论是航海中的船舶导航、绘制飞行路线，还是简单地阅读地图，理解方位角都能让你准确地表达从一点到另一点的角度。本指南将带你全面学习方位角的知识，从定义到解题技巧，完全契合剑桥KS3教学大纲。

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1. What Are Bearings? | 什么是方位角？

A bearing is a way of describing the direction of one point relative to another point. It is measured as an angle from the north direction, moving clockwise. Bearings are always given as a three-figure number, such as 045° or 210°.

方位角是描述一个点相对于另一个点的方向的一种方式。它以北方向为基准，按顺时针方向测量的角度来表示。方位角始终用三位数字表示，例如045°或210°。

In geometry and navigation, bearings allow us to specify direction uniquely. For example, if you are told to walk on a bearing of 120°, you would face north, turn clockwise through 120°, and walk forward. This method avoids confusion between directions like ‘north-east’ and ‘east-north-east’.

在几何和导航中，方位角使我们能够唯一地指定方向。例如，如果你被告知以120°的方位角行走，你应面向北方，顺时针旋转120°，然后直行。这种方法避免了类似“东北”和“东北偏东”这样的方向混淆。

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2. The Three Golden Rules of Bearings | 方位角的三大黄金法则

To work with bearings correctly, you must follow three key rules. First, always start measuring from the north line (a line pointing directly upwards on a diagram). Second, measure the angle in a clockwise direction. Third, write the bearing using three digits; for example, 60° becomes 060°, and 5° becomes 005°.

要正确使用方位角，你必须遵循三条关键规则。第一，始终从正北线（图中指向正上方的线）开始测量。第二，沿顺时针方向测量角度。第三，用三位数字书写方位角；例如，60°要写成060°，5°要写成005°。

• Rule 1: Always measure from North.
• 规则1：始终从北方向测量。
• Rule 2: Always measure clockwise.
• 规则2：始终沿顺时针方向测量。
• Rule 3: Always give bearings as three-figure numbers.
• 规则3：始终以三位数字表示方位角。

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3. Measuring Bearings on a Diagram | 在图上测量方位角

To find the bearing of point B from point A on a diagram, draw a north line at A. Place a protractor with its centre at A and the 0° mark aligned with the north line. Measure the angle clockwise from north to the line AB. Record this angle as a three-figure bearing. If the angle is less than 100°, remember to add leading zeros.

要在图中求出从点A到点B的方位角，需在A处画一条向北的线。将量角器的中心对准A，0°刻度线与北线对齐。顺时针测量从北线到线段AB的角度。将此角度记为三位数的方位角。若角度小于100°，切记在前面补零。

Always double-check whether the measured angle lies inside the protractor’s inner or outer scale. Many students misread the scale when measuring clockwise, so it is helpful to mark the direction with an arrow before reading the angle.

务必仔细确认所测角度位于量角器的内圈还是外圈刻度。许多学生在顺时针测量时会读错刻度，因此在读数前用箭头标出方向会很有帮助。

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4. Drawing Bearings Accurately | 准确绘制方位角

To draw a bearing accurately, start by marking the point and drawing a vertical north line. Place the protractor with the centre at the point and the baseline along the north line. Mark the required angle clockwise, then draw a line from the point through your mark. Indicate the bearing and label the line. This skill is vital for scale drawings involving bearings.

要准确绘制一个方位角，首先标出点并画一条垂直的北线。将量角器的中心放在该点上，基线与北线对齐。顺时针标记所需角度，然后从该点出发穿过标记画一条射线。注明方位角并标记线段。这项技能对于涉及方位角的比例图绘制至关重要。

When drawing bearings for journeys with several legs, keep your pencil sharp and use a ruler to maintain precision. A slight error in angle can cause a large displacement over long distances in scale drawings.

在绘制包含多段行程的方位角时，保持铅笔尖锐并使用直尺以保持精度。在比例图中，微小的角度误差在长距离上可能造成巨大的位移偏差。

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5. Back Bearings – The Reverse Direction | 反方位角——反向方向

In many practical problems, you need the bearing for the return journey. If the bearing from A to B is known, the back bearing (bearing from B to A) is found by adding or subtracting 180°. If the original bearing is less than 180°, add 180°; if it is 180° or more, subtract 180°. This ensures the result remains a valid three-figure bearing.

在许多实际问题中，你需要返程的方位角。若已知从A到B的方位角，反方位角（从B到A的方位角）可通过加减180°求得。如果原始方位角小于180°，则加上180°；若大于或等于180°，则减去180°。这样可以保证结果仍是有效的三位数方位角。

Back Bearing = (Bearing + 180°) mod 360°

反方位角 = (方位角 + 180°) mod 360°

The ‘mod 360°’ operation simply means that if the sum exceeds 360°, subtract 360° to bring the bearing back into the range 000° to 360°. For example, a bearing of 270° gives a back bearing of (270°+180°) – 360° = 090°.

“mod 360°”运算的含义是，若和超过360°，则减去360°，使方位角回到000°到360°的范围内。例如，方位角270°的反方位角为(270°+180°) – 360° = 090°。

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6. Bearings and Scale Drawings | 方位角与比例图

Bearings often appear alongside distances in scale drawing questions. You may be asked to interpret a map or construct a diagram where positions are given by bearings and distances. For instance, a ship sails 5 km on a bearing of 070°, then 3 km on a bearing of 150°. To find its distance from the start, you draw the paths accurately using a ruler and protractor.

方位角通常与距离一起出现在比例图问题中。你或许需要解读地图，或根据方位角和距离构造图形。例如，一艘船以070°的方位角航行5公里，再以150°的方位角航行3公里。要找到它与起点的距离，就需要用尺规和量角器精确绘制路径。

Once the diagram is drawn to scale, you can measure the straight-line distance from start to finish with a ruler and convert it back using the scale factor. This method combines measuring skills with bearing knowledge.

一旦按比例画出图形，你可用直尺量出起点到终点的直线距离，再利用比例尺换算回实际距离。这种方法将测量技能与方位角知识结合起来。

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7. Solving Problems with Bearings | 解决方位角问题

Many bearing problems combine angle facts with parallel lines. North lines drawn at different points are parallel, so alternate angles and co-interior angles can be used to find unknown bearings. For example, if you know the bearing from A to B and need the bearing from B to C, you can construct a triangle and apply the fact that angles around a point sum to 360°.

许多方位角问题将角度性质与平行线结合起来。在不同点绘制的北线是平行的，因此可以利用内错角和同旁内角来求未知方位角。例如，若已知A到B的方位角，需求B到C的方位角，你可以构造三角形，并运用一点周围角度之和为360°的原理。

Consider a problem: The bearing of B from A is 110°, and the bearing of C from B is 240°. Find the bearing of A from C. By drawing north lines and identifying alternate angles, you can deduce the required angle using the geometry of the triangle and parallel lines.

考虑一个问题：B在A的方位角为110°，C在B的方位角为240°。求A在C的方位角。通过画北线、识别内错角，可利用三角形和平行线的几何性质推导出所需角度。

It often helps to sketch the scenario and label all known bearings and angles. Break the problem into small steps: find interior angles of the triangle, then use the fact that the angle sum is 180° to discover missing angles, finally convert them back into bearings.

绘制示意图并标出所有已知的方位角和角度通常会很有帮助。将问题分解为小步骤：求出三角形的内角，然后利用角度和为180°的性质找出缺失的角度，最后再将其转换回方位角。

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8. Bearings in Real‑Life Contexts | 现实生活中的方位角

Bearings are widely used in aviation, shipping, and outdoor activities. Pilots use bearings to navigate between waypoints, sailors use them to chart courses, and hikers use compass bearings to stay on track. Mastering bearings not only boosts your geometry skills but also prepares you for practical map reading.

方位角广泛应用于航空、航运和户外活动中。飞行员利用方位角在航点之间导航，水手用它规划航线，徒步旅行者使用指南针方位角保持行进方向。掌握方位角不仅能提升你的几何技能，还能让你具备实际的地图阅读能力。

Even in modern GPS systems, bearings are computed behind the scenes. Understanding the underlying mathematics enriches your spatial awareness and is an excellent foundation for further study in trigonometry and vector geometry.

即使在现代GPS系统中，方位角也是在后台计算出来的。理解其背后的数学知识能够丰富你的空间感知能力，并为后续的三角学和矢量几何学习打下坚实的基础。

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9. Common Mistakes to Avoid | 常见错误及避免

Many students forget to use three figures, writing 60° instead of 060°. Others measure anticlockwise or misinterpret the north line. A frequent error is confusing the bearing from A to B with the bearing from B to A; remember the 180° rule. Always check your protractor alignment and ensure the angle is measured from the north clockwise.

许多学生忘记使用三位数字，写了60°而不是060°。还有人按逆时针测量，或误解北线的位置。一个常见错误是混淆从A到B与从B到A的方位角；请记住180°规则。务必检查量角器的对齐情况，确保角度是从北顺时针测量的。

Also, watch out for diagrams where the north line is not vertical on the page – the north direction remains the reference regardless of how the map is rotated. Keep the protractor’s baseline aligned with that north line, not with the edge of the paper.

此外，要留意图中北线并不垂直于纸面的情形——无论地图如何旋转，北方向始终是基准。让量角器的基线与那条北线对齐，而不是与纸边对齐。

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10. Practice Questions and Tips | 练习题与技巧

To excel in bearing questions, practice drawing and measuring bearings from diagrams. Start with simple two-point bearings, progress to three-point journey problems, and finally tackle combined scale drawing questions. In exams, always write bearings as three-figure numbers and show your construction lines. Remember, a neat diagram is half the answer.

要在方位角题目中取得优异成绩，请多练习从图中绘制和测量方位角。从简单的两点方位角开始，逐步过渡到三点行程问题，最后攻克综合比例图题目。在考试中，务必以三位数字书写方位角，并展示作图辅助线。记住，一张清晰的图就是一半答案。

When revising, create flashcards with different bearings and back bearings, and practise converting between them quickly. Also, try to explain the rules to a friend – teaching is one of the most effective ways to reinforce your own understanding.

复习时，可以制作一些写有不同方位角和反方位角的卡片，并练习快速转换。此外，尝试向朋友讲解这些规则——教别人是巩固自己理解的最有效方法之一。

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📚 Cambridge KS3 Mathematics Revision: Core Topics from Page 95 Exercise 1 | 剑桥KS3数学复习：第95页练习1核心专题

This article revisits the key concepts typically encountered in a Cambridge KS3 mathematics practice sheet such as page 95, exercise 1. It covers whole number operations, negative numbers, fractions, decimals, percentages, and the first steps into algebra. Working through these ideas will help you tackle mixed problem sets and prepare confidently for the Checkpoint exam.

本文重新梳理在剑桥 KS3 数学练习册（如第 95 页练习 1）中经常出现的关键概念，包括整数运算、负数、分数、小数、百分数以及代数入门。通过逐一攻破这些知识点，你将能从容应对混合练习题组，并为 Checkpoint 考试做好充分准备。

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1. Whole Number Operations | 整数运算

Addition, subtraction, multiplication, and division of whole numbers form the backbone of KS3 mathematics. When adding 478 + 356, align the digits by place value: units, tens, hundreds. Begin with the units column: 8 + 6 = 14, write 4 and carry 1 to the tens column. Then 7 + 5 + 1 = 13, write 3 and carry 1 to the hundreds column. Finally, 4 + 3 + 1 = 8, giving 834. This column method eliminates confusion when numbers have different digits.

整数的加、减、乘、除是 KS3 数学的根基。计算 478 + 356 时，要按个位、十位、百位对齐数字。先从个位开始：8+6=14，写下 4，向十位进 1。然后 7+5+1=13，写下 3，向百位进 1。最后 4+3+1=8，得到 834。这种列竖式的方法避免了因数字位数不同而产生的混乱。

For multiplication, such as 24 × 36, break it into parts using the grid method. Multiply 20 × 30 = 600, 20 × 6 = 120, 4 × 30 = 120, and 4 × 6 = 24. Adding 600 + 120 + 120 + 24 gives 864. Alternatively, long multiplication with careful carrying works equally well. Always check your result by estimating: 24 × 36 is about 25 × 35 = 875, so 864 is reasonable.

对于乘法，如 24 × 36，可用网格法拆分计算：20×30=600，20×6=120，4×30=120，4×6=24。把 600+120+120+24 相加得 864。也可以使用进位长乘法，同样可靠。最后要用估算检验结果：24×36 大约等于 25×35=875，因此 864 是合理的。

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2. Negative Numbers | 负数

Working with negative numbers requires understanding the number line. Adding a negative is equivalent to moving left: 3 + (−5) = −2. Subtracting a negative means moving right: 4 − (−3) = 4 + 3 = 7. When two signs appear together, remember the rule: same signs make a positive, different signs make a negative. So −(−4) = +4, and +(−2) = −2.

处理负数需要理解数轴。加上一个负数相当于向左移动：3 + (−5) = −2。减去一个负数相当于向右移动：4 − (−3) = 4 + 3 = 7。当两个符号碰到一起时，记住“同号得正，异号得负”的规则。因此 −(−4)=+4，+(−2)=−2。

Multiplying and dividing with negatives follows a similar pattern. (−6) × (−4) = 24, while (−6) × 4 = −24. 15 ÷ (−3) = −5. These rules are vital when substituting negative values into algebraic expressions later.

负数的乘除法也遵循类似规律：（−6）×（−4）= 24，而（−6）× 4 = −24。15 ÷（−3）= −5。这些规则在后续向代数式中代入负数值时至关重要。

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3. Order of Operations (BIDMAS) | 运算顺序 (BIDMAS)

The BIDMAS rule tells us the correct sequence when a calculation involves more than one operation: Brackets, Indices, Division and Multiplication (left to right), Addition and Subtraction (left to right). For example, evaluate 10 − 2 × 3². First, calculate the index: 3² = 9. Then multiply: 2 × 9 = 18. Finally, subtract: 10 − 18 = −8. If we had simply worked left to right, we would have obtained (10 − 2) × 9 = 72, which is incorrect.

当一个算式包含多种运算时，BIDMAS 规则告诉我们正确的顺序：先算括号，再算指数（乘方），接着算乘法和除法（从左到右），最后算加法和减法（从左到右）。例如计算 10 − 2 × 3²。先算指数：3² = 9，再算乘法：2×9=18，最后算减法：10−18=−8。如果只是从左往右机械计算，会得到 (10−2)×9=72，那就错了。

When brackets are present, always evaluate what is inside them first. In 24 ÷ (4 + 2) × 3, compute 4 + 2 = 6 first. Then do 24 ÷ 6 = 4, and finally 4 × 3 = 12. Inserting brackets changes the meaning of an expression, so it is important to write them clearly in your own working.

若式子中含有括号，一定要先算括号内的部分。如 24 ÷ (4 + 2) × 3，先算 4+2=6，再算 24÷6=4，最后 4×3=12。增加括号会改变整个算式的含义，因此自己写过程时也要把括号标得清清楚楚。

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4. Fractions: Simplifying and Equivalent | 分数：化简与等值分数

A fraction represents a part of a whole. To simplify a fraction such as 15/35, find the greatest common factor of the numerator and denominator. Both 15 and 35 are divisible by 5, so 15 ÷ 5 = 3 and 35 ÷ 5 = 7, giving 3/7. Simplifying makes fractions easier to compare and operate with.

分数表示整体的一部分。化简如 15/35 这样的分数，需要找到分子与分母的最大公因数。15 和 35 都能被 5 整除，因此 15÷5=3，35÷5=7，得到 3/7。化简后的分数更便于比较和计算。

Equivalent fractions can be created by multiplying or dividing the numerator and denominator by the same non-zero number. For example, 1/2, 2/4, 3/6, and 5/10 are all equivalent. You can check this by cross-multiplying: 1 × 4 equals 2 × 2, both giving 4. Equivalent fractions are useful when adding and subtracting fractions with different denominators.

将分子和分母同时乘或除以同一个非零数，就能得到等值分数。例如 1/2、2/4、3/6 和 5/10 都是等值的。你可以用交叉相乘来验证：1×4 等于 2×2，都得 4。在为异分母分数进行加减运算时，等值分数非常有用。

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5. Adding and Subtracting Fractions | 分数加减法

To add or subtract fractions, they must have the same denominator. For 3/8 + 1/4, convert 1/4 to an equivalent fraction with denominator 8: multiply the numerator and denominator by 2 to get 2/8. Then add: 3/8 + 2/8 = 5/8. The denominator stays the same while the numerators are added.

要进行分数加减，必须先化为同分母。以 3/8 + 1/4 为例，把 1/4 化成以 8 为分母的等值分数：分子分母同时乘 2，得 2/8。然后相加：3/8 + 2/8 = 5/8。分母不变，只把分子相加。

When subtracting, apply the same rule. For 7/10 − 2/5, change 2/5 to 4/10, then subtract: 7/10 − 4/10 = 3/10. If the result is an improper fraction, you may need to convert it to a mixed number. For example, 9/4 = 2 1/4. Always simplify your final answer where possible.

做减法时规则相同。计算 7/10 − 2/5，先把 2/5 化成 4/10，然后相减：7/10−4/10=3/10。如果结果为假分数，可能需要化为带分数，例如 9/4 = 2 ¼。最后一定要尽可能化简答案。

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6. Multiplying and Dividing Fractions | 分数乘除法

Multiplying fractions is straightforward: multiply the numerators together and multiply the denominators together. So 2/3 × 4/5 = (2×4)/(3×5) = 8/15. If possible, simplify before multiplying by cancelling any common factors between numerators and denominators across the fractions. For instance, 3/8 × 4/9: cancel 3 and 9 (both can be divided by 3) and 4 and 8 (divide by 4), giving 1/2 × 1/3 = 1/6.

分数乘法很直接：分子乘分子，分母乘分母。所以 2/3 × 4/5 = (2×4)/(3×5)=8/15。如果可以，在乘法运算前先约分会更简便，跨分数寻找分子分母之间的公因数。例如 3/8 × 4/9：将 3 和 9 约掉（除以 3），将 4 和 8 约掉（除以 4），得到 1/2 × 1/3 = 1/6。

Dividing fractions involves multiplying by the reciprocal of the divisor. To calculate 5/6 ÷ 2/3, flip the second fraction and multiply: 5/6 × 3/2 = 15/12. Simplify 15/12 to 5/4, which is 1 1/4. Always remember to invert only the divisor, not the first fraction.

分数除法则需乘上除数的倒数。计算 5/6 ÷ 2/3，把第二个分数翻转后相乘：5/6 × 3/2 = 15/12。化简 15/12 得 5/4，也就是 1 ¼。一定要记住，只翻转除数，不动第一个分数。

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7. Decimals and Place Value | 小数与位值

Decimals are an extension of the place value system. The columns to the right of the decimal point represent tenths, hundredths, thousandths, and so on. In the number 35.607, the digit 6 is in the tenths place (6/10), 0 in the hundredths place, and 7 in the thousandths place (7/1000). Understanding place value is essential when ordering, adding, and subtracting decimals.

小数实质上是位值体系的延伸。小数点右边的数位依次表示十分位、百分位、千分位等。在 35.607 中，数字 6 在十分位（6/10），0 在百分位，7 在千分位（7/1000）。理解位值对比较、加减小数至关重要。

When adding decimals, align the decimal points vertically. For 12.8 + 3.45, write 12.80 above 3.45, then add column by column: 0 + 5 = 5, 8 + 4 = 12 (write 2, carry 1), 2 + 3 + 1 = 6, and bring down the 1 in the tens place. The result is 16.25. Adding zeros as placeholders prevents misalignment.

做小数加法时，要把小数点上下对齐。计算 12.8 + 3.45 时，写成 12.80 对齐 3.45，再按列相加：0+5=5，8+4=12（写 2 进 1），2+3+1=6，最后把十位的 1 落下来，结果为 16.25。用零占位可以防止错位。

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

Fractions, decimals and percentages are three ways of representing the same idea. To change a fraction to a decimal, divide the numerator by the denominator: 3/8 = 3 ÷ 8 = 0.375. To convert a decimal to a percentage, multiply by 100 and add the % sign: 0.375 × 100 = 37.5%. Reversing the process, from a percentage to a decimal, divide by 100: 62% = 0.62.

分数、小数和百分数是同一概念的三种不同表示法。分数化小数，用分子除以分母：3/8 = 3÷8 = 0.375。小数化百分数，乘 100 并加上 % 号：0.375×100=37.5%。反过来，百分数化小数除以 100：62% = 0.62。

Some conversions are so common that they should be memorised. The table below lists a few key equivalents. Being able to recall these instantly makes problem-solving much faster, especially in ratio and proportion questions.

有些转换实在太常用，最好直接记住。下表列出了几个关键等价关系。能够立刻反应出这些值，会大大提高解题速度，尤其是在比例和比率问题中。

Fraction	Decimal	Percentage
1/2	0.5	50%
1/4	0.25	25%
3/4	0.75	75%
1/5	0.2	20%
1/10	0.1	10%

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9. Introduction to Algebra: Substitution | 代数入门：代入求值

Algebra uses letters to represent unknown or changing numbers. Substitution means replacing a letter with a given value and calculating the result. If a = 3 and b = 4, evaluate 2a + 3b. Substitute: 2 × 3 + 3 × 4. Following BIDMAS, multiply first: 6 + 12 = 18. Always put the value in brackets when substituting if there is any chance of sign confusion, e.g. when b = −2, 3b becomes 3 × (−2) = −6.

代数用字母表示未知或可变的数。代入即用给定的数值替换字母然后计算出结果。已知 a=3，b=4，求 2a + 3b 的值：代入 2×3 + 3×4，遵照运算顺序先乘得 6+12=18。如果代入的数值是负数或易造成符号混淆，要用括号框好，例如 b=−2 时，3b 变成 3×(−2)=−6。

Another example: evaluate (x² + y) / 2 when x = 5 and y = 7. Replace x and y: (5² + 7) / 2. Calculate the index: 25 + 7 = 32, then divide by 2 to get 16. Accurate substitution is crucial before moving on to solving equations.

再如：已知 x=5，y=7，求 (x² + y) / 2。替换后得 (5² + 7) / 2，先算指数 25+7=32，再除以 2 得 16。在进一步学习解方程之前，准确的代入是基础。

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10. Solving One-step Equations | 解一步方程

An equation shows that two expressions are equal. To solve a one-step equation, perform the inverse operation to isolate the variable. For x + 7 = 15, subtract 7 from both sides: x + 7 − 7 = 15 − 7, giving x = 8. For y − 3 = 9, add 3 to both sides: y = 12. The golden rule is: whatever you do to one side, you must do to the other to keep the equation balanced.

方程表示两个表达式相等。解一步方程需要运用逆运算把变量孤立出来。以 x + 7 = 15 为例，两边同时减 7：x+7−7=15−7，得 x=8。对于 y − 3 = 9，两边同时加 3 得 y=12。黄金法则是：对一边做了什么，另一边也必须做同样的事，以保持天平平衡。

Multiplication and division equations work similarly. If 4n = 20, divide both sides by 4: n = 5. If m / 3 = 6, multiply both sides by 3: m = 18. Always check your answer by substituting it back into the original equation: for 4n = 20, 4 × 5 = 20, which is true.

乘法和除法方程同理。若 4n = 20，两边同除以 4 得 n=5。若 m / 3 = 6，两边同乘 3 得 m=18。养成好习惯，把答案代回原方程检验：4n=20 中，4×5=20，成立。

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11. Introduction to Co-ordinates | 坐标系入门

Co-ordinates describe a point’s position on a grid using an ordered pair (x, y). The x-value comes first and tells you how far to move horizontally from the origin (0,0); the y-value tells you how far to move vertically. Plot the point (2, 5) by starting at the origin, moving 2 units right and 5 units up. Negative x-values move left, negative y-values move down.

坐标用一对有序数 (x, y) 来描述网格中点的位置。x 值在前，表示从原点 (0,0) 出发沿水平方向移动的距离；y 值表示竖直移动的距离。画点 (2,5) 时从原点出发，向右移动 2 格再向上移动 5 格。x 值为负向左移，y 值为负向下移。

Being able to read and plot co-ordinates is the foundation of graphing lines and shapes. In a practice exercise like page 95, you may be asked to identify the co-ordinates of vertices of a rectangle or to complete a shape when given three vertices. Always count the grid lines carefully and double-check the order of x and y.

正确读、画坐标是后续画直线和图形的基础。在类似第 95 页的练习中，你可能会被要求写出长方形各顶点的坐标，或根据已知的三个顶点完成图形。务必仔细数网格线，并始终确认 x 和 y 的顺序。

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12. Practical Problem Solving | 实际应用问题

KS3 mathematics often combines skills in word problems. For example: “A book costs $12.50. A school buys 15 copies with a budget of $200. How much money is left?” First, find the total cost: 12.50 × 15 = $187.50. Then subtract from $200: 200 − 187.50 = $12.50 remaining. Break the problem into clear steps and decide which operations to use.

KS3 数学常通过文字题把多种技能融合在一起。比如：“一本书售价 $12.50，学校预算 $200 购买 15 本，还剩多少钱？”先求总额：12.50×15=$187.50，再拿 $200 去减：200−187.50=$12.50。把问题分解成清晰的步骤，并选择恰当的运算。

Another common type involves fractions and percentages: “In a class of 30 students, 2/5 are boys. How many girls are there?” Since 2/5 are boys, 1 − 2/5 = 3/5 are girls. 3/5 of 30 = 3/5 × 30 = 18 girls. Reading the question carefully and identifying what fraction or percentage refers to the whole is the key.

另一常见题型融合分数与百分数：“一个有 30 名学生的班级，2/5 是男生，请问有多少女生？”因为 2/5 是男生，则 1−2/5=3/5 是女生。30 的 3/5 是 3/5×30=18 名女生。仔细读题，明确分数或百分数所对应的“整体”是什么，是解题关键。

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📚 Solving Equations with Brackets (Cambridge KS3 p92 q1) | 求解带括号的方程（剑桥KS3数学第92页第1题）

Welcome to this KS3 Cambridge Mathematics revision article. Today we focus on solving linear equations that include brackets, a key skill tested in the Cambridge Checkpoint curriculum. We will work through a specific example from your practice book: Page 92, Question 1. This question asks us to solve 2(x+3)=14. Following the steps carefully will help you build confidence in algebra and ace your assessments.

欢迎阅读这篇KS3剑桥数学复习文章。今天我们重点学习求解带括号的一元一次方程，这是剑桥Checkpoint课程中考查的重要技能。我们将通过练习册中的一个具体问题进行讲解：第92页第1题。题目要求解方程 2(x+3)=14。认真跟随解题步骤将有助于你建立代数的信心，并在考试中取得好成绩。

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1. The Problem from Page 92 | 第92页的问题

Let’s take a close look at the exact problem from your Cambridge Checkpoint Mathematics Practice Book 8. On page 92, the first question reads: “Solve the equation 2(x + 3) = 14.” This is a linear equation containing a bracket, so we must first deal with the bracket before isolating the variable x. The goal is to find the value of x that makes both sides of the equation equal.

让我们仔细看看剑桥Checkpoint数学练习册8第92页的具体问题。第92页的第一题是：“解方程 2(x + 3) = 14。”这是一个含有括号的线性方程，因此我们必须先处理括号，然后再分离出变量x。我们的目标是找到使等式两边相等的x值。

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2. What is an Equation? | 什么是方程？

An equation is a mathematical statement that shows two expressions are equal, using the equals sign ‘=’. For example, 2(x + 3) = 14 is an equation. The left-hand side is 2(x + 3) and the right-hand side is 14. Solving an equation means finding the value of the unknown (usually x) that makes the statement true. In KS3, we focus on linear equations where the variable has an exponent of 1, meaning no x² or higher powers.

方程是使用等号’=’表示两个表达式相等的数学陈述。例如，2(x + 3) = 14 就是一个方程。左边是2(x + 3)，右边是14。解方程意味着找出使该陈述成立的未知数（通常是x）的值。在KS3阶段，我们重点学习线性方程，其中变量的指数为1，意味着没有x²或更高次幂。

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3. The Balancing Method | 平衡法

To solve any equation, you must always keep it balanced. Imagine an old-fashioned balance scale. Whatever operation you do to one side of the equation, you must do exactly the same to the other side. This maintains the equality. If you add, subtract, multiply or divide on one side, do the identical operation on the other side. This principle is the foundation of solving all linear equations.

求解任何方程，你必须始终保持其平衡。想象一台旧式天平秤。无论你对等式的一边做什么操作，都必须对另一边做完全相同的操作。这保持了等式的成立。如果你在一边加、减、乘或除，就要在另一边执行相同的运算。这条原则是解决所有线性方程的基础。

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4. Expanding Brackets | 去括号

When an equation contains brackets, the first step is usually to expand them using the distributive property. The distributive property states that a(b + c) = ab + ac. In our problem, the bracket is 2(x + 3). We multiply the 2 by every term inside the bracket: 2 × x gives 2x, and 2 × 3 gives 6. Therefore, 2(x + 3) expands to 2x + 6. The original equation 2(x + 3) = 14 now becomes 2x + 6 = 14. This is a simpler equation without brackets.

当方程含有括号时，第一步通常是用分配律去括号。分配律指出 a(b + c) = ab + ac。在我们的问题中，括号是2(x + 3)。我们将2乘以括号内的每一项：2 × x 得到 2x，2 × 3 得到 6。因此，2(x + 3) 展开为 2x + 6。原方程 2(x + 3) = 14 现在变为 2x + 6 = 14。这是一个没有括号的更简单方程。

A common mistake is to multiply only the first term. Always ensure the outside number multiplies both the x-term and the constant term inside the brackets. For practice, try expanding 3(2x – 5) on your own – you should get 6x – 15.

一个常见的错误是只乘第一项。务必确保括号外的数同时乘以括号内的x项和常数项。自己尝试展开 3(2x – 5)，应该得到 6x – 15。

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5. Step-by-Step Solution: p92 Question 1 | 逐步求解：第92页第1题

Now let’s solve the bracket-free equation 2x + 6 = 14 using the balancing method. We want to isolate x on one side. Observe the operations attached to x: it is multiplied by 2, and then 6 is added. We reverse these operations in the opposite order – first deal with the addition, then the multiplication. Subtract 6 from both sides to remove the constant term. Then divide both sides by 2 to undo the multiplication.

现在让我们用平衡法求解无括号的方程 2x + 6 = 14。我们希望将x单独留在等式一边。观察与x相关的运算：x先乘以2，然后加上6。我们按相反顺序逆操作——先处理加法，再处理乘法。两边减去6以消去常数项，然后两边除以2以撤销乘法。

Step	Equation	Operation
After expanding	2x + 6 = 14	Given: 2(x+3)=14
Subtract 6 from both sides	2x = 8	2x + 6 − 6 = 14 − 6
Divide both sides by 2	x = 4	2x ÷ 2 = 8 ÷ 2

The solution to the equation is x = 4. This means that if you substitute 4 back into the original bracket form, you will get a true statement. The step-by-step clearing of operations turns a seemingly tricky equation into two simple arithmetic moves.

方程的解是 x = 4。这意味着如果你把4代回原来的括号形式，你会得到一个成立的等式。逐步消去运算将看似复杂的方程变成了两个简单的算术步骤。

2(x + 3) = 14 → x = 4

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6. Checking Your Answer | 检验答案

Always verify your solution by substituting it into the original equation. This habit not only catches mistakes but deepens your understanding. For x = 4, the left-hand side becomes 2(4 + 3) = 2 × 7 = 14. The right-hand side is 14. Since both sides are equal, our solution is correct. If they did not match, you would know to retrace your steps.

一定要将解代回原方程进行验证。这个习惯不仅能捕捉错误，还能加深理解。当x=4时，左边变为 2(4 + 3) = 2 × 7 = 14。右边是14。因为两边相等，所以我们的解正确。如果不相等，你就能知道需要重新检查步骤。

You can also check mentally: ‘What number plus 3 gives 7 when multiplied by 2?’ Working backwards from 14: half of 14 is 7, and 7 minus 3 is 4. That matches our solution.

你也可以心算检验：“什么数加3后再乘以2等于14？”从14倒推：14的一半是7，7减3等于4。这与我们的解相符。

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7. Common Mistakes to Avoid | 常见错误

Even careful students slip up. Here are the most frequent errors when solving bracket equations, and how to steer clear of them.

即使是细心的学生也会犯错误。以下是求解括号方程时最常见的错误，以及如何避免它们。

• Forgetting to multiply both terms inside the bracket: Some write 2(x+3) = 2x+3 instead of 2x+6. Always apply the outside factor to every term inside. 忘记乘以括号内的所有项：有些人会把 2(x+3) 写成 2x+3，而不是 2x+6。一定要将外面的因数乘以括号内的每一项。
• Incorrect balancing: Subtracting 6 from only one side breaks the equality. Whatever you do to one side, you must do to the other. 平衡步骤错误：仅从一边减去6会破坏等式。对一边做什么，就必须对另一边做同样的操作。
• Division mishandling: When dividing 2x = 8, students might divide only the 8 by 2 and write x = 8, or miscalculate 8 ÷ 2. Double-check your arithmetic. 除法操作失误：当对 2x = 8 除以2时，学生可能只把8除以2而写成 x=8，或者算错 8÷2。再次检查你的算术。
• Not checking the answer: Skipping verification allows simple errors to go undetected. Always plug your answer back in. 不检验答案：跳过验证会让简单错误无法被发现。始终将答案代入检验。

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8. Equations with Variables on Both Sides | 两边含有变量的方程

In KS3, you will soon encounter equations that have x on both sides, such as 3(x − 1) = 2x + 4. The method is an extension of what we have just practised. First, expand the bracket: 3(x − 1) becomes 3x − 3. Now the equation reads 3x − 3 = 2x + 4. The next step is to collect the variable terms on one side. Subtract 2x from both sides to move all x-terms to the left: 3x − 2x − 3 = 4, which simplifies to x − 3 = 4. Then add 3 to both sides, giving x = 7. Always keep the balance and check: 3(7 − 1) = 3×6 = 18, and 2×7+4 = 18. It works!

在KS3阶段，你很快会遇到两边都含有x的方程，例如 3(x − 1) = 2x + 4。解题方法是我们刚才练习的延伸。首先去括号：3(x − 1) 变为 3x − 3。现在方程读作 3x − 3 = 2x + 4。下一步是将含有变量的项移到一边。两边减去2x，把所有x项移到左边：3x − 2x − 3 = 4，简化为 x − 3 = 4。然后两边加3，得到 x = 7。始终维持平衡并检验：3(7 − 1) = 3×6 = 18，而 2×7+4 = 18。完美解出！

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9. Real-World Application | 现实世界应用

Equations with brackets model many real-life situations. Imagine a rectangular garden where the length is 3 metres more than the width. If the perimeter is 14 metres, we can set up an equation. Let the width be x, then length is x + 3. The perimeter formula is 2(length + width) = 2(x + (x+3)) = 2(2x+3). Setting this equal to 14 gives 2(2x+3)=14. Divide both sides by 2 to get 2x+3=7, subtract 3, then divide by 2: x=2. So the width is 2 m and the length is 5 m. This is exactly the same structure as our page 92 problem.

带括号的方程可以模拟许多现实情形。想象一个矩形花园，其长度比宽度多3米。如果周长为14米，我们可以建立方程。设宽为x，则长为 x+3。周长公式为 2(长 + 宽) = 2(x + (x+3)) = 2(2x+3)。令其等于14，得到 2(2x+3)=14。两边除以2得 2x+3=7，减3，再除以2：x=2。所以宽为2米，长为5米。这与我们第92页的问题结构完全相同。

Seeing algebra in context helps you appreciate why these skills matter. Whether you are designing a floor plan or splitting a restaurant bill, linear equations are powerful tools.

在具体情境中

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📚 Mastering Linear Equations: A KS3 Cambridge Guide | 掌握线性方程：KS3剑桥数学指南

Equations are at the heart of algebra, and in the KS3 Cambridge Mathematics curriculum, they appear in many forms — from simple puzzles to multi-step problems involving brackets and fractions. Building a solid understanding of how to solve linear equations now will give you the tools you need for graphs, simultaneous equations, quadratic equations and real-world problem solving. This revision guide takes you through every essential skill, step by step, with clear examples, common pitfalls and plenty of opportunities to check your understanding.

方程是代数的核心。在 KS3 剑桥数学课程中，它们以多种形式出现——从简单的谜题到涉及括号和分数的多步骤问题。现在扎实地掌握如何解线性方程，将为你日后学习图形、联立方程、二次方程以及解决实际问题提供必要的工具。本复习指南将一步步带你掌握所有关键技能，配有清晰的例子、常见错误分析以及大量检查理解的机会。

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

A linear equation is a mathematical statement where two expressions are equal and the highest power of the variable is 1. You may see letters such as x, y or a representing the unknown number. The aim is always to find the value of the variable that makes the equation true. In KS3, you mostly work with one unknown, though later you will meet equations involving two variables.

线性方程是一个数学陈述，其中两个表达式相等，并且变量的最高次幂为 1。你可能会看到像 x、y 或 a 这样的字母代表未知数。目标始终是求出使方程成立的变量的值。在 KS3 阶段，你主要处理一个未知数，不过之后你会遇到含有两个变量的方程。

Example: 2x + 3 = 11

示例：2x + 3 = 11

The expression on the left (2x + 3) has the same value as the number on the right (11). We need to discover what x must be for this to be true. Linear equations can look different — sometimes the variable appears on both sides, sometimes brackets are involved — but the underlying principle is always the same.

左边的表达式 (2x + 3) 与右边的数字 (11) 具有相同的值。我们需要找出 x 必须是多少才能使该式成立。线性方程的形式可能各不相同——有时变量出现在两边，有时会包含括号——但其基本原理始终相同。

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2. The Balance Method | 平衡法

Imagine an old-fashioned pair of scales. When the scales are balanced, the contents of the left pan equal the contents of the right pan. An equation works in exactly the same way. The equality sign (=) is the balancing point. If we add, subtract, multiply or divide something on one side, we must do exactly the same to the other side, otherwise the equation will no longer be true.

想象一架老式天平。当天平平衡时，左边托盘中的内容物等于右边托盘中的内容物。方程就是以完全相同的方式运作的。等号 (=) 就是平衡点。如果我们在等式的一边进行加、减、乘或除某种操作，就必须在另一边执行完全相同的操作，否则方程将不再成立。

This idea is often called the ‘balance method’ and it is the foundation of all equation solving. Instead of guessing values, you systematically ‘undo’ operations around the variable until it is alone on one side. Keeping the balance visual in your mind helps avoid the common mistake of only performing an operation on one side.

这一想法通常被称为“平衡法”，它是所有解方程方法的基础。与其猜测数值，不如系统性地“撤销”围绕变量的运算，直到变量单独出现在等式的一边。在脑海中保持平衡的直观图像有助于避免只在一边执行操作的常见错误。

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3. Solving One-Step Equations | 解一步方程

The simplest linear equations require just one step to isolate the variable. For instance:

最简单的线性方程只需一步就能分离出变量。例如：

x + 5 = 12

We need to remove the ‘+ 5’ from the left-hand side. The inverse of adding 5 is subtracting 5. So we subtract 5 from both sides:

我们需要将左边的 “+ 5” 移除。加 5 的逆运算是减 5。因此我们从两边同时减去 5：

x + 5 − 5 = 12 − 5 → x = 7

For multiplication, consider:

对于乘法，考虑：

3x = 15

The variable is multiplied by 3. The inverse of multiplying by 3 is dividing by 3. Divide both sides by 3:

变量被乘以 3。乘以 3 的逆运算是除以 3。两边同时除以 3：

3x ÷ 3 = 15 ÷ 3 → x = 5

A subtraction example: x − 4 = 9. Add 4 to both sides: x = 13. A division example: x ÷ 2 = 6. Multiply both sides by 2: x = 12. Once you are comfortable with these, you can move on to equations that combine operations.

一个减法的例子：x − 4 = 9。两边加 4：x = 13。一个除法的例子：x ÷ 2 = 6。两边乘以 2：x = 12。一旦你熟练掌握了这些，就可以继续学习结合了多种运算的方程。

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4. Using Inverse Operations | 使用逆运算

Understanding inverse operations is the key to unlocking any equation. The table below summarises the most common pairs you will use throughout KS3.

理解逆运算是解开任何方程的关键。下表总结了你将在整个 KS3 阶段使用的最常见的运算对。

Operation	Inverse Operation
+ a	− a
− a	+ a
× a	÷ a
÷ a	× a

Notice that addition and subtraction undo each other, as do multiplication and division. When you face a more complex equation, identify the operations that have been applied to the variable, then undo them in the reverse order — just like unwrapping a parcel.

请注意，加法和减法互为逆运算，乘法和除法也是如此。当你面对一个更复杂的方程时，先识别出对变量施加了哪些运算，然后按照相反的顺序将它们撤销——就像拆开包裹一样。

This ‘reverse order’ becomes especially important in two-step equations. For instance, in 2x + 3 = 11, the variable is first multiplied by 2, then 3 is added. To solve, you first undo the addition (subtract 3), then undo the multiplication (divide by 2).

这种“相反的顺序”在两步方程中尤为重要。例如，在 2x + 3 = 11 中，变量先被乘以 2，然后加上 3。要解这个方程，你首先撤销加法（减去 3），然后撤销乘法（除以 2）。

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5. Solving Two-Step Equations | 解两步方程

Two-step equations involve two operations affecting the variable. A classic example is:

两步方程涉及对变量施加的两种运算。一个经典的例子是：

2x + 3 = 11

Step 1: Undo the addition of 3 by subtracting 3 from both sides. This leaves 2x = 8.

第 1 步：通过从两边减去 3 来撤销加 3。这样得到 2x = 8。

Step 2: Undo the multiplication by 2 by dividing both sides by 2. This gives x = 4.

第 2 步：通过两边除以 2 来撤销乘 2。得到 x = 4。

Always follow the order opposite to the one used when building the expression. If the equation is 5x − 7 = 13, the variable was multiplied by 5 and then 7 was subtracted. So first add 7 to both sides (5x = 20), then divide by 5 (x = 4).

始终按照与构建表达式时相反的顺序操作。如果方程是 5x − 7 = 13，变量先被乘以 5，然后减去 7。因此首先两边加 7（5x = 20），然后除以 5（x = 4）。

Practise with different numbers and signs, including negative coefficients. For example, 10 − 2y = 4 can be rewritten as −2y + 10 = 4; subtract 10, then divide by −2, giving y = 3. Keeping the steps precise and written clearly will prevent sign errors.

用不同的数字和符号进行练习，包括负系数。例如，10 − 2y = 4 可以重写为 −2y + 10 = 4；减去 10，然后除以 −2，得到 y = 3。保持步骤精确并书写清晰将防止符号错误。

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6. Equations with Variables on Both Sides | 两边都有变量的方程

In many KS3 problems, the variable appears on both sides of the equals sign. The strategy is to collect all variable terms on one side and all constant terms on the other. For example:

在许多 KS3 问题中，变量会出现在等号的两边。策略是将所有含有变量的项移到等式的一边，将所有常数项移到另一边。例如：

5x + 2 = 3x + 10

Start by eliminating the smaller variable term. Subtract 3x from both sides so that the variable terms are on the left:

首先消去较小的变量项。两边同时减去 3x，使变量项集中在左边：

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

Now you have a two-step equation. Subtract 2 from both sides to get 2x = 8, then divide by 2 to get x = 4.

现在你得到了一个两步方程。两边减去 2 得到 2x = 8，然后除以 2 得到 x = 4。

If you prefer, you could subtract 5x from both sides and work with negative coefficients, but choosing the side that keeps the variable positive is usually simpler. Always remember to carry the sign in front of the term when moving it.

如果你愿意，也可以从两边减去 5x 并处理负系数，但选择能使变量保持为正的一边通常更简单。始终记住，移动某一项时要带上它前面的符号。

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7. Expanding Brackets Before Solving | 先展开括号再求解

When an equation contains brackets, your first job is to multiply out those brackets using the distributive law. For example:

当方程含有括号时，你的首要任务是用分配律将括号展开。例如：

3(x + 2) = 21

Multiply the term outside the bracket by each term inside: 3 × x and 3 × 2. This gives:

将括号外的项与括号内的每一项相乘：3 × x 和 3 × 2。得到：

3x + 6 = 21

Now solve the resulting two-step equation. Subtract 6 from both sides: 3x = 15, then divide by 3: x = 5.

现在解所得的两步方程。两边减去 6：3x = 15，然后除以 3：x = 5。

If the equation has brackets on both sides, expand both first. For instance, 2(x − 4) = 3(x + 1) becomes 2x − 8 = 3x + 3. Then bring variable terms to one side: subtract 2x from both sides to get −8 = x + 3, then subtract 3 to obtain x = −11.

如果方程两边都有括号，则先两边都展开。例如，2(x − 4) = 3(x + 1) 变为 2x − 8 = 3x + 3。然后将含变量的项移到一边：两边减去 2x 得 −8 = x + 3，再减去 3 得到 x = −11。

Be careful with negative signs outside brackets: −2(y + 5) = −2y − 10. Expanding correctly is just as important as the solving steps that follow.

注意括号外的负号：−2(y + 5) = −2y − 10。正确地展开与后续的求解步骤同样重要。

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8. Equations with Fractions | 含分数的方程

Fractions in equations can look intimidating, but they follow the same balance rules. Consider:

方程中出现分数可能看起来令人生畏，但它们遵循同样的平衡法则。考虑：

x/4 + 1 = 3

First, subtract 1 from both sides to isolate the fractional term: x/4 = 2. Then multiply both sides by 4 to clear the denominator: x = 8.

首先，两边减去 1 以分离出含有分数的项：x/4 = 2。然后两边乘以 4 以消去分母：x = 8。

Alternatively, you can multiply every term in the equation by the denominator straight away. For the same equation, multiply everything by 4: 4 × (x/4) + 4 × 1 = 4 × 3 → x + 4 = 12. Then subtract 4 to get x = 8. Both methods work; choose the one that makes more sense to you.

或者，你也可以一开始就将方程中的每一项都乘以分母。对于同一个方程，将所有项都乘以 4：4 × (x/4) + 4 × 1 = 4 × 3 → x + 4 = 12。然后减去 4 得到 x = 8。两种方法都有效；选择对你来说更直观的那一种。

When there are two fractions with different denominators, the technique is to find a common denominator or multiply through by the lowest common multiple. For instance, (x/3) + (x/4) = 7. Multiply every term by 12: 4x + 3x = 84 → 7x = 84 → x = 12.

当有两个分母不同的分数时，技巧是找到公分母或乘以最小公倍数。例如，(x/3) + (x/4) = 7。将每一项乘以 12：4x + 3x = 84 → 7x = 84 → x = 12。

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9. Checking Your Solution | 检查你的答案

Finding a value for x does not mean the work is finished. You should always substitute your answer back into the original equation to verify that both sides produce the same number. This habit will catch many small mistakes with signs or arithmetic.

求出 x 的值并不意味着工作结束了。你应该始终将答案代回到原方程中，以验证两边得出相同的数值。这个习惯能帮你发现许多符号或算术上的小错误。

Using the earlier example 2x + 3 = 11 with x = 4: Left side = 2(4) + 3 = 8 + 3 = 11. Right side = 11. Both sides match, so the solution is correct.

用前面的例子 2x + 3 = 11，x = 4：左边 = 2(4) + 3 = 8 + 3 = 11。右边 = 11。两边相等，因此解是正确的。

If substitution gives different numbers, retrace your steps. Check whether you subtracted when you should have added, or whether a sign was lost when expanding. Often the error is in the first or second step. Checking is a powerful way to build confidence in your answers during a test.

如果代入后得到不同的数字，就重新检查你的步骤。看看你是否在本该加的时候减了，或者在展开时丢掉了符号。错误往往出现在第一步或第二步。检查是一种强有力的方法，能让你在考试中对答案充满信心。

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

Even students who understand the balance method can slip up on details. Here are four frequent errors:

即使是理解了平衡法的学生也可能在细节上出错。以下是四种常见错误：

• Forgetting to apply an operation to both sides: When you add 5 to the left, you must add 5 to the right as well.
  忘记对两边同时进行操作： 当你给左边加上 5 时，右边也必须加上 5。
• Ignoring negative signs: In 8 − x = 3, subtracting 8 gives −x = −5, so x = 5. Be careful when dividing by a negative.
  忽略负号： 在 8 − x = 3 中，减去 8 得到 −x = −5，因此 x = 5。除以负数时要小心。
• Expanding brackets incorrectly: −2(x − 3) is −2x + 6, not −2x − 6.
  括号展开错误： −2(x − 3) 是 −2x + 6，而不是 −2x − 6。
• Mishandling fractions: When multiplying (2/3)x = 8, multiply by the reciprocal 3/2, not just 2.
  分数处理不当： 当 (2/3)x = 8 时，要乘以其倒数 3/2，而不只是 2。

Slow down, show every step on a new line, and read the equation aloud if it helps. Precision matters more than speed when you are learning.

放慢速度，每一步都另起一行，并在必要时把方程读出声来。

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📚 KS3 Cambridge Maths: Page 74 Q1 – Solving Equations with Variables on Both Sides | KS3 剑桥数学：第74页第1题 – 解含双边变量的方程

In the Cambridge KS3 curriculum, pupils encounter algebraic equations where the unknown appears on both sides of the equals sign. A typical example is Question 1 on page 74 of the practice book: Solve 5x – 2 = 3x + 6. This article walks through the method in detail and builds up to more challenging variations, helping students develop confidence in solving linear equations.

在剑桥 KS3 课程中，学生会遇到等号两边都含有未知数的代数方程。练习册第 74 页第 1 题就是一个典型例子：解方程 5x – 2 = 3x + 6。本文将详细讲解求解方法，并逐步深入到更具挑战性的变体，帮助学生建立解方程的信心。

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1. Understanding the Problem | 理解问题

Before we start, observe that the left side contains the term 5x and a constant –2, while the right side contains 3x and +6. The goal is to find the single value of x that makes both sides equal. We need to move the x-terms together, keeping the equation balanced at every step.

在开始之前，先观察左侧有 5x 和常数项 –2，右侧有 3x 和 +6。目标是找出使两边相等的唯一 x 值。我们需要把含 x 的项移到一起，同时每一步都要保持等式平衡。

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2. Step 1: Eliminate the Variable from One Side | 第1步：将变量从一边消去

Subtract 3x from both sides to remove the x-term from the right. This uses the balance method:
5x – 2 – 3x = 3x + 6 – 3x
This simplifies to 2x – 2 = 6. Now the variable appears only on the left.

两边同时减去 3x，消去右边的含 x 项。这就运用了天平法：
5x – 2 – 3x = 3x + 6 – 3x
化简得 2x – 2 = 6。此时变量只出现在左边。

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3. Step 2: Remove the Constant Term | 第2步：移走常数项

Add 2 to both sides to isolate the term containing x. The equation becomes:
2x – 2 + 2 = 6 + 2
which simplifies to 2x = 8.

两边同时加 2，将含 x 的项单独留在左边。方程变为：
2x – 2 + 2 = 6 + 2
化简得 2x = 8。

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4. Step 3: Solve for x | 第3步：解出 x

Divide both sides by the coefficient of x. Here the coefficient is 2, so we divide by 2:
2x ÷ 2 = 8 ÷ 2
Therefore, x = 4.

两边同时除以 x 的系数。这里系数是 2，所以两边除以 2：
2x ÷ 2 = 8 ÷ 2
得到 x = 4。

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5. Checking Your Answer | 检验答案

Always substitute your solution back into the original equation. Left side: 5(4) – 2 = 20 – 2 = 18. Right side: 3(4) + 6 = 12 + 6 = 18. Both sides equal 18, so x = 4 is correct.

一定要把解代回原方程检验。左边：5(4) – 2 = 20 – 2 = 18。右边：3(4) + 6 = 12 + 6 = 18。两边都等于 18，所以 x = 4 是正确解。

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6. Equations with Negative Coefficients | 系数为负的方程

Consider –2x + 5 = x + 11. To avoid errors, add 2x to both sides to keep the coefficient of x positive: 5 = 3x + 11. Then subtract 11 from both sides: –6 = 3x, so x = –2.

考虑 –2x + 5 = x + 11 这样的方程。为了避免出错，可以同时在两边加 2x，使 x 的系数为正：5 = 3x + 11。然后两边减 11：–6 = 3x，得 x = –2。

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7. Equations Involving Brackets | 含括号的方程

If the equation includes brackets, expand them first. For example, 3(2x – 1) = 4x + 5 becomes 6x – 3 = 4x + 5. Then subtract 4x from both sides: 2x – 3 = 5. Add 3: 2x = 8, so x = 4.

如果方程含有括号，先展开。例如，3(2x – 1) = 4x + 5 变成 6x – 3 = 4x + 5。然后两边减 4x：2x – 3 = 5。加 3：2x = 8，得 x = 4。

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

When fractions appear, multiply every term by the lowest common denominator. For (x/2) + 3 = (x/4) + 5, multiply through by 4: 2x + 12 = x + 20. Then solve: 2x – x = 20 – 12, giving x = 8.

当出现分数时，可以用最小公分母乘以每一项。对于 (x/2) + 3 = (x/4) + 5，两边乘 4：2x + 12 = x + 20。再解：2x – x = 20 – 12，得 x = 8。

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

Watch out for these frequent pitfalls when solving equations with variables on both sides.

解答变量在两侧的方程时，要注意以下常见错误。

Common Mistake 常见错误	Why It Happens 原因	How to Avoid 如何避免
Forgetting to change the sign when moving a term across the equals sign. 移项时忘记变号	Inverse operations are not applied consistently. 逆运算运用不一致	Always perform the same operation (add, subtract, multiply, divide) on both sides. 始终对等式两边执行相同的运算。
Losing a negative sign when collecting like terms. 合并同类项时丢掉负号	Subtracting a negative term is confused. 减去一个负数项容易混淆	Re-write the step clearly: e.g. –2x + 7 = 3x becomes 7 = 5x. 清晰地写出步骤：如 –2x + 7 = 3x 变为 7 = 5x。
Only moving the variable, but not the constant. 只移变量项，不动常数	Learners focus only on the x-terms and forget to balance constants. 学生只关注 x 项而忘记平衡常数	Treat the equation like a balanced scale; every term must be moved correctly. 把方程当作天平，每一项都要正确移动。

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10. Practice Questions with Worked Solutions | 练习题及解答步骤

Try these equations to strengthen your skills.

尝试解下列方程，强化你的技巧。

• Question 1: 7x + 4 = 3x + 20
  Subtract 3x: 4x + 4 = 20. Subtract 4: 4x = 16. x = 4.

  第1题：7x + 4 = 3x + 20
  减 3x：4x + 4 = 20。减 4：4x = 16，x = 4。

• Question 2: 2(x – 5) = x + 3
  Expand: 2x – 10 = x + 3. Subtract x: x – 10 = 3. Add 10: x = 13.

  第2题：2(x – 5) = x + 3
  展开：2x – 10 = x + 3。减 x：x – 10 = 3。加 10：x = 13。

• Question 3: 5x – 9 = –2x + 12
  Add 2x: 7x – 9 = 12. Add 9: 7x = 21. x = 3.

  第3题：5x – 9 = –2x + 12
  加 2x：7x – 9 = 12。加 9：7x = 21，x = 3。

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11. Why Mastering These Equations Matters | 掌握这些方程的重要性

Solving equations with variables on both sides is a cornerstone of algebra. It trains logical thinking and prepares you for more advanced topics such as simultaneous equations, inequalities, and functions. In KS3 assessments, you will often see these questions in both non-calculator and calculator papers.

解两边都有变量的方程是代数的基石。它训练逻辑思维，并为后续更高级的课题——如联立方程、不等式和函数——打下基础。在 KS3 考核中，无论是使用计算器还是不允许使用计算器的试卷里，这类题目都经常出现。

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12. Tips for Exam Success | 考试通关建议

• Always write each step clearly. Clear working helps you spot mistakes and earns method marks.

  每一步都要写清楚。清晰的演算有助于发现错误，也能拿到步骤分。

• Check your answer by substitution, especially if time allows. It is the only way to be certain.

  时间允许的话，一定要用代入法检验答案。这是唯一能确定的办法。

• If you get stuck, try adding or subtracting x-terms so that the variable ends up on the side where the coefficient is positive. This often reduces sign errors.

  如果卡住了，可以尝试把 x 项移到系数为正的一边。这常常能减少符号错误。

• Practice with negatives and fractions regularly—they appear frequently in exam papers.

  经常练习带负数和分数的方程——它们在试卷中出现频率很高。

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📚 Solving Linear Equations: KS3 Cambridge Maths p36 Worked Example | 解线性方程：剑桥KS3数学第36页例题解析

Linear equations are a fundamental part of the Cambridge KS3 mathematics curriculum. Being able to solve equations like the one on page 36, question 1, builds the foundation for algebra, problem solving, and later topics such as graphs and inequalities. In this article, we will work through the step-by-step solution of the equation 3(2x – 1) = 4x + 5, check the answer, discuss common mistakes, and explore related problems.

线性方程是剑桥初中数学课程的核心内容。能够解出类似第36页第1题的方程，为代数、问题解决以及后续图像与不等式等课题打下坚实基础。本文将逐步解析方程 3(2x – 1) = 4x + 5，验证答案，讨论常见错误，并探讨相关问题。

1. The Problem | 问题呈现

We are asked to solve the equation 3(2x – 1) = 4x + 5 from page 36, exercise 1. The target is to find the value of x that makes both sides equal.

题目要求解第36页练习1中的方程 3(2x – 1) = 4x + 5。目标是找出使等式两边相等的 x 值。

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2. Expanding Brackets | 展开括号

The first step is to expand the bracket on the left-hand side. Multiply 3 by each term inside: 3 × 2x = 6x and 3 × (–1) = –3. This gives 6x – 3. So the equation becomes 6x – 3 = 4x + 5.

第一步是展开左侧的括号。将 3 乘以括号内的每一项：3 × 2x = 6x，3 × (–1) = –3，得到 6x – 3。于是方程变为 6x – 3 = 4x + 5。

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📚 Mastering Linear Equations: KS3 Cambridge Mathematics | 掌握一元一次方程：KS3剑桥数学

Linear equations are the first major step into algebra for most KS3 students following the Cambridge curriculum. These equations allow us to find unknown values and are used in countless real‑life situations, from calculating mobile phone bills to working out travel times. In this revision guide, we will cover all the key skills you need: understanding what an equation is, mastering the balancing method, and tackling equations with brackets, fractions, and variables on both sides. By working through each section carefully, you will build a solid foundation for your Checkpoint tests and beyond.

线性方程是大多数遵循剑桥课程的KS3学生迈入代数的第一大步。这些方程让我们能够求出未知数，并用于无数现实生活情境中，从计算手机账单到算出路途时间。在本复习指南中，我们会涵盖你需要掌握的所有关键技能：理解方程是什么，掌握天平法，并解决带括号、分数以及变量在方程式两边的方程。通过仔细完成每一节的学习，你将为Checkpoint考试以及更高年级打下扎实的基础。

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

A linear equation is a mathematical statement that shows two expressions are equal. In KS3, we focus on equations with one unknown, usually written as a letter such as x or y. The equation will only contain powers of the unknown no greater than 1, which means we do not see terms like x². For example, 3x + 2 = 14 is a linear equation. The goal is to find the value of the unknown that makes the equation true.

线性方程是一个表明两个表达式相等的数学陈述。在KS3阶段，我们聚焦于含有一个未知数的方程，通常用x或y这样的字母表示。方程中未知数的幂次最高只能是1，这意味着我们不会看到像x²这样的项。例如，3x + 2 = 14是一个线性方程。我们的目标就是找出使方程成立的未知数的值。

It is important to recognise that the equals sign acts as a balance: whatever is on the left side must have exactly the same value as whatever is on the right side. When solving, we must keep this balance at all times. If we add a number to one side, we must add the same number to the other.

要认识到等号起着天平的作用：左边的一切必须与右边的一切具有完全相同的值。在解方程时，我们必须始终保持这一平衡。如果我们给一边加上一个数，就必须给另一边也加上同样的数。

2. The Balancing Method | 天平法

The balancing method is the most reliable way to solve linear equations. Imagine a set of old‑fashioned balance scales. Whatever operation we perform on one side (adding, subtracting, multiplying, or dividing), we must perform the same operation on the other side to keep the scales level. This idea is often expressed as ‘do the same to both sides’.

天平法是解线性方程最可靠的方法。想象一副老式天平秤。无论我们对一边进行什么运算（加、减、乘或除），我们都必须对另一边进行完全相同的运算，以保持秤的平衡。这个思路通常表述为“对两边做同样的事”。

Let’s start with a simple example: x + 5 = 13. To isolate x, we need to remove the ‘+5’ from the left side. The inverse operation of adding 5 is subtracting 5. So we subtract 5 from both sides:

让我们从一个简单的例子开始：x + 5 = 13。为了把x分离出来，我们需要从左边去掉“+5”。加5的逆运算是减5。于是我们从两边减去5：

x + 5 − 5 = 13 − 5 → x = 8

Always check your answer by substituting it back into the original equation: 8 + 5 = 13, which is correct.

一定要将答案代回原方程进行检验：8 + 5 = 13，正确。

3. Solving One‑Step Equations | 解单步方程

One‑step equations require just one operation to isolate the variable. The operation depends on what is joined to the variable. For an equation such as x − 7 = 9, the variable has 7 subtracted from it. The inverse is addition, so we add 7 to both sides: x − 7 + 7 = 9 + 7, giving x = 16. For multiplication, such as 5x = 30, the inverse is division, so we divide both sides by 5, giving x = 6.

单步方程只需要进行一次运算就能把变量分离出来。进行什么运算取决于变量与什么相连接。对于像x − 7 = 9这样的方程，变量被减去了7。逆运算是加法，所以我们给两边加上7：x − 7 + 7 = 9 + 7，得出x = 16。对于乘法，比如5x = 30，逆运算是除法，所以两边同时除以5，得到x = 6。

When the variable is divided, as in x / 4 = 8, the inverse operation is multiplication. Multiply both sides by 4: (x / 4) × 4 = 8 × 4, so x = 32. It is essential to remember that division can also be written with a fraction bar: x/4 means x divided by 4.

当变量被除时，如x / 4 = 8，逆运算是乘法。给两边同乘4：(x / 4) × 4 = 8 × 4，得到x = 32。记住，除法也可以用分数线表示：x/4表示x除以4，这一点很重要。

4. Solving Two‑Step Equations | 解两步方程

A two‑step equation requires two inverse operations. Consider 2x + 3 = 11. The operations applied to x are ‘multiply by 2’ and then ‘add 3’. To solve, we reverse the order: first undo the addition, then undo the multiplication. So subtract 3 from both sides: 2x + 3 − 3 = 11 − 3, leaving 2x = 8. Then divide both sides by 2: 2x / 2 = 8 / 2, so x = 4.

两步方程需要进行两次逆运算。考虑2x + 3 = 11。对x进行的运算是“乘以2”然后“加3”。解方程时，我们倒过来做：先抵消加法，再抵消乘法。因此，先从两边减去3：2x + 3 − 3 = 11 − 3，剩下2x = 8。然后两边除以2：2x / 2 = 8 / 2，得到x = 4。

If the subtraction comes first, such as 5y − 7 = 18, start by adding 7 to both sides: 5y = 25, then divide by 5 to get y = 5. The order of undoing operations is crucial: sometimes students mistakenly divide first before dealing with the addition or subtraction, which leads to errors.

如果先出现减法，例如5y − 7 = 18，先从两边加7：5y = 25，然后除以5得到y = 5。解除运算的顺序至关重要：有时学生会错误地先做除法，再处理加减法，这就会导致错误。

5. Equations with Brackets | 带括号的方程

When brackets appear, we usually expand (multiply out) them first using the distributive law. For example, 3(x + 2) = 21. Expand the bracket to 3x + 6 = 21. Now it is a two‑step equation: subtract 6 from both sides to get 3x = 15, then divide by 3 to find x = 5.

当出现括号时，我们通常先用分配律展开（乘开）。例如，3(x + 2) = 21。把括号展开成3x + 6 = 21。现在它成了一个两步方程：从两边减6得到3x = 15，然后除以3求出x = 5。

Sometimes you may see a bracket with a subtraction, such as 4(2a − 3) = 20. Expand it as 8a − 12 = 20, then add 12 to both sides to make 8a = 32, and finally divide by 8 to get a = 4. Always be careful with negative signs inside brackets: a common mistake is to forget to multiply the negative term correctly.

有时你可能会看到带有减法的括号，如4(2a − 3) = 20。把它展开为8a − 12 = 20，然后两边加12得到8a = 32，最后除以8得出a = 4。一定要小心括号里的负号：一个常见错误是忘记正确乘上负数项。

If the equation has a negative multiplier outside the bracket, such as −2(p + 5) = 8, expand to −2p − 10 = 8, add 10 to get −2p = 18, then divide by −2 to obtain p = −9. The signs can be tricky, so work step by step.

如果方程在括号外有负的乘数，例如−2(p + 5) = 8，展开得−2p − 10 = 8，加10得到−2p = 18，然后除以−2得到p = −9。符号可能有些棘手，所以要一步步来做。

6. Equations with the Variable on Both Sides | 变量在方程式两边的方程

So far, the variable has been on only one side. When we have equations like 5x + 4 = 3x + 12, we need to collect the variable terms on one side and the numbers on the other. Aim to have the larger coefficient of x on the left‑hand side to keep numbers positive. Subtract 3x from both sides: 5x − 3x + 4 = 3x − 3x + 12, giving 2x + 4 = 12. Then subtract 4: 2x = 8, so x = 4.

到目前为止，变量都只出现在方程的一边。当我们遇到像5x + 4 = 3x + 12这样的方程时，需要把含变量的项集中到一边，把数字集中到另一边。争取让x系数较大的一边留在左边，以保持数字为正。从两边减去3x：5x − 3x + 4 = 3x − 3x + 12，得到2x + 4 = 12。然后减去4：2x = 8，所以x = 4。

If the variable terms are on the right and the constant on the left, such as 7 = 2y − 3, you can either add 3 to both sides and then divide, or swap the sides entirely: 2y − 3 = 7 is easier to solve. Remember that equations are symmetrical: if a = b then b = a.

如果变量项在右边而常数在左边，比如7 = 2y − 3，你可以先给两边加3再除以，或者直接把两边交换：2y − 3 = 7会更容易解。要记住方程是对称的：若a = b，则b = a。

Another example: 6n + 1 = 4n + 9. Subtract 4n from both sides to get 2n + 1 = 9, subtract 1 to get 2n = 8, and divide to find n = 4. Always check: 6×4 + 1 = 25, 4×4 + 9 = 25, both match.

另一个例子：6n + 1 = 4n + 9。从两边减4n得到2n + 1 = 9，减1得到2n = 8，除以后得出n = 4。始终检验：6×4 + 1 = 25, 4×4 + 9 = 25，两边相等。

7. Equations Involving Fractions | 涉及分数的方程

Fractions can make equations look more complicated, but they can be cleared using the lowest common denominator (LCD). For example, in x/3 + 2 = 5, you could subtract 2 first to get x/3 = 3, then multiply by 3 to find x = 9. However, for something like (2x + 1)/5 = 7, multiply both sides by 5 to clear the fraction: 2x + 1 = 35, then subtract 1 and divide by 2 to get x = 17.

分数可能会使方程看起来更复杂，但可以通过最小公分母（LCD）来去掉分母。例如，在x/3 + 2 = 5中，你可以先减去2得到x/3 = 3，然后乘以3求出x = 9。然而，对于像(2x + 1)/5 = 7这样的方程，可以两边同乘5来去掉分数：2x + 1 = 35，然后减1再除以2得到x = 17。

When there are multiple fractions, such as x/2 + x/4 = 9, find the LCD of 2 and 4, which is 4. Multiply every term by 4: 4(x/2) + 4(x/4) = 4×9, giving 2x + x = 36. Simplify to 3x = 36, so x = 12. Always reduce the equation to a form without fractions before using the balancing method.

当有多个分数时，比如x/2 + x/4 = 9，找出2和4的最小公分母，即4。给每一项都乘以4：4(x/2) + 4(x/4) = 4×9，得到2x + x = 36。化简为3x = 36，所以x = 12。在使用天平法之前，总是先把方程转化为没有分母的形式。

8. Testing and Verifying Solutions | 检验答案

Substituting your solution back into the original equation is not just good practice – it is a vital step to catch arithmetic mistakes. Take your value for x and replace every x in the equation with that value. Evaluate both sides separately. If they are equal, your solution is correct. For instance, solving 3(x − 2) = x + 10 gives x = 8. Check: left side 3(8 − 2) = 3×6 = 18; right side 8 + 10 = 18. They match.

将你的答案代回原方程不仅是一种好习惯——它还是能发现计算错误的关键步骤。把你求出的x值带入方程中的每一个x。分别计算两边。如果两边相等，你的答案就是正确的。例如，解3(x − 2) = x + 10得到x = 8。检验：左边3(8 − 2) = 3×6 = 18；右边8 + 10 = 18。两边一致。

If the two sides do not give the same number, you have made a mistake. Go back through your working line by line. Look for common errors such as forgetting to multiply both terms inside a bracket, mixing up the sign of a term when moving it across the equals sign, or an arithmetic slip while simplifying.

如果两边得出的数字不同，就说明你出错了。从头逐行检查你的解题过程。寻找常见错误，比如忘记乘上括号里的每一项、在把项移到等号另一边时弄错符号，或者化简时出现的简单算术错误。

9. Common Mistakes to Avoid | 需要避免的常见错误

One of the biggest mistakes is forgetting to apply an operation to the whole side. For example, when solving x/5 = 4, some students incorrectly subtract 5 instead of multiplying by 5. Remember the inverse of division is multiplication, not subtraction. Another frequent error is mishandling negative signs: when expanding −2(3 − y), you must get −6 + 2y, not −6 − 2y, because −2 × −y = +2y.

最大的一个错误是忘记对整边进行运算。例如，在解x/5 = 4时，有些学生错误地减去5而不是乘5。要记住，除法的逆运算是乘法，而不是减法。另一个常见错误是符号处理不当：展开−2(3 − y)时，你必须得到−6 + 2y，而不是−6 − 2y，因为−2 × −y = +2y。

Many students also change the order of operations incorrectly. In the equation 2x/3 = 10, you should multiply by 3 first (giving 2x = 30) before dividing by 2. Multiplying both sides by 3 and then dividing by 2 at the same time is fine, but writing 2x = 10 × 3 and then forgetting to divide is a slip. Careful, systematic working prevents such errors.

很多学生还会弄错运算的顺序。在方程2x/3 = 10中，你应该先乘以3（得到2x = 30），再除以2。同时乘3再除以2是可以的，但写下2x = 10 × 3后忘记除以就是疏忽。仔细、有条不紊的解题能避免此类错误。

10. Solving Problems that Lead to Equations | 解应用题并转化为方程

Many real‑world problems can be turned into linear equations. The key is to read the question carefully, define the unknown, and build the equation from the description. For example: ‘I think of a number, multiply it by 4, add 7, and get 31.’ Let the number be n. The equation is 4n + 7 = 31. Solve: 4n = 24, so n = 6.

很多现实世界的问题都可以转化为线性方程。关键在于仔细读题，设出未知数，然后根据描述建立方程。例如：“我想一个数，把它乘以4，再加7，得到31。”设这个数为n。方程就是4n + 7 = 31。求解：4n = 24，所以n = 6。

Word problems often involve age, money, or measurements. For instance, ‘A rectangle has length (2x + 3) cm and width 5 cm. Its perimeter is 36 cm. Find x.’ The perimeter is 2(length + width): 2((2x + 3) + 5) = 36. Simplify inside: 2x + 8, then 2(2x + 8) = 36, so 4x + 16 = 36, 4x = 20, x = 5. Always check the dimensions make sense.

文字题经常会涉及年龄、金钱或度量。例如：“一个矩形的长为(2x + 3) cm，宽为5 cm。它的周长是36 cm。求x。”周长等于2×(长+宽)：2((2x + 3) + 5) = 36。化简括号内：2x + 8，然后2(2x + 8) = 36，所以4x + 16 = 36，4x = 20，x = 5。始终检查各边尺寸是否合理。

11. Building Confidence with Practice | 通过练习建立信心

Becoming fluent in solving linear equations requires regular practice. Start with one‑step equations until you can do them mentally, then gradually include two‑step, brackets, and unknowns on both sides. Use a notebook to organise your working neatly – this makes it easier to spot errors. Many Cambridge Checkpoint questions mix these types, so practising a variety is essential.

要熟练地解线性方程，需要定期练习。从单步方程开始，直到能心算出来，然后再逐渐加入两步方程、带括号以及变量在方程两边的类型。用笔记本把解题过程整理得工整一些——这样更容易发现错误。很多剑桥Checkpoint考题会混合这些题型，所以练习多种类型十分关键。

When revising, try to create your own problems or explain your working aloud. If you can teach the method to someone else, you have truly understood it. Use online interactive quizzes or worksheets to test yourself under timed conditions. Remember, speed will come with confidence – accuracy should always be your first goal.

复习时，尝试自己出题或者把解题步骤大声讲出来。如果你能把方法教给别人，就说明你真的理解了。利用在线互动测验或练习卷在限时条件下进行自测。记住，速度会随着信心而来——准确性应当始终是你的首要目标。

12. Summary of Key Steps | 关键步骤总结

To solve any linear equation: first, simplify each side if possible by expanding brackets and combining like terms. Second, collect all the variable terms on one side and all the constant terms on the other using inverse operations. Third, isolate the variable by using inverse operations in the correct order (multiplication/division last). Finally, substitute your solution back into the original equation to verify.

解任何线性方程：首先，尽可能化简两边，展开括号并合并同类项。第二，通过逆运算把所有含变量的项集中到一边，把所有常数项集中到另一边。第三，按正确顺序运用逆运算（乘除放到最后）将变量分离出来。最后，将答案代回原方程进行检验。

Keep this checklist handy whenever you are solving equations. It works for all KS3 linear equations, from the simplest to the most complex. With consistent effort and the methods outlined in this guide, you will master linear equations and be ready for the algebra challenges ahead in the Cambridge curriculum.

无论何时解方程，都把这份核查清单放在手边。它适用于所有KS3线性方程，从最简单到最复杂的。通过持续的努力和本指南所总结的方法，你将掌握一元一次方程，并为剑桥课程中后续的代数学挑战做好准备。

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📚 Cambridge KS3 Maths – Page 179: Probability and Tree Diagrams | 剑桥初中数学 – 第179页：概率与树状图

Probability helps us measure how likely an event is to happen. On page 179 of the Cambridge KS3 Mathematics course, you will explore how to calculate probabilities, use probability scales, and draw tree diagrams to represent multiple outcomes clearly. Tree diagrams are powerful tools for solving combined event problems, such as flipping two coins or pulling coloured socks from a drawer. Mastering these skills will build a strong foundation for IGCSE and beyond.

概率帮助我们衡量事件发生的可能性。在剑桥初中数学课程的第 179 页中，你将学习如何计算概率、使用概率尺度以及绘制树状图来清晰地表示多重结果。树状图是解决组合事件问题（例如抛两枚硬币或从抽屉里取袜子）的强大工具。掌握这些技能将为 IGCSE 及更高阶段的学习打下坚实基础。

1. What Is Probability? | 什么是概率？

Probability is a number between 0 and 1 that describes the chance of an event occurring. A probability of 0 means the event is impossible, while a probability of 1 means it is certain. Most events have probabilities somewhere in between.

概率是一个介于 0 和 1 之间的数字，描述某个事件发生的可能性。概率为 0 表示事件不可能发生，概率为 1 表示事件必然发生。大多数事件的概率介乎两者之间。

We can write probability as a fraction, decimal, or percentage. For example, when flipping a fair coin, the probability of getting heads is 1/2, 0.5, or 50%. The notation P(heads) = 1/2 is commonly used.

我们可以用分数、小数或百分数来表示概率。例如，抛一枚公平硬币时，得到正面的概率是 1/2、0.5 或 50%。常用记法为 P(正面) = 1/2。

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

The sample space is the set of all possible outcomes of an experiment. For a single dice roll, the sample space is {1, 2, 3, 4, 5, 6}. Each individual result is called an outcome. Understanding the sample space is the first step in calculating any probability.

样本空间是某个实验所有可能结果的集合。对于掷一个骰子，样本空间为 {1, 2, 3, 4, 5, 6}。每一个单独的结果称为一个“结果”。理解样本空间是计算任何概率的第一步。

For combined events, such as rolling two dice, you can create a sample space diagram (grid) to list all 36 possible pairs. This organized approach prevents missing outcomes and ensures accurate probability calculations.

对于组合事件，例如掷两个骰子，你可以绘制样本空间图（网格）列出全部 36 个可能的数对。这种有条理的方法可以防止遗漏结果，确保概率计算准确。

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3. The Probability Scale | 概率尺度

The probability scale is a visual line from 0 to 1. Marking events on this line helps you compare their likelihoods. Words like ‘impossible’, ‘unlikely’, ‘evens’, ‘likely’, and ‘certain’ correspond to ranges on the scale.

概率尺度是一条从 0 到 1 的视觉化线段。将事件标注在该线上有助于比较它们的可能性大小。“不可能”、“不太可能”、“等可能”、“很可能”、“必然”等词语对应尺度上的不同区间。

For instance, the chance of the sun rising tomorrow is almost 1 (certain), while the chance of rolling a 7 on a standard dice is 0 (impossible). A 50% chance lies exactly in the middle, often called an even chance.

例如，明天太阳升起的概率几乎为 1（必然），而掷标准骰子得到 7 的概率为 0（不可能）。50% 的概率恰好位于中间，常被称为等可能的概率。

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4. Calculating Basic Probabilities | 计算基本概率

The basic probability formula is: P(event) = number of favourable outcomes / total number of possible outcomes. This only applies when all outcomes are equally likely. It is essential to count outcomes systematically.

基本的概率公式为：P(事件) = 有利结果的数量 / 所有可能结果的总数。这仅在所有结果等可能时成立。系统性地清点结果是十分重要的。

In a bag with 3 red pens and 5 blue pens, the probability of picking a red pen at random is 3 / 8. The total outcomes are 8, and favourable ones are 3. Always simplify fractions if possible: 3/8 is already in simplest form.

在一个装有 3 支红笔和 5 支蓝笔的袋子里，随机抽到红笔的概率是 3/8。总结果数为 8，有利结果数为 3。如果可能，记得化简分数：3/8 已是最简形式。

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5. Complementary Events | 互补事件

The complement of an event A is the event that A does not happen, written as A’ or not A. The probabilities of an event and its complement always add up to 1: P(A) + P(not A) = 1. This rule saves time when calculating ‘at least one’ style problems.

事件 A 的补事件是指 A 不发生的那个事件，记作 A’ 或非 A。任一事件与其补事件的概率之和总是 1：P(A) + P(非 A) = 1。在计算“至少一个”这类问题时，这一规则能节省时间。

If the probability of rain tomorrow is 0.3, then the probability of no rain is 1 – 0.3 = 0.7. Using complements often avoids adding many separate probabilities.

如果明天下雨的概率是 0.3，那么不下雨的概率就是 1 – 0.3 = 0.7。利用补事件常常可以避免将多个单独概率相加。

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6. Introducing Tree Diagrams | 树状图入门

A tree diagram is a branching structure that shows all possible outcomes of a sequence of events. Each branch represents an outcome and is labelled with its probability. Tree diagrams are especially useful for two or more stages.

树状图是一种分支结构，展示一连串事件的所有可能结果。每一分支代表一个结果，并标有其概率。树状图对于涉及两个或两个以上阶段的问题尤其有用。

To draw a tree diagram, start with a single point, then draw branches for each possible outcome of the first event. From the end of each of those branches, draw branches for the second event, and continue if there are more stages.

绘制树状图时，从一个点出发，为第一个事件的每种可能结果画出分支。再从这些分支末端，为第二个事件画出分支；如有更多阶段则继续延伸。

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7. Tree Diagram for Independent Events | 独立事件的树状图

Two events are independent if the outcome of one does not affect the outcome of the other. Flipping a coin and rolling a dice are independent events. On a tree diagram, the probabilities on the second set of branches remain the same regardless of the first outcome.

如果一件事的结果不影响另一件事的结果，那么这两个事件就是独立的。抛硬币和掷骰子是独立事件。在树状图上，第二组分支的概率不会因第一组结果的不同而改变。

For a coin flip followed by a dice roll, the coin has branches H (1/2) and T (1/2). From each, draw six dice branches, each with probability 1/6. The probability of any combined outcome, like H and 5, is 1/2 × 1/6 = 1/12.

对于先抛硬币再掷骰子的情况，硬币有两支分支 H（1/2）和 T（1/2）。从每支再画出六支骰子分支，每支概率 1/6。任何组合结果（例如 H 和 5）的概率为 1/2 × 1/6 = 1/12。

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8. Tree Diagram for Dependent Events | 相关事件的树状图

Events are dependent when the outcome of the first event changes the probability of the second event. Picking items from a bag without replacing them is a classic example. The probabilities on the second branches must be updated according to what has already happened.

如果第一件事的结果改变了第二件事的概率，那么这些事件是相关的。不放回地从袋子中抽取物品就是一个典型的例子。第二组分支的概率必须根据已经发生的情况进行调整。

Imagine a bag with 2 red and 3 green marbles. If you pick a red first and do not replace it, the bag now has 1 red and 3 green left, so the probability of red on the second pick becomes 1/4, not 2/5.

假设一个袋子里有 2 个红色弹珠和 3 个绿色弹珠。如果你先抽到一个红色弹珠且不放回，袋子里就剩下 1 红 3 绿，因此第二次抽到红色弹珠的概率变为 1/4，而不是 2/5。

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9. Using Tree Diagrams to Find Probabilities | 利用树状图求概率

To find the probability of a sequence of outcomes, multiply the probabilities along the path. To find the probability of an event that can happen in more than one way, calculate each path probability and then add them together. This is the ‘multiply along, add across’ rule.

要找到一系列结果的概率，将路径上的各概率相乘。若一个事件可以通过多种路径发生，则分别计算每条路径的概率，然后将它们相加。这就是“沿路径相乘，路径间相加”的规则。

For example, if you flip two coins, the probability of getting exactly one head is found by looking at the paths HT and TH. Each path probability is 1/2 × 1/2 = 1/4, so total P(exactly one head) = 1/4 + 1/4 = 1/2.

例如，抛两枚硬币时，恰好出现一次正面的概率通过观察 HT 和 TH 两条路径求得。每条路径概率为 1/2 × 1/2 = 1/4，因此总 P(恰好一次正面) = 1/4 + 1/4 = 1/2。

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10. Common Mistakes and Tips | 常见错误与提示

Students often forget to update probabilities for dependent events, or they add all the endpoint probabilities instead of multiplying along branches. Always double‑check whether the problem involves replacement or no replacement.

学生常忘记为相关事件更新概率，或者一味地将所有端点概率全部相加，而不是先沿分支相乘。务必仔细检查题目中是否涉及“放回”还是“不放回”。

Another error is assuming outcomes are equally likely when they are not. If a spinner is biased, the sections do not have equal probabilities. Read the question text carefully to identify given probabilities. Drawing a tree diagram clearly and labelling every branch with a fraction or decimal is the best safeguard.

另一个常见错误是在结果并非等可能时却假定它们等可能。如果转盘是偏心的，各区域就不具有等概率。仔细阅读题目信息以识别给定的概率。清晰地画出树状图，并用分数或小数标注每一分支，是最佳的防错手段。

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11. Practice Question Walkthrough | 典型例题讲解

Question: A box contains 4 black pens and 2 green pens. Two pens are taken out at random without replacement. Draw a tree diagram and find the probability that at least one of the pens is black.

题目：一个盒子里有 4 支黑笔和 2 支绿笔。随机取出两支且不放回。画出树状图，并求至少有一支是黑笔的概率。

Step 1: First pick – P(black) = 4/6 = 2/3, P(green) = 2/6 = 1/3. Step 2: If black was taken first, remaining pens: 3 black, 2 green – so P(black second) = 3/5, P(green second) = 2/5. If green was taken first, remaining: 4 black, 1 green – so P(black second) = 4/5, P(green second) = 1/5.

步骤 1：第一次抽取 – P(黑) = 4/6 = 2/3，P(绿) = 2/6 = 1/3。步骤 2：若第一次抽到黑笔，剩余 3 黑 2 绿 – 所以 P(第二次黑) = 3/5，P(第二次绿) = 2/5。若第一次抽到绿笔，剩余 4 黑 1 绿 – 所以 P(第二次黑) = 4/5，P(第二次绿) = 1/5。

Now calculate probability of at least one black: this is the complement of getting no black (i.e., both green). Path for both green: 1/3 × 1/5 = 1/15. Therefore P(at least one black) = 1 – 1/15 = 14/15.

现在计算至少一支黑笔的概率：这等价于未抽到黑笔（即两支全绿）的补事件。两支全绿的路径：1/3 × 1/5 = 1/15。因此 P(至少一支黑) = 1 – 1/15 = 14/15。

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12. Summary and Beyond | 总结与拓展

Probability and tree diagrams are essential tools for organising outcomes logically. Always remember to set up your sample space, label branches with correct probabilities, and apply the multiply‑and‑add rules carefully. These techniques are directly tested in Cambridge Checkpoint and will be expanded in IGCSE topics such as conditional probability and Venn diagrams.

概率与树状图是逻辑组织结果的重要工具。务必记住建立样本空间，用正确的概率标注分支，并谨慎应用相乘与相加规则。这些技巧会在剑桥 Checkpoint 考试中直接考查，并将在 IGCSE 中有条件概率、文氏图等主题中进一步拓展。

Keep practising with different structures – three branches, biased dice, or picking sweets from a bag. The more you draw and label, the more intuitive these diagrams become. Page 179 is just the beginning; the logic you learn here will support statistical reasoning for years to come.

请用不同的结构多加练习——三支分支、偏心骰子、或从袋中取糖果等。画得越多、标注得越多，这些图就会越直观。第 179 页仅仅是个开始；你在这里学到的逻辑思维将支持你未来多年的统计推理。

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📚 Simplifying Algebraic Expressions (Cambridge KS3 p113) | 化简代数表达式（剑桥KS3数学第113页）

Welcome to this revision guide based on page 113 of the Cambridge KS3 Mathematics textbook. Here we explore the fundamentals of algebraic expressions, learning how to simplify them by combining like terms, using the distributive law, and substituting values. Mastering these skills is crucial for your success in algebra.

欢迎来到这篇基于剑桥KS3数学教材第113页的复习指南。我们将探索代数表达式的基础知识，学习如何通过合并同类项、运用分配律和代入数值来化简表达式。掌握这些技能对你的代数学习至关重要。

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1. What is an Algebraic Expression? | 什么是代数表达式？

An algebraic expression is a combination of numbers, variables (letters representing unknown values), and operation symbols (+, −, ×, ÷). For example, 3x + 2, 5a − 7b, and x² + 2x − 4 are all algebraic expressions. Unlike an equation, an expression does not contain an equals sign. Think of it as a phrase in mathematics that describes a quantity.

代数表达式是由数字、变量（代表未知值的字母）和运算符号（+、−、×、÷）组合而成的。例如，3x + 2、5a − 7b 和 x² + 2x − 4 都是代数表达式。与方程不同，表达式不包含等号。可以把它看作是数学中描述数量的短语。

3x + 2y − 5

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2. Understanding Terms, Coefficients, and Constants | 理解项、系数和常数

Every algebraic expression is made up of terms. A term can be a single number, a variable, or a number multiplied by a variable. For instance, in 4x² + 3x − 7, the terms are 4x², 3x, and −7. The number in front of a variable is called the coefficient (e.g., 4 is the coefficient of x², 3 is the coefficient of x). A term without a variable is a constant (here, −7).

每个代数表达式都由项组成。项可以是单个数字、变量或数字与变量的乘积。例如，在 4x² + 3x − 7 中，项为 4x²、3x 和 −7。变量前面的数字称为系数（例如，4 是 x² 的系数，3 是 x 的系数）。没有变量的项是常数（此处为 −7）。

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3. Like Terms and Unlike Terms | 同类项与异类项

Like terms are terms that have exactly the same variable part, including the same power. For example, 2x and 5x are like terms because both contain x to the power of 1. Similarly, 3y² and −y² are like terms. Unlike terms have different variable parts: 2x and 3x² are not like terms, and 4a and 5b are also not like terms. Only like terms can be combined through addition or subtraction.

同类项是指变量部分完全相同（包括相同的指数）的项。例如，2x 和 5x 是同类项，因为两者都含有 x 的一次方。同样，3y² 和 −y² 也是同类项。异类项的变量部分不同：2x 和 3x² 不是同类项，4a 和 5b 也不是同类项。只有同类项才能通过加法或减法合并。

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4. Simplifying Expressions by Collecting Like Terms | 通过合并同类项化简

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📚 Solving Linear Equations (Cambridge KS3 p115) | 一元一次方程求解（剑桥初中数学第115页）

Linear equations are one of the most important building blocks in Key Stage 3 algebra. On page 115 of the Cambridge Mathematics course, you will meet a range of techniques designed to help you solve equations confidently and accurately. This article walks you through each method, from simple one-step equations to those involving brackets, fractions and variables on both sides.

一元一次方程是 KS3 代数中最重要的基石之一。在剑桥数学课程的第115页，你将学习一系列旨在帮助你自信且准确地解方程的方法。本文将从简单的一步方程到含有括号、分数和未知数在两边的情况，逐步讲解每一种解法。

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

A linear equation in one variable is an equality that contains a variable, usually x, raised only to the power 1. There are no terms like x², x³ or 1/x. The solution is the value that makes the equation true when substituted for the variable.

一元一次方程是含有一个变量（通常是 x）且变量的指数仅为 1 的等式。式中没有 x²、x³ 或 1/x 这样的项。方程的解是代入变量后使等式成立的数值。

Example: 2x + 3 = 11

示例：2x + 3 = 11

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2. Solving One-Step Equations | 解一步方程

One-step equations can be solved by performing a single inverse operation. If the equation is x + a = b, subtract a from both sides. If it is x – a = b, add a to both sides.

一步方程可以通过执行单一的逆运算来求解。如果方程是 x + a = b，则两边同时减去 a；如果是 x – a = b，则两边同时加上 a。

For multiplication and division, if the equation is ax = b, divide both sides by a. If it is x/a = b, multiply both sides by a.

对于乘法和除法，若方程为 ax = b，则两边除以 a；若方程为 x/a = b，则两边乘以 a。

x + 7 = 15 → x = 8

4x = 20 → x = 5

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3. Solving Two-Step Equations | 解两步方程

Two-step equations require two inverse operations, usually undoing addition/subtraction first, then multiplication/division. For instance, in 2x + 3 = 11, subtract 3 from both sides to get 2x = 8, then divide by 2 to obtain x = 4.

两步方程需要两次逆运算，通常先处理加法/减法，再处理乘法/除法。例如，对于 2x + 3 = 11，两边先减 3 得到 2x = 8，再除以 2 得到 x = 4。

Always remember the order of operations is reversed when solving: we undo any addition or subtraction before dealing with the coefficient of x.

始终记住，求解时运算顺序是相反的：先解除加减，再处理 x 的系数。

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4. Equations with Brackets | 带括号的方程

When brackets appear, expand them first using the distributive law. For example, 3(x + 2) = 15 becomes 3x + 6 = 15. Then solve as a two-step equation: subtract 6 to get 3x = 9, and divide by 3 to find x = 3.

当出现括号时，首先用分配律展开。例如，3(x + 2) = 15 变为 3x + 6 = 15。然后当作两步方程求解：减 6 得到 3x = 9，除以 3 得到 x = 3。

If there is a negative sign in front of a bracket, be careful to multiply every term inside by -1: -(2x – 5) = -2x + 5.

如果括号前有负号，要小心将里面的每一项都乘以 -1：-(2x – 5) = -2x + 5。

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5. Equations with Variables on Both Sides | 含未知数在两边

When the variable appears on both sides of the equation, collect all variable terms on one side and constant terms on the other. For 5x + 2 = 3x + 10, subtract 3x from both sides to obtain 2x + 2 = 10. Then subtract 2: 2x = 8, so x = 4.

当未知数出现在方程两边时，将所有含变量的项移到一边，常数项移到另一边。对于 5x + 2 = 3x + 10，两边减去 3x 得 2x + 2 = 10。再减 2：2x = 8，因此 x = 4。

Always aim to have a positive coefficient for x. If you end up with -x, multiply both sides by -1.

始终争取让 x 的系数为正。如果最终得到 -x，就将两边乘以 -1。

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6. Equations with Fractions | 分数方程

To solve equations containing fractions, eliminate the denominators by multiplying every term by the lowest common denominator. For x/3 + 1/2 = 5/6, multiply through by 6: 2x + 3 = 5, then 2x = 2, so x = 1.

要解含有分数的方程，可以通过将每一项都乘以最小公分母来消去分母。对于 x/3 + 1/2 = 5/6，两边乘以 6：2x + 3 = 5，然后 2x = 2，得 x = 1。

Another common type is when a fraction equals a constant: (2x)/5 = 4. Multiply both sides by 5 to give 2x = 20, then x = 10.

另一种常见类型是分数等于一个常数：(2x)/5 = 4。两边乘以 5 得 2x = 20，x = 10。

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7. Checking Your Solution | 验算你的解

After finding a value for x, always substitute it back into the original equation to check both sides are equal. This habit catches arithmetic mistakes and reinforces your understanding of what a solution means.

求出 x 的值后，一定要将其代回原方程，检查两边是否相等。这个习惯可以捕捉算术错误，并加深你对解的含义的理解。

For 2x + 3 = 11, if x = 4, then left side = 2(4) + 3 = 8 + 3 = 11 = right side. Correct!

对于 2x + 3 = 11，如果 x = 4，左边 = 2(4) + 3 = 8 + 3 = 11 = 右边。正确！

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8. Real-life Applications | 实际应用

Linear equations model many everyday situations. If a taxi charges a fixed fee of pound 3 plus pound 2 per mile, and the total fare is pound 15, the equation is 2m + 3 = 15. Solving gives m = 6 miles.

一元一次方程可以模拟许多日常情景。如果出租车收取 3 英镑固定费用加上每英里 2 英镑，总车费为 15 英镑，方程就是 2m + 3 = 15。解得 m = 6 英里。

Interpreting the solution in context ensures you answer the original problem, not just manipulate symbols.

在情境中解释解的含义，能确保你回答的是最初的问题，而不仅仅是进行符号运算。

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

A frequent error is forgetting to perform an operation on both sides. If you subtract 5 from the left, you must subtract 5 from the right as well. Another is mishandling negative signs when expanding brackets.

一个常见错误是忘记对两边执行相同的运算。如果你在左边减 5，右边也必须减 5。另一个错误是展开括号时处理负号不当。

When solving two-step equations, some students divide before subtracting. Always reverse the order of operations: do addition/subtraction first, then multiplication/division.

在解两步方程时，有些学生先除后减。一定要逆转运算顺序：先进行加/减，再进行乘/除。

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10. Summary | 总结

To solve any linear equation reliably, follow a clear sequence: eliminate brackets, clear fractions, collect variable terms on one side and constants on the other, then isolate the variable by inverse operations. Check your answer by substitution. Page 115 of the Cambridge course gives you plenty of practice to turn these steps into a confident routine.

要可靠地求解任何一元一次方程，请遵循明确的顺序：去括号，消分数，将变量项和常数项分别集中到两边，然后通过逆运算求出变量。通过代入检查答案。剑桥课程的第115页为你提供了大量练习，帮助你将这些步骤变成自信的习惯。

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