📚 Cambridge KS3 Maths: Solving Linear Equations from Exercise 2 (p.280) | 剑桥 KS3 数学：求解一元一次方程（第280页练习2）
Welcome to this TutorHao revision lesson for Cambridge KS3 Mathematics. In this article we work through the style of questions found in Exercise 2 on page 280 of the Cambridge Lower Secondary Maths course. The focus is linear equations, balancing steps, expanding brackets and checking answers.
欢迎来到 TutorHao 的剑桥 KS3 数学复习课。本文讲解剑桥初中数学教材第280页练习2中常见的一元一次方程题型，重点包括等式平衡步骤、去括号、未知数在两边以及代入检验。
1. Understanding the Exercise 2 Context | 理解练习2的背景
Page 280 Exercise 2 in the Cambridge KS3 course usually contains a mixed set of linear equations. The questions are designed to test whether you can keep an equation balanced while isolating the unknown.
剑桥 KS3 教材第280页练习2通常包含一组混合的一元一次方程。题目旨在考查你是否能在分离未知数的同时保持等式两边平衡。
In KS3, the unknown is usually written as x, y or n, and the solution must make the original equation true. This means every step should preserve equality.
在 KS3 阶段，未知数通常写作 x、y 或 n，解必须能使原方程成立。这意味着每一步都要保持等式的相等关系。
2. Key Vocabulary and Notation | 核心词汇与符号
Before solving, make sure you know these key terms: equation, expression, variable, coefficient, constant, left-hand side (LHS), right-hand side (RHS), solution and root.
解题之前，请确保你掌握以下术语：方程、表达式、变量、系数、常数项、左边（LHS）、右边（RHS）、解和根。
ient is 3, the constant is 5, and x is the variable. The left-hand side is 3x + 5 and the right-hand side is 14.
例如，在 3x + 5 = 14 中，系数是 3，常数项是 5，x 是变量。左边是 3x + 5，右边是 14。
| English term | 中文术语 | Example |
|---|---|---|
| Equation | 方程 | 2x + 1 = 9 |
| Variable | 变量 | x |
| Coefficient | 系数 | 2 |
| Constant | 常数项 | 1 |
3. The Balancing Method | 平衡法
The balancing method is the golden rule: whatever you do to one side of an equation, you must do to the other side. This keeps the equation true.
平衡法是黄金法则：对方程一边进行任何运算，另一边也必须进行相同运算。这样才能保持方程成立。
Think of an equation as a set of scales. If you add 2 kg to one side, you must add 2 kg to the other side to keep it balanced. The same idea applies to subtraction, multiplication and division.
可以把方程想象成一台天平。如果在一侧加 2 kg，就必须在另一侧也加 2 kg 才能保持平衡。减法、乘法和除法也同样适用。
3x + 5 − 5 = 14 − 5, giving 3x = 9. Then divide both sides by 3: 3x ÷ 3 = 9 ÷ 3, so x = 3.
对于 3x + 5 = 14，两边同时减去 5：3x + 5 − 5 = 14 − 5，得到 3x = 9。然后两边同时除以 3：3x ÷ 3 = 9 ÷ 3，所以 x = 3。
3x + 5 = 14 → 3x = 9 → x = 3
4. Solving Two-Step Equations | 求解两步方程
A two-step equation requires exactly two inverse operations to isolate the variable. The usual order is: undo addition or subtraction first, then undo multiplication or division.
两步方程需要正好两个逆运算来分离变量。通常顺序是：先消去加法或减法，再消去乘法或除法。
5x − 7 = 18. Step 1: add 7 to both sides: 5x = 25. Step 2: divide both sides by 5: x = 5.
解 5x − 7 = 18。第一步：两边同时加 7，得 5x = 25。第二步：两边同时除以 5，得 x = 5。
Always show both steps clearly. If you skip a step, it is easy to lose track of the operations and make a sign error.
一定要清楚地写出两步。如果省略步骤，就很容易忘记运算顺序并出现符号错误。
5x − 7 = 18 → 5x = 25 → x = 5
5. Equations with Brackets | 含括号的方程
When an equation has brackets, expand them first using the distributive law. Then solve the simpler equation using the balancing method.
当方程含有括号时，先用分配律展开括号。然后用平衡法解简化后的方程。
x + 3) = 16. Expand: 2x + 6 = 16. Subtract 6 from both sides: 2x = 10. Divide by 2: x = 5.
解 2(x + 3) = 16。展开得 2x + 6 = 16。两边同时减 6 得 2x = 10，再除以 2 得 x = 5。
2) = 21. Expand carefully: −3x + 6 = 21. Then subtract 6: −3x = 15. Divide by −3: x = −5.
有时括号外是负号，例如 −3(x − 2) = 21。展开时要仔细：−3x + 6 = 21。然后减 6 得 −3x = 15，再除以 −3 得 x = −5。
6. Unknowns on Both Sides | 未知数在两边
If the unknown appears on both sides of the equation, collect the variable terms on one side and the constants on the other side.
如果未知数出现在方程两边，就先把含未知数的项移到一边，把常数项移到另一边。
x + 20. Subtract 3x from both sides: 4x + 4 = 20. Subtract 4: 4x = 16. Divide by 4: x = 4.
解 7x + 4 = 3x + 20。两边同时减去 3x，得 4x + 4 = 20。再减 4 得 4x = 16，除以 4 得 x = 4。
If the variable terms cancel completely, the equation may have no solution or infinitely many solutions. For KS3, focus on the usual unique-solution case.
如果含未知数的项完全抵消，方程可能无解或有无穷多个解。KS3 阶段主要掌握通常有唯一解的情况。
7. Forming Equations from Word Problems | 根据文字题列方程
Many Exercise 2 questions ask you to form an equation from a written problem. Read the question carefully, define the unknown with a letter, and translate each phrase into algebraic language.
练习2中的许多题目要求你根据文字题列出方程。仔细读题，用字母表示未知数，并把每个短语转化为代数语言。
Example: "Three more than twice a number is 19." Let the number be n. Then 2n + 3 = 19. Subtract 3: 2n = 16. Divide by 2: n = 8.
例题："一个数的两倍加 3 等于 19。" 设这个数为 n，则 2n + 3 = 19。减 3 得 2n = 16，除以 2 得 n = 8。
Example: " perimeter of a rectangle is 26 cm. Its length is 3 cm more than its width. Find the width." Perimeter: 2(w + w + 3) = 26, so 4w + 6 = 26. Subtract 6: 4w = 20. Divide by 4: w = 5 cm.
例题："一个长方形的周长是 26 cm，长比宽多 3 cm。求宽。" 设宽为 w，则长为 w + 3。周长：2(w + w + 3) = 26，所以 4w + 6 = 26。减 6 得 4w = 20，除以 4 得 w = 5 cm。
8. Checking Your Solution | 检验你的解
Always substitute your answer back into the original equation to check. The left-hand side must equal the right-hand side.
一定要把答案代回原方程检验。左边必须等于右边。
For 5x − 7 = 18 with x = 5: LHS = 5(5) − 7 = 25 − 7 = 18 = RHS. Correct.
对于 5x − 7 = 18 且 x = 5：左边 = 5(5) − 7 = 25 − 7 = 18 = 右边。正确。
Checking also helps catch sign errors and arithmetic mistakes before you move to the next question. It is a quick way to boost your accuracy.
检验还能帮助你在做下一题之前发现符号错误和计算错误。这是提高准确率的快速方法。
9. Common Mistakes to Avoid | 常见错误与避免方法
Mistake 1: Applying an operation to only one side. Remember, every operation must be applied to both sides.
错误1：只对一边进行运算。记住，每一运算都必须同时作用于两边。
Mistake 2: Incorrect sign when expanding brackets, especially with negative multipliers.
错误2：展开括号时符号错误，尤其是括号前为负号时。
Mistake 3: Confusing the order of inverse operations. Undo addition or subtraction before multiplication or division.
错误3：逆运算顺序混淆。应先消去加减法，再消去乘除法。
Mistake 4: For
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