Solving Linear Equations and Inequalities | 一元一次方程与不等式的解法

📚 Solving Linear Equations and Inequalities | 一元一次方程与不等式的解法

Linear equations and inequalities are central to algebraic reasoning in the Cambridge Lower Secondary KS3 mathematics course. The p159_2 practice sheet focuses on balancing equations, using inverse operations, expanding brackets, and interpreting inequality symbols on a number line. This revision guide explains each method step by step, with worked examples and common mistakes to avoid.

一元一次方程和不等式是剑桥初中 KS3 数学课程中代数推理的核心内容。p159_2 练习页重点关注方程的平衡、逆运算、括号展开，以及在数轴上表示不等式解集。本复习指南逐步讲解每种方法，配有例题和常见错误提醒。

1. Expressions, Equations and Inequalities | 代数式、方程与不等式

An expression such as 3x + 5 does not contain an equals sign, so it cannot be solved. An equation states that two expressions are equal, such as 3x + 5 = 17. An inequality uses the symbols <, >, ≤ or ≥ to compare two expressions, for example 2x − 3 < 7. Knowing the difference helps you decide whether to simplify, solve, or shade a solution set.

代数式（如 3x + 5）不包含等号，因此不能“求解”。方程表示两个代数式相等，例如 3x + 5 = 17。不等式使用 <、>、≤ 或 ≥ 来比较两个代数式，例如 2x − 3 < 7。分清三者有助于判断应当化简、求解还是用数轴表示解集。

3x + 5 = 17 and 2x − 3 < 7

2. The Balance Method | 等式平衡法

An equation behaves like a balanced scale. If you add, subtract, multiply or divide both sides by the same non-zero number, the equation remains true. This is called the balance method. For example, to solve x + 7 = 15, subtract 7 from both sides to obtain x = 8.

方程就像一个平衡的天平。只要对等式两边同时加、减、乘或除以同一个非零数，等式仍然成立，这就是平衡法。例如，解 x + 7 = 15 时，两边同时减去 7，得到 x = 8。

x + 7 = 15 → x = 8

3. Solving Two-Step Linear Equations | 解两步一元一次方程

A two-step equation contains two operations applied to the variable. The key is to reverse the order of operations: undo addition or subtraction first, then undo multiplication or division. For 2x + 3 = 11, subtract 3 from both sides to get 2x = 8, then divide both sides by 2 to get x = 4.

两步方程指的是对变量进行了两次运算。解题关键是逆着运算顺序：先消去加法或减法，再消去乘法或除法。对于 2x + 3 = 11，先两边减 3 得到 2x = 8，再两边除以 2 得到 x = 4。

2x + 3 = 11 → 2x = 8 → x = 4

4. Equations with Brackets | 含括号的方程

When a bracket has a number or sign in front, multiply each term inside the bracket by that factor. In 3(x + 2) = 18, expand the left side to give 3x + 6 = 18. Then subtract 6 and divide by 3 to find x = 4. You can also divide both sides by 3 first, which is quicker: x + 2 = 6, so x = 4.

当括号前有数字或符号时，要用括号外的因数乘以括号内的每一项。在 3(x + 2) = 18 中，展开左边得到 3x + 6 = 18，然后减 6、除以 3，得到 x = 4。也可以先两边同时除以 3，得到 x + 2 = 6，所以 x = 4，这种方法更快。

3(x + 2) = 18 → 3x + 6 = 18 → x = 4

5. Variables on Both Sides | 含有两侧变量的方程

If the variable appears on both sides, collect the variable terms on one side and the constant terms on the other. For 5x = 2x + 9, subtract 2x from both sides to obtain 3x = 9, then divide by 3 to get x = 3. Always choose the smaller variable term to remove so that the coefficient stays positive.

如果变量同时出现在方程两边，应把变量项集中到一边，常数项集中到另一边。对于 5x = 2x + 9，两边同时减去 2x，得到 3x = 9，再除以 3 得到 x = 3。通常先消去较小的变量项，这样系数可以保持为正。

5x = 2x + 9 → 3x = 9 → x = 3

6. Forming Equations from Word Problems | 由文字题列方程

Word problems require you to translate English or everyday statements into algebraic form. For example, ‘A number doubled and then increased by 5 gives 21’ becomes 2n + 5 = 21. Identify the unknown, assign a letter, write the operations in order, and then solve the equation.

文字题要求把日常语言转化为代数形式。例如，“一个数乘以 2 后再加 5 得到 21”可以写成 2n + 5 = 21。首先要确定未知数并用字母表示，然后按顺序写出运算过程，最后解方程。

2n + 5 = 21 → 2n = 16 → n = 8

7. Inequality Symbols and the Number Line | 不等式符号与数轴表示

Inequality symbols show the relationship between two quantities. The symbol < means ‘less than’, > means ‘greater than’, ≤ means ‘less than or equal to’, and ≥ means ‘greater than or equal to’. On a number line, strict inequalities use an open circle, while ≤ and ≥

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