📚 Solving Linear Equations and Inequalities: An IGCSE Unit 6 Teacher's Guide | 求解线性方程与不等式：IGCSE 第 6 单元教师指南
Linear equations and inequalities form the backbone of the IGCSE Algebra syllabus. They underpin more advanced topics such as simultaneous equations, quadratic functions and linear programming. This teacher's guide focuses on Unit 6 core skills: solving equations with unknowns on both sides, handling brackets, working with fractions, interpreting inequalities, and applying these tools to word problems.
线性方程与不等式是 IGCSE 代数课程的主干。它们为后续更高级的专题（如联立方程、二次函数和线性规划）奠定基础。本教师指南聚焦第 6 单元的核心技能：求解未知数在等号两边的方程、处理括号与分数、理解不等式，以及将这些工具应用到文字题中。
1. Learning Objectives and Prerequisite Knowledge | 学习目标与前置知识
Before teaching Unit 6, check that students can add, subtract, multiply and divide directed numbers; collect like terms such as 3x + 5x − 2x; apply the order of operations (BIDMAS/BODMAS); and substitute values into simple expressions. The main objectives for this unit are: solve linear equations with one unknown algebraically, solve equations involving brackets and fractions, represent inequalities on a number line, solve linear inequalities and interpret word problems by forming equations or inequalities.
在教授第 6 单元之前，请确认学生已经掌握：正负数的加减乘除；合并同类项，如 3x + 5x − 2x；使用运算顺序（BIDMAS/BODMAS）；以及将数值代入简单表达式。本单元的主要目标是：用代数方法求解一元一次方程，求解含括号和分数的方程，在数轴上表示不等式，求解一元一次不等式，并通过建立方程或不等式来理解文字题。
2. Solving Linear Equations with One Unknown | 求解一元一次方程
side. For example, to solve 3x + 5 = 17, subtract 5 from both sides to obtain 3x = 12, then divide both sides by 3 to obtain x = 4.
3x + 5 = 17 → 3x = 12 → x = 4
线性方程是未知数只出现一次幂的方程。核心策略是按照 BIDMAS 的逆顺序使用逆运算来分离未知数。始终要保持方程平衡：对一边所做的任何运算，必须对另一边也做同样的运算。例如，求解 3x + 5 = 17，先两边减去 5 得到 3x = 12，然后两边除以 3 得到 x = 4。
3. Equations with Unknowns on Both Sides | 未知数在等号两边的方程
e the variable on the side that gives a positive coefficient, as this reduces sign errors.
当未知数出现在方程两边时，应把所有含未知数的项集中到一边，把所有常数项集中到另一边。对于 5x − 3 = 2x + 9，先从两边减去 2x 得到 3x − 3 = 9，再两边加 3 得到 3x = 12，因此 x = 4。要鼓励学生把未知数放在系数为正的一边，这样可以减少符号错误。
4. Brackets and Fractions in Linear Equations | 方程中的括号与分数
− 3x = 2 + 12, giving x = 14. Check by substitution: LHS = 2(11) = 22, RHS = 21 + 1 = 22.
如果方程中有括号，先用分配律展开；如果含有分数，则两边同乘以最小公分母（LCM）来去分母。求解 2(x − 3) = 3x/2 + 1。两边乘以 2：4(x − 3) = 3x + 2。展开得 4x − 12 = 3x + 2。合并同类项：4x − 3x = 2 + 12，得到 x = 14。代入检验：左边 = 2(11) = 22，右边 = 21 + 1 = 22。
5. Introduction to Inequalities and Their Symbols | 不等式及其符号简介
han, less than, at least, or at most another. The symbols are >, <, ≥ and ≤; ≠ means 'not equal to'. On a number line, strict inequalities > or <
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