📚 Linear Equations and Inequalities | 线性方程与不等式
Linear equations and inequalities are essential algebra topics in IGCSE Mathematics. They appear in many different forms, from simple one-step equations to word problems and graphical inequalities. In this article, we will revise the key methods, common mistakes, and exam techniques needed to solve these questions confidently.
线性方程与不等式是 IGCSE 数学中的重要代数主题。它们以多种形式出现，包括简单的一步方程、应用题以及图像中的不等式。本文将复习关键方法、常见错误以及解答此类题型所需的考试技巧。
1. What Is a Linear Equation? | 什么是线性方程？
hat can be written in the form ax + b = 0, where a and b are constants and a ≠ 0. The word 'linear' means the variable appears only to the first power, so there are no x², x³, √x or 1/x terms. In IGCSE mathematics, solving linear equations is a core skill used in algebra, graphs, and word problems.
一元线性方程是可以写成 ax + b = 0 形式的方程，其中 a 与 b 为常数且 a ≠ 0。'线性' 一词表示变量只出现一次方，因此不会出现 x²、x³、√x 或 1/x 等项。在 IGCSE 数学中，解线性方程是代数、图像与应用题中的核心技能。
ppear as part of simultaneous equations or graphing questions. Recognising the form helps you choose the right method.
在 IGCSE 中，线性方程可能以不同形式出现，如 4x - 7 = 13 或 2(x - 3) = 10。它们也可能作为联立方程或图像题的一部分出现。识别方程形式有助于选择正确的方法。
2. Solving One-Step and Two-Step Equations | 解一步和两步方程
3x = 18, divide both sides by 3: x = 6. The key is to do the same operation on both sides.
一步方程需要一次逆运算。对于 x + 5 = 12，两边同时减 5：x = 7。对于 3x = 18，两边同时除以 3：x = 6。关键是等式两边进行相同运算。
4. Work backwards through the order of operations: undo addition or subtraction before multiplication or division.
两步方程包含两种运算。对于 2x + 3 = 11，先减 3 得到 2x = 8，再除以 2 得到 x = 4。按照运算顺序逆推：先消去加减，再消去乘除。
| Equation | First step | Second step | Solution |
|---|---|---|---|
| x + 5 = 12 | Subtract 5 | - | x = 7 |
| 3x = 18 | Divide by 3 | - | x = 6 |
| 2x + 3 = 11 | Subtract 3 | Divide by 2 | x = 4 |
The table above shows how to identify the inverse operations in the correct order. Always check your final answer by substituting it back into the original equation.
上表展示了如何按正确顺序确定逆运算。始终将最终答案代回原方程进行检验。
3. Equations with Brackets | 带括号的方程
to get 3x + 12 = 27. Subtract 12 from both sides to get 3x = 15, then divide by 3: x = 5.
当出现括号时，使用乘法分配律展开。对于 3(x + 4) = 27，将 3 乘以括号内的两项，得到 3x + 12 = 27。两边同时减 12 得到 3x = 15，再除以 3：x = 5。
For 2(x - 1) = 3(x + 2), expand both sides first: 2x - 2 = 3x + 6. Subtract
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