📚 IGCSE Maths: Linear Equations and Inequalities | IGCSE 数学：线性方程与不等式
Linear equations and inequalities are among the most important algebra skills in the IGCSE mathematics syllabus. They appear in almost every exam paper, both as stand-alone questions and within larger problems such as sequences, functions, and geometry. This article explains how to solve linear equations and inequalities step by step, with worked examples, common mistakes, and practice questions.
线性方程和不等式是 IGCSE 数学大纲中最重要的代数技能之一。它们几乎出现在每一份试卷中，既会单独考查，也会出现在数列、函数和几何等更大的问题中。本文逐步讲解如何解线性方程和不等式，并配有例题、常见错误和练习题。
1. What Is a Linear Equation? | 什么是线性方程？
IGCSE exams, you will often need to solve equations such as 3x + 5 = 20 or 2(x − 4) = 10.
线性方程是变量最高次数为 1 的代数等式。它通常可以写成 ax + b = c，其中 a、b、c 是常数，x 是未知数。方程的解就是使等式成立的 x 值。IGCSE 考试中常需要解诸如 3x + 5 = 20 或 2(x − 4) = 10 的方程。
ax + b = c
3x + 5 = 20
2. Solving One-Step Equations | 解一步方程
to obtain x = 28 ÷ 4 = 7. The key principle is to keep the equation balanced at all times.
解一步方程时，要对等式两边同时进行逆运算。如果 x + 7 = 12，两边同时减去 7，得到 x = 12 − 7 = 5。如果 4x = 28，两边同时除以 4，得到 x = 28 ÷ 4 = 7。关键原则是始终保持等式平衡。
x + 7 = 12 → x = 12 − 7 = 5
4x = 28 → x = 28 ÷ 4 = 7
3. Solving Two-Step Equations | 解两步方程
ows the reverse of BIDMAS: undo addition or subtraction before multiplication or division.
两步方程需要进行两次逆运算。例如，在 2x − 3 = 11 中，先两边同时加 3，得到 2x = 14，然后两边同时除以 2，得到 x = 7。运算顺序与 BIDMAS 相反：先处理加减，再处理乘除。
2x − 3 = 11
2x = 14 → x = 7
4. Equations with Brackets | 含括号的方程
ative sign outside the bracket, multiply all terms inside carefully: −2(x + 5) = −2x − 10.
当方程中含有括号时，要先用分配律展开括号。对于 3(x − 2) = 12，展开得到 3x − 6 = 12。然后两边同时加 6，得到 3x = 18，再除以 3，得到 x = 6。如果括号外有负号，要仔细乘以括号内的每一项：−2(x + 5) = −2x − 10。
3(x − 2) = 12
3x − 6 = 12 → 3x = 18 → x = 6
5. Equations with Variables on Both Sides | 变量在两侧的方程
by 2 to get x = 3. Always check by substituting the value back into the original equation.
当变量出现在方程的两侧时，要把所有含变量的项移到一边，所有常数项移到另一边。对于 5x + 2 = 3x + 8，两边同时减去 3x，得到 2x + 2 = 8。然后两边同时减去 2，得到 2x = 6，再除以 2，得到 x = 3。一定要把解代入原方程进行检验。
5x + 2 = 3x + 8
2x + 2 = 8 → 2x = 6 → x = 3
6. Introduction to Inequalities | 不等式入门
han or equal to −2. Unlike equations, inequalities usually have infinitely many solutions, so they are often shown on a number line or written using interval notation.
不等式使用 <、>、≤ 或 ≥ 来比较两个表达式。例如，x > 3 表示 x 大于 3，而 x ≤ −2 表示 x 小于或等于 −2。与方程不同，不等式通常有无穷多个解，因此常用数轴表示或用区间记号书写。
x > 3, x ≤ −2, −1 < x ≤ 4
7. Solving Linear Inequalities | 解线性不等式
For 3x + 1 ≥ 10, subtract 1 from both sides to get 3x ≥ 9, then divide by 3 to get x ≥ 3.
解不等式的方法与解方程相同，但当两边同时乘以或除以一个负数时，需要改变不等号的方向。对于 −2x < 6，两边同时除以 −2，并改变不等号方向，得到 x > −3。对于 3x + 1 ≥ 10，两边同时减 1，得到 3x ≥ 9，再除以 3，得到 x ≥ 3。
−2x < 6 → x > −3
3x + 1 ≥ 10 → 3x ≥ 9 → x ≥ 3
8. Representing Inequalities on a Number Line | 在数轴上表示不等式
; x ≤ 3, draw an open circle at −2, a closed circle at 3, and shade the line between them.
对于严格不等式（如 < 或 >）使用空心圆点，对于包含等于的不等式（如 ≤ 或 ≥）使用实心圆点。涂黑满足不等式的区域。对于 −2 < x ≤ 3，在 −2 处画空心圆点，在 3 处画实心圆点，并涂黑两者之间的线段。
−2 < x ≤ 3
9. Common Mistakes and Exam Tips | 常见错误与应试技巧
Students often forget to apply an operation to both sides of the equation, which destroys the balance. Another common error is mishandling negative signs when expanding brackets. In inequalities, many candidates forget to reverse the inequality sign when dividing by a negative number. To avoid these errors, always show clear working and check your final answer by substitution.
学生经常忘记对等式两边同时进行运算，从而破坏了等式的平衡。另一个常见错误是在展开括号时错误处理负号。在不等式中，许多考生在除以负数时忘记改变不等号方向。为了避免这些错误，一定要写出清晰的步骤，并通过代入检验最终答案。
- Always maintain balance: do the same operation to both sides.
- Expand brackets carefully, especially with negative coefficients.
- Reverse the inequality sign only when multiplying or dividing by a negative.
- Check your solution by substituting it into the original equation or inequality.
始终维持平衡：对等式两边进行相同的运算。仔细展开括号，尤其是系数为负数时。只有在乘以或除以负数时，才改变不等号方向。将解代入原方程或不等式进行检验。
10. Practice Questions and Answers | 练习题与答案
= 3x + 4, subtract 2x from both sides to get −10 = x + 4, then subtract 4 to get x = −14.
尝试以下 IGCSE 风格的题目。首先解 4x + 7 = 31。两边同时减 7，得到 4x = 24，再除以 4，得到 x = 6。第二，解 2(x − 5) = 3x + 4。左边展开得到 2x − 10 = 3x + 4，两边同时减 2x，得到 −10 = x + 4，再减 4，得到 x = −14。
4x + 7 = 31 → x = 6
2(x − 5) = 3x + 4 → x = −14
present x ≥ −1 on a number line by drawing a closed circle at −1 and shading to the right.
第三，解 −3x ≤ 12。两边同时除以 −3，并改变不等号方向，得到 x ≥ −4。第四，在数轴上表示 x ≥ −1：在 −1 处画实心圆点，并向右涂黑。
−3x ≤ 12 → x ≥ −4
x ≥ −1: closed circle at −1, shade to the right
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