IGCSE Mathematics: Solving Linear Equations and Inequalities | IGCSE 数学：解线性方程与不等式

📚 IGCSE Mathematics: Solving Linear Equations and Inequalities | IGCSE 数学：解线性方程与不等式

Linear equations and inequalities are among the most important building blocks in IGCSE Mathematics. They appear in nearly every paper, often hidden inside word problems, geometry tasks, and function questions. In this revision guide, you will learn how to solve linear equations step by step, how to handle inequalities safely, and how to present your solutions clearly.

线性方程和不等式是 IGCSE 数学中最重要的基础模块之一。它们几乎出现在每份试卷中，经常隐藏在应用题、几何题和函数题里。在这份复习指南中，你将学会如何一步一步解线性方程，如何正确处理不等式，以及如何清晰地呈现答案。

---

1. What is a Linear Equation? | 什么是线性方程？

A linear equation is an equation in which the highest power of the variable is 1. It can usually be written in the form ax + b = c, where a, b, and c are constants and a ≠ 0. The word ‘linear’ tells you that the graph of the equation is a straight line.

线性方程是指变量的最高次数为 1 的方程。它通常可以写成 ax + b = c 的形式，其中 a、b、c 是常数，且 a ≠ 0。’线性’这个词说明方程的图像是一条直线。

ax + b = c, a ≠ 0

For example, 3x + 5 = 17 is a linear equation. The variable x appears only to the first power. The goal of solving is to find the value of x that makes the statement true, which is called the solution or root of the equation.

例如，3x + 5 = 17 就是一个线性方程。变量 x 只出现一次方。解方程的目标就是找到使等式成立的 x 值，这个值称为方程的解或根。

In IGCSE exams, linear equations usually have one solution, but you may also meet special cases where the equation has no solution or infinitely many solutions after simplification.

在 IGCSE 考试中，线性方程通常有一个解，但你也可能遇到化简后无解或有无穷多个解的特殊情况。

---

2. The Balancing Method | 等式平衡法

To solve a linear equation, use the balancing method: whatever you do to one side, you must do to the other side. This keeps the equation balanced and unchanged in meaning.

解线性方程时要使用平衡法：你对一边做什么运算，就必须对另一边做同样的运算。这样可以保持方程平衡，意义不变。

The table below shows common equation operations and their inverse operations. When solving, apply the inverse operation in the reverse order of BIDMAS.

下表列出了常见的方程运算及其逆运算。解方程时，要按照 BIDMAS 的相反顺序进行逆运算。

Operation 运算	Inverse operation 逆运算
Addition 加法	Subtraction 减法
Subtraction 减法	Addition 加法
Multiplication 乘法	Division 除法
Division 除法	Multiplication 乘法

Example: Solve 3x + 5 = 17. Step 1: Subtract 5 from both sides: 3x = 12. Step 2: Divide both sides by 3: x = 4. Check: 3 × 4 + 5 = 12 + 5 = 17.

例子：解 3x + 5 = 17。步骤 1：两边同时减去 5：3x = 12。步骤 2：两边同时除以 3：x = 4。检验：3 × 4 + 5 = 12 + 5 = 17。

---

3. Two-Step and Multi-Step Equations | 两步方程和多步方程

Two-step equations require two inverse operations, such as addition or subtraction followed by multiplication or division. Always undo addition or subtraction before multiplication or division when isolating the variable.

两步方程需要两次逆运算，例如先做加减，再做乘除。在分离变量时，通常先消去加减，再消去乘除。

Example: Solve 5x – 7 = 18. Add 7 to both sides: 5x = 25. Divide by 5: x = 5.

例子：解 5x – 7 = 18。两边同时加 7：5x = 25。两边同时除以 5：x = 5。

For multi-step equations, collect like terms on each side first, then move variable terms to one side and constant terms to the other side, and finally divide by the coefficient.

对于多步方程，应先在每一边合并同类项，然后把含变量项移到一边，常数项移到另一边，最后除以未知数的系数。

Example: Solve 7x + 3 = 2x + 18. Subtract 2x from both sides: 5x + 3 = 18. Subtract 3 from both sides: 5x = 15. Divide by 5: x = 3.

例子：解 7x + 3 = 2x + 18。两边同时减去 2x：5x + 3 = 18。两边同时减去 3：5x = 15。两边同时除以 5：x = 3。

---

4. Equations with Brackets | 含括号的方程

If an equation contains brackets, expand them first using the distributive law. Then simplify both sides before using the balancing method. This makes the equation easier to solve.

如果方程含有括号，应先用分配律展开。然后在两边化简，再进行平衡法求解。这样会使方程更容易解。

Example: Solve 2(x + 3) = 14. Expand: 2x + 6 = 14. Subtract 6: 2x = 8. Divide by 2: x = 4.

例子：解 2(x + 3) = 14。展开：2x + 6 = 14。两边减 6：2x = 8。两边除以 2：x = 4。

Be careful when a negative sign appears before a bracket: – (x – 2) = -x + 2. The negative sign changes every sign inside the bracket.

当括号前有负号时要特别小心： – (x – 2) = -x + 2。负号会改变括号内每一项的符号。

Example: Solve 4 – (x – 2) = 1. Expand: 4 – x + 2 = 1. Simplify: 6 – x = 1. Subtract 6: -x = -5. Multiply by -1: x = 5.

例子：解 4 – (x – 2) = 1。展开：4 – x + 2 = 1。化简：6 – x = 1。两边减 6：-x = -5。两边乘以 -1：x = 5。

---

5. Equations with Fractions | 含分数的方程

To remove fractions, multiply every term on both sides by the lowest common denominator, often called the LCD. This simplifies the equation to an equivalent equation without denominators.

要去掉分数，可以将方程两边每一项都乘以最小公分母，通常称为 LCD。这样就能把方程化为没有分母的等价方程。

Example: Solve x/3 + 2 = 5. Multiply every term by 3: x + 6 = 15. Subtract 6: x = 9.

例子：解 x/3 + 2 = 5。每一项乘以 3：x + 6 = 15。两边减 6：x = 9。

For equations with two fractions, such as (2x + 1)/5 = (x – 3)/2, multiply both sides by 10, the LCD of 5 and 2.

对于含有两个分式的方程，例如 (2x + 1)/5 = (x – 3)/2，两边同时乘以 10，即 5 和 2 的最小公分母。

(2x + 1)/5 = (x – 3)/2 → 2(2x + 1) = 5(x – 3) → 4x + 2 = 5x – 15 → x = 17

Always check by substituting the answer back into the original equation, especially when fractions are involved.

尤其在涉及分数时，始终要把答案代回原方程进行检验。

---

6. Forming Equations from Word Problems | 根据应用题列方程

Word problems require translating written statements into algebraic equations. Use a variable, usually x, to represent the unknown quantity, then build the equation step by step. Look for key words such as ‘more than’, ‘less than’, ‘twice’, ‘sum’ and ‘difference’.

应用题要求把文字语句转化为代数方程。通常用变量 x 表示未知量，然后一步一步建立方程。要注意关键词，例如 ‘多于’、’少于’、’两倍’、’和’ 和 ‘差’。

Example: Three more than twice a number is 19. Let the number be x. Twice the number is 2x. Three more is 2x + 3. Equation: 2x + 3 = 19, so x = 8.

例子：一个数的两倍再加 3 等于 19。设这个数为 x。

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）