📚 IGCSE Grade 8 Maths Series 600: Linear Equations and Inequalities | IGCSE 8年级数学系列600：线性方程与不等式
In IGCSE Mathematics, linear equations are the foundation of algebra. This article covers solving equations step by step, forming equations from real-life problems, and an introduction to linear inequalities. Mastering these skills will help you tackle a wide range of questions in the IGCSE exam.
在IGCSE数学中，线性方程是代数的基础。本文涵盖逐步解方程、从实际问题建立方程以及线性不等式入门。掌握这些技能将帮助你在IGCSE考试中应对各种题型。
1. Equations and Expressions: Know the Difference | 方程与表达式的区别
An expression is a combination of numbers, variables, and operations but has no equals sign. Examples include 3x + 2, 5y - 7, and 2a/3. You can simplify an expression by collecting like terms, but you cannot solve it because there is no statement of equality.
表达式由数字、变量和运算组合而成，但不含等号。例如 3x + 2、5y - 7 和 2a/3。表达式可以通过合并同类项来化简，但不能求解，因为它没有表示相等关系。
ng an equation means finding the value or values of the variable that make the statement true. This distinction is essential for setting up correct working in IGCSE exams.
方程表示两个表达式相等，例如 3x + 2 = 14。解方程就是找出使等式成立的变量值。这一区别对于在IGCSE考试中正确解题至关重要。
2. The Balancing Principle | 天平原理
The balancing principle is the foundation of solving equations. Think of an equation as a perfect balance scale. If you add 5 to the left side, you must add 5 to the right side; if you divide the left side by 3, you must divide the right side by 3. This keeps the equation balanced.
天平原理是解方程的基础。把方程想象成一架完美的天平。如果你在左边加5，右边也必须加5；如果你把左边除以3，右边也必须除以3。这样方程才能保持平衡。
s c is not zero. Using inverse operations in this balanced way isolates the variable step by step.
例如，如果 a = b，那么 a + c = b + c，a - c = b - c，a × c = b × c，以及 a ÷ c = b ÷ c，其中 c 不等于零。以这种平衡的方式使用逆运算可以逐步把变量分离出来。
If a = b, then a + c = b + c and a × c = b × c (c ≠ 0)
3. Solving One-Step Equations | 解一步方程
divide both sides by 4 to get x = 5. For x/6 = 2, multiply both sides by 6 to get x = 12.
一步方程只需一步逆运算即可解出。对于 x + 5 = 12，两边同时减5得到 x = 7。对于 x - 4 = 9，两边同时加4得到 x = 13。对于 4x = 20，两边同时除以4得到 x = 5。对于 x/6 = 2，两边同时乘以6得到 x = 12。
, then 7 + 5 = 12, which is true. This checking habit catches arithmetic mistakes quickly.
始终将解代回原方程进行检验。例如，如果 x = 7 代入 x + 5 = 12，则 7 + 5 = 12，成立。这个检验习惯能快速发现算术错误。
4. Solving Two-Step Equations | 解两步方程
th sides to get 2x = 8. Then undo the multiplication: divide both sides by 2 to get x = 4.
两步方程中变量经历两次运算。例如，在 2x + 3 = 11 中，变量 x 先乘以2，再加3。求解时，先消去加法：两边同时减3得到 2x = 8。再消去乘法：两边同时除以2得到 x = 4。
The general rule is to perform inverse operations in the reverse order of the original operations. If the equation follows the order multiply then add, you solve by subtract then divide. This reverse order unwinds the equation layer by layer.
一般规则是逆运算按原始运算的相反顺序进行。如果方程按先乘后加的顺序构建，求解时就应先减后除。这种相反顺序能一层一层地拆解方程。
2x + 3 = 11 → 2x = 8 → x = 4
5. Expanding Brackets First | 先去括号再解方程
21 expands to 3x + 6 = 21. Then subtract 6 from both sides and divide by 3 to find x = 5.
当方程含有括号时，先展开括号再求解。使用分配律 a(b + c) = ab + ac。例如，3(x + 2) = 21 展开得到 3x + 6 = 21。然后两边减6，再除以3，得到 x = 5。
is forgetting to multiply the negative sign by every term inside the bracket. Write each term separately to avoid sign mistakes.
注意负号：-2(x - 3) = -2x + 6，而不是 -2x - 6。一个常见错误是忘记将负号乘以括号内的每一项。将每一项单独写出来可以避免符号错误。
6. Equations with Fractions | 含分数的方程
by 3 to get x = 9. This step-by-step method works well when only one fraction is present.
解 x/3 + 2 = 5 时，两边先减2得到 x/3 = 3。然后两边乘以3得到 x = 9。当只有一个分数时，这种分步方法非常有效。
mple, in x/2 + x/3 = 5, multiply each term by 6 to get 3x + 2x = 30, so 5x = 30 and x = 6.
对于含有多个分数的方程，可以每一项都乘以最小公分母，一次性去分母。例如，在 x/2 + x/3 = 5 中，每项乘以6得到 3x + 2x = 30，所以 5x = 30，x = 6。
x/2 + x/3 = 5 → 3x + 2x = 30 → x = 6
7. Forming Equations from Word Problems | 从应用题建立方程
Word problems require translating natural language into algebraic equations. First define the variable: let x represent the unknown quantity. Then identify the operations described by words such as more than, less than, times, and total.
应用题要求将自然语言转化为代数方程。首先定义变量：设 x 表示未知量。然后确定“比……多”“比……少”“乘以”“总共”等词语所描述的运算。
For example, "Three times a number plus five equals twenty" Always answer the original question in context, including units if the question gives any.
例如，“三倍一个数加五等于二十”可以写成 3n + 5 = 20。求解得到 n = 5。始终结合题意回答原问题，如果题目给出单位，答案中也要包含单位。
- English: "The sum of twice a number and 7 is 19" → 2x + 7 = 19
- 中文: “一个数的两倍与7的和是19” → 2x + 7 = 19
8. Introduction to Linear Inequalities | 线性不等式入门
er than or equal to. Inequalities often have infinitely many solutions rather than a single value.
不等式使用以下符号比较两个表达式：< 表示小于，> 表示大于，≤ 表示小于或等于，≥ 表示大于或等于。不等式通常有无穷多个解，而不是单个值。
does not include 3 itself. By contrast, x ≤ -1 includes -1 and every number less than -1.
例如，x > 3 表示所有大于 3 的实数，如 3.1、4 和 100。它不包括 3 本身。相反，x ≤ -1 包括 -1 以及所有小于 -1 的数。
| Symbol | 符号 | Meaning | 含义 | Example | 示例 |
|---|---|---|
| < | less than | 小于 | x < 4 |
| > | greater than | 大于 | x > -2 |
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