📚 Solving Linear Equations and Inequalities for IGCSE Mathematics | IGCSE 数学：解一元一次方程与不等式
Linear equations and inequalities form one of the most important foundation topics in IGCSE Mathematics. They appear in almost every paper, either as standalone questions or as steps inside larger problems involving graphs, sequences, area, or speed. This revision guide explains the key methods step by step, from simple equations to compound inequalities, and includes common errors and exam-style practice. Work through each section carefully and always show clear algebraic working.
一元一次方程与不等式是 IGCSE 数学中最基础也最重要的主题之一。它们几乎出现在每份试卷中，既可能独立出题，也可能作为图形、数列、面积或速度等综合题中的解题步骤。本复习指南逐步讲解从简单方程到复合不等式的主要方法，并包含常见错误和考试风格练习。请仔细学习每一节，并始终写出清晰的代数过程。
1. What is a Linear Equation? | 什么是一元一次方程？
uadratic and therefore not linear. Linear equations in one variable usually have exactly one solution. Other typical linear forms include 5 − 2y = 1, x/4 = 3, and 2(x + 5) = 14, because after any brackets are expanded the unknown is still only to the power 1.
一元一次方程是指未知数只出现一次幂的方程。例如 3x + 7 = 19 是一元一次方程，而 x² − 4x + 3 = 0 是二次方程，因此不是一元一次方程。一元一次方程通常恰好有一个解。其他常见的一元一次形式包括 5 − 2y = 1、x/4 = 3 以及 2(x + 5) = 14，因为展开括号后未知数仍然只是一次幂。
To solve a linear equation, we find the value of x that makes the left-hand side equal to the right-hand side. This value is called the solution or the root of the equation. A quick substitution check should always be performed at the end: if both sides give the same number, the answer is correct.
要解一元一次方程，我们需要找到使左边等于右边的 x 的值。这个值称为方程的解，也称为根。最后一定要做一次代入检验：如果两边算出同一个数，答案就是正确的。
| Equation | Linear? | Reason |
| 3x + 7 = 19 | Yes | x appears to power 1 |
| x² − 4x + 3 = 0 | No | Highest power is 2 (quadratic) |
| 5 − 2y = 1 | Yes | Single power of y |
| 2(x + 3) = 10 | Yes | Expands to 2x + 6 = 10 |
| 1/x + 2 = 5 | No | Unknown in the denominator |
| √x = 4 | No | Unknown under a root sign |
2. The Balance Method: One-Step and Two-Step Equations | 天平法：解一步与两步方程
Think of an equation as a balanced pair of scales. Whatever you do to one side, you must do exactly the same to the other side. To undo an operation we use its inverse: addition is undone by subtraction, multiplication is undone by division, and vice versa. This single idea solves almost every linear equation you will meet at IGCSE.
把方程想象成一台平衡的天平。你对一边做什么，就必须对另一边做完全相同的事。要抵消一个运算，就用它的逆运算：加法用减法抵消，乘法用除法抵消，反之亦然。这个简单的思想可以解决 IGCSE 中几乎所有的线性方程。
3x + 7 = 19 → 3x = 12 → x = 4
2: divide both sides by 3 to leave x alone. Step 3: check by substitution, 3 × 4 + 7 = 19. ✓ The value x = 4 is the root of the equation.
第一步：两边同时减去 7，消去常数项。第二步：两边同时除以 3，使 x 单独留在一边。第三步：代入检验，3 × 4 + 7 = 19。✓ 因此 x = 4 是该方程的根。
= 40. Notice that dividing by 5 is undone by multiplying by 5, not by dividing by 5 again.
第二个带分数的例子是 x/5 − 2 = 6。两边同时加 2 得到 x/5 = 8，再两边同时乘 5 得到 x = 40。注意：除以 5 要用乘以 5 来抵消，而不是再除以 5。
3. Equations with Brackets | 含括号的方程
x − 3) = 20 expands to 8x − 12 = 20, so 8x = 32 and x = 4. Expanding first keeps the arithmetic readable and reduces sign mistakes.
当方程中含有括号时，应先展开括号，再合并同类项，最后才移项。例如 4(2x − 3) = 20 展开为 8x − 12 = 20，所以 8x = 32，x = 4。先展开可以让运算更清晰，也能减少符号错误。
because every term inside must be multiplied by −2. The equation then gives 3x − 12 − 2x − 2 = 5, so x − 14 = 5 and x = 19.
要特别小心括号前的负号或负数。在 3(x − 4) − 2(x + 1) = 5 中，第二个括号展开为 −2x − 2，而不是 −2x + 2，因为括号内每一项都要乘以 −2。于是方程变为 3x − 12 − 2x − 2 = 5，即 x − 14 = 5，所以 x = 19。
- Expand every bracket completely. | 完整展开每一个括号。
- Collect like terms on each side. | 在每一边合并同类项。
- Move all unknown terms to one side and all numbers to the other. | 把所有含未知数的项移到一边，所有数字移到另一边。
- Divide by the coefficient of the unknown. | 除以未知数的系数。
4. Equations with Fractions | 含分数的方程
+ (x − 2)/4 = 2, the LCM of 3 and 4 is 12. Multiplying every term by 12 gives 4(x + 1) + 3(x − 2) = 24, which expands to 4x + 4 + 3x − 6 = 24, so 7x − 2 = 24, 7x = 26 and x = 26/7 ≈ 3.71.
分数看起来很吓人，但只要把整个方程乘以各分母的最小公倍数，分数就会立刻消失。对于 (x + 1)/3 + (x − 2)/4 = 2，3 和 4 的最小公倍数是 12。每一项乘以 12 得到 4(x + 1) + 3(x − 2) = 24，展开为 4x + 4 + 3x − 6 = 24，即 7x − 2 = 24，所以 7x = 26，x = 26/7 ≈ 3.71。
er that the denominator can never be zero, and leave final answers as exact fractions unless the question asks for a decimal.
当两边各是一个分数时，交叉相乘更快：由 (2x + 1)/5 = (x − 3)/2 可得 2(2x + 1) = 5(x − 3)，即 4x + 2 = 5x − 15，所以 x = 17。请记住分母永远不能为零，并且除题目要求小数外，最终答案应保留精确分数形式。
5. Equations with the Unknown on Both Sides | 未知数在两边的方程
ubtract 2x from both sides to get 3x − 3 = 9, then add 3 and divide by 3 to obtain x = 4. Checking: 5 × 4 − 3 = 17 and 2 × 4 + 9 = 17. ✓
当未知数出现在两边时，选择系数较大的那一边，通过相减使系数保持为正。对于 5x − 3 = 2x + 9，两边同时减去 2x 得到 3x − 3 = 9，再加 3 并除以 3，得到 x = 4。检验：5 × 4 − 3 = 17，2 × 4 + 9 = 17。✓
x + 1) = 2x + 2, both sides simplify to the same expression, so the equation is an identity and every value of x satisfies it. Recognising these two cases earns marks in extended papers.
有些方程无解，或者有无穷多个解。在 2x + 5 = 2x + 9 中，两边减去 2x 得到 5 = 9，这是错误的，所以方程无解。在 2(x + 1) = 2x + 2 中，两边化简后完全相同，因此这是一个恒等式，任何 x 值都满足它。能识别这两种情况在进阶试卷中可以得分。
6. Forming and Solving Equations from Word Problems | 由应用题列方程并求解
de a simple linear equation inside a word problem. Translate the words into algebra first: "sum" means +, "difference" means −, "product" means ×, "is" or "gives" means =, and "more than" or "less than" . Then define your unknown clearly, usually as x, and write down what each sentence means.
很多考题会把一个简单的线性方程藏在应用题里。先把文字翻译成代数："和"表示 +，"差"表示 −，"积"表示 ×，"是"或"等于"表示 =，而"多"或"少"告诉了你哪个量更大。然后清楚地定义未知数（通常设为 x），并逐句写出它对应的代数含义。
, so 3n = 69 and n = 23. The three integers are 23, 24 and 25. Always finish by answering the question in context and stating the units.
例如：三个连续整数之和为 72。设最小的为 n，则另外两个是 n + 1 和 n + 2。于是 n + (n + 1) + (n + 2) = 72，得到 3n + 3 = 72，所以 3n = 69，n = 23。三个整数是 23、24 和 25。最后一定要结合题目情境作答，并写出单位。
th is w + 3 and 2(w + w + 3) = 42, giving 2w + 3 = 21, 2w = 18 and w = 9. The width is 9 cm and the length is 12 cm.
几何与运动问题同样常见。一个长方形的长比宽多 3 cm，周长为 42 cm。设宽为 w，则长为 w + 3，于是 2(w + w + 3) = 42，得到 2w + 3 = 21，2w = 18，w = 9。因此宽是 9 cm，长是 12 cm。
7. Rearranging Formulas (Changing the Subject) | 公式变形（换主元）
ke x the subject of 2x + 3y = 12, subtract 3y from both sides to get 2x = 12 − 3y, then divide by 2 to obtain x = (12 − 3y)/2, or equivalently x = 6 − 1.5y.
公式变形所用的规则与解方程完全相同，只是答案是一个表达式而不是一个数。要把 2x + 3y = 12 变为以 x 为主元，两边同时减去 3y 得到 2x = 12 − 3y，再除以 2 得到 x = (12 − 3y)/2，等价地写作 x = 6 − 1.5y。
his factorising step is a favourite of examiners, so practise it until it feels automatic.
如果未知数出现两次，先把它们集中到一边再提取公因式。由 ax + b = cx + d 可得 ax − cx = d − b，所以 x(a − c) = d − b，即 x = (d − b)/(a − c)。这一步提取公因式是考官非常喜欢的考点，请练到能自动反应为止。
8. What is a Linear Inequality? | 什么是一元一次不等式
han), ≤ (less than or equal to) and ≥ (greater than or equal to). There is also ≠, meaning "not equal to" ty contains the unknown only to the first power, for example 3x + 7 < 19 or 5 − 2y ≥ 1.
不等式是把两个表达式进行比较，而不是说明它们相等。四个主要符号是：<（小于）、>（大于）、≤（小于或等于）和 ≥（大于或等于），另外还有 ≠，表示"不等于"。一元一次不等式中的未知数只出现一次幂，例如 3x + 7 < 19 或 5 − 2y ≥ 1。
The big difference from an equation is the number of solutions. Instead of one value, an inequality normally has an infinite set of solutions, which is why answers are often shown on a number line or given as a range such as x < 4. When the question asks for integer solutions, you simply list the whole numbers that fit inside the range.
与方程最大的区别在于解的个数。不等式通常有无穷多个解，而不是一个值，因此答案常画在数轴上，或写成如 x < 4 这样的范围。当题目要求整数解时，只需列出落在该范围内的整数即可。
9. Solving Linear Inequalities: The Flip Rule | 解一元一次不等式：翻转法则
Solving a linear inequality works in exactly the same way as solving a linear equation: add, subtract, multiply or divide both sides by the same value, and collect terms as usual. For 3x + 7 < 19 we subtract 7 to get 3x < 12, then divide by 3 to obtain x < 4. The only extra rule concerns negative numbers.
解一元一次不等式的方法与解一元一次方程完全相同：两边同时加、减、乘、除同一个数，并照常合并同类项。对于 3x + 7 < 19，两边减去 7 得到 3x < 12，再除以 3 得到 x < 4。唯一额外的规则与负数有关。
When you multiply or divide both sides of an inequality by a negative number, the inequality sign must be reversed. For example, −2x > 8 becomes x < −4, not x > −4. Forgetting this reversal is one of the most common mistakes in the whole IGCSE course, so check the sign of the number you are dividing by every single time.
当不等式两边同时乘以或除以一个负数
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