📚 IGCSE Grade 8 Maths Topic 794: Solving Linear Equations | IGCSE 八年级数学主题 794：解一元一次方程
Linear equations are one of the most important building blocks in IGCSE Mathematics. They appear in algebra, problem solving, graphs and even geometry, and almost every later topic - simultaneous equations, straight-line graphs, sequences and formulae - assumes that you can rearrange and solve them quickly. This revision guide covers the essential methods you need to solve one-variable linear equations confidently in Grade 8, from one-step equations to brackets, fractions, word problems and exam-style practice.
一元一次方程是 IGCSE 数学中最重要的基础内容之一。它们出现在代数、应用题、图像甚至几何中，而之后几乎每一个主题——联立方程、直线图像、数列与公式变形——都假定你能快速、熟练地解出这类方程。本复习指南将涵盖八年级需要掌握的求解一元一次方程的核心方法，从一步方程到含括号、含分数、应用题以及考试风格的练习。
1. What Is a Linear Equation? | 什么是一元一次方程？
A linear equation in one variable is an equation in which the unknown appears only to the power of 1. It contains an equals sign, and it can usually be written in the standard form ax + b = c, where a, b and c are numbers and a is not zero.
一元一次方程是指未知数的次数为 1 的方程。它含有等号，通常可以写成标准形式 ax + b = c，其中 a、b、c 是数字，且 a 不等于 0。
ax + b = c, a ≠ 0
For example, x + 5 = 12, 3x − 2 = 10 and 2(x + 1) = 8 are all linear equations. The expression x² + 3 = 7 is not linear because the power of x is 2. Similarly, 1 ÷ x = 4 and √x = 3 are not linear equations, because the unknown is not simply multiplied by a number.
例如，x + 5 = 12、3x − 2 = 10 和 2(x + 1) = 8 都是一元一次方程。而 x² + 3 = 7 不是一元一次方程，因为 x 的次数是 2。同样，1 ÷ x = 4 和 √x = 3 也不是一元一次方程，因为未知数并不只是被一个数字相乘。
The word 'linear' tells us that the graph of the equation is a straight line. In Grade 8, we focus on finding the value of the unknown that makes the equation true, and on using that skill to answer word problems and check answers by substitution.
“线性”一词表示方程的图像是一条直线。在八年级，我们重点学习如何求使方程成立的未知数的值，并运用这一技能解决应用题、通过代入检验答案。
2. Key Vocabulary and Notation | 关键术语与符号
Before solving equations, it is useful to know the main words used in IGCSE questions. A variable is a letter that stands for an unknown number; a coefficient is the number multiplied by the variable; a constant is a number on its own. Recognising these words quickly saves time in the exam.
在解方程之前，了解 IGCSE 题目中的主要术语非常有用。变量是表示未知数的字母；系数是乘以变量的数；常数是单独出现的数字。快速识别这些词能在考试中节省时间。
- Equation | 方程：a statement that two expressions are equal, such as 2x + 3 = 9。
- LHS / RHS | 左边 / 右边：the left-hand side and the right-hand side of the equals sign。
- Solution or root | 解或根：the value of the unknown that makes the equation true。
- Like terms | 同类项：terms with exactly the same variable part, such as 3x and 5x。
- Simplifying | 化简：collecting like terms before solving。
- Inverse operation | 逆运算：the operation that undoes another, e.g. addition and subtraction, multiplication and division。
Understanding these terms will help you follow instructions such as 'write an equation', 'solve for x' or 'show that x = 7'. A useful habit is to underline the key instruction word before you start working.
理解这些术语将帮助你理解“列方程”“求解 x”或“证明 x = 7”等指令。一个良好的习惯是：动笔前先把关键的指令词划出来。
3. The Golden Rule: Keep the Equation Balanced | 黄金法则：保持方程两边平衡
An equation is like a balance scale. Whatever you do to one side, you must do exactly the same to the other side. If you add 5 to the left-hand side, you must add 5 to the right-hand side; if you divide the left by 3, you must divide the right by 3. This single rule justifies every step you will use in Grade 8.
方程就像一架天平。你对一边做什么，就必须对另一边做完全相同的事。如果左边加 5，右边也必须加 5；如果左边除以 3，右边也必须除以 3。这一条法则可以解释八年级要用到的每一个解题步骤。
Solving therefore means undoing operations using inverse operations, usually working in reverse order to the way the expression was built. Undo addition or subtraction first, then undo multiplication or division. Remember: a term that moves across the equals sign changes its sign, which is simply the balance rule written in a shortcut form.
因此，解方程就是用逆运算把运算一步步“撤销”，通常按与式子构造相反的顺序进行：先撤销加减，再撤销乘除。记住：项从等号一边移到另一边时要变号，这其实就是平衡法则的简写形式。
| Operation in the equation | 方程中的运算 | Inverse operation | 逆运算 |
| + 7 | 加 7 | − 7 | 减 7 |
| − 4 | 减 4 | + 4 | 加 4 |
| × 5 | 乘 5 | ÷ 5 | 除以 5 |
| ÷ 3 | 除以 3 | × 3 | 乘 3 |
4. Solving One-Step Equations | 解一步方程
The simplest equations need only one inverse operation. If the unknown is added to a number, subtract that number from both sides. If the unknown is multiplied by a number, divide both sides by that number.
最简单的方程只需要一次逆运算。如果未知数与一个数相加，就从两边同时减去这个数；如果未知数乘以一个数，就把两边同时除以这个数。
Example 1: Solve x + 7 = 15. Subtract 7 from both sides: x = 15 − 7 = 8.
例 1：解 x + 7 = 15。两边同时减 7：x = 15 − 7 = 8。
Example 2: Solve x − 4 = 9. Add 4 to both sides: x = 9 + 4 = 13.
例 2：解 x − 4 = 9。两边同时加 4：x = 9 + 4 = 13。
Example 3: Solve 5x = 35. Divide both sides by 5: x = 35 ÷ 5 = 7.
例 3：解 5x = 35。两边同时除以 5：x = 35 ÷ 5 = 7。
Example 4: Solve x ÷ 4 = 6. Multiply both sides by 4: x = 6 × 4 = 24.
例 4：解 x ÷ 4 = 6。两边同时乘 4：x = 6 × 4 = 24。
Example 5: Solve −3x = 12. Divide both sides by −3: x ÷ (−3) = −4. Watch the sign carefully; this is the most common slip in one-step questions.
例 5：解 −3x = 12。两边同时除以 −3：x = 12 ÷ (−3) = −4。务必小心正负号，这是一步方程中最常见的失误。
5. Solving Two-Step Equations | 解两步方程
Most IGCSE equations at this level need two steps. The reliable order is: undo the addition or subtraction first, then undo the multiplication or division. Do not divide first unless the coefficient is a factor of every term.
这个阶段的大多数 IGCSE 方程需要两步。可靠的顺序是：先撤销加减，再撤销乘除。除非系数是每一项的因数，否则不要先做除法。
Example 1: Solve 3x + 5 = 20. Subtract 5: 3x = 15. Divide by 3: x = 5.
例 1：解 3x + 5 = 20。先减 5：3x = 15；再除以 3：x = 5。
Example 2: Solve 2x − 7 = 11. Add 7: 2x = 18. Divide by 2: x = 9.
例 2：解 2x − 7 = 11。先加 7：2x = 18；再除以 2：x = 9。
Example 3: Solve 5 − 2x = 1. Subtract 5: −2x = −4. Divide by −2: x = 2. When the coefficient is negative, many students prefer to add 2x to both sides first: 5 = 1 + 2x, so 2x = 4 and x = 2. Both routes are acceptable.
例 3：解 5 − 2x = 1。先减 5：−2x = −4；再除以 −2：x = 2。当系数为负时，很多同学喜欢先在两边同时加 2x：5 = 1 + 2x，所以 2x = 4，x = 2。两种做法都可以。
Example 4: Solve x ÷ 3 + 4 = 10. Subtract 4: x ÷ 3 = 6. Multiply by 3: x = 18.
例 4：解 x ÷ 3 + 4 = 10。先减 4：x ÷ 3 = 6；再乘 3：x = 18。
6. Equations with Brackets | 含括号的方程
When an equation contains brackets, you have two valid strategies: expand the brackets first, or divide both sides by the number in front of the bracket if it divides exactly. Choose the method that keeps the numbers simple.
当方程含有括号时，有两种可行的策略：先展开括号，或者当括号前的数能整除时，先把两边同时除以这个数。选择让数字更简单的方法。
Method A - Expand first | 方法 A：先展开： Solve 3(x + 4) = 27. Expanding gives 3x + 12 = 27, so 3x = 15 and x = 5.
方法 A：解 3(x + 4) = 27。展开得 3x + 12 = 27，所以 3x = 15，x = 5。
Method B - Divide first | 方法 B：先除： The same equation can be solved by dividing both sides by 3 immediately: x + 4 = 9, so x = 5. This is usually faster when the number outside divides the other side exactly.
方法 B：同一个方程也可以两边先除以 3：x + 4 = 9，所以 x = 5。当括号外的数能整除另一边时，这种方法通常更快。
Careful with negative signs | 小心负号： Solve 2(3x − 1) = 16. Expanding: 6x − 2 = 16, so 6x = 18 and x = 3. Expanding −4(x − 3) gives −4x + 12, not −4x − 12. The minus sign multiplies every term inside the bracket.
小心负号：解 2(3x − 1) = 16。展开得 6x − 2 = 16，所以 6x = 18，x = 3。展开 −4(x − 3) 得到的是 −4x + 12，而不是 −4x − 12。负号要乘到括号内的每一项。
Longer chain | 更长的式子： Solve 5(2x + 3) − 4 = 31. Add 4: 5(2x + 3) = 35. Divide by 5: 2x + 3 = 7. Subtract 3: 2x = 4. Divide by 2: x = 2.
更长的式子：解 5(2x + 3) − 4 = 31。先加 4：5(2x + 3) = 35；除以 5：2x + 3 = 7；减 3：2x = 4；除以 2：x = 2。
7. Variables on Both Sides | 两边都有未知数的方程
When the unknown appears on both sides, first collect all the variable terms on one side and all the constant terms on the other. It helps to move the smaller variable term so that the coefficient stays positive.
当未知数出现在两边时，先把所有含未知数的项集中到一边，把所有常数项集中到另一边。把较小的未知数项移走，能让系数保持为正，更不容易出错。
Example 1: Solve 5x + 3 = 2x + 12. Subtract 2x from both sides: 3x + 3 = 12. Subtract 3: 3x = 9, so x = 3.
例 1：解 5x + 3 = 2x + 12。两边同时减 2x：3x + 3 = 12；再减 3：3x = 9，所以 x = 3。
Example 2: Solve 3x − 4 = 7x + 8. Subtract 3x: −4 = 4x + 8. Subtract 8: −12 = 4x. Divide by 4: x = −3.
例 2：解 3x − 4 = 7x + 8。两边减 3x：−4 = 4x + 8；再减 8：−12 = 4x；除以 4：x = −3。
Example 3 with brackets: Solve 4(x + 1) = 2x + 10. Expand: 4x + 4 = 2x + 10. Subtract 2x: 2x + 4 = 10. Subtract 4: 2x = 6, so x = 3.
例 3（含括号）：解 4(x + 1) = 2x + 10。展开：4x + 4 = 2x + 10；减 2x：2x + 4 = 10；减 4：2x = 6，所以 x = 3。
Special cases | 特殊情况： The equation 2(x + 3) = 2x + 6 is true for every value of x - it is an identity with infinitely many solutions. The equation x + 5 = x + 2 has no solution, because subtracting x which is false. Higher-tier questions sometimes ask you to state the number of solutions.
特殊情况：方程 2(x + 3) = 2x + 6 对任何 x 都成立——它是恒等式，有无穷多个解。方程 x + 5 = x + 2 无解，因为两边同时减 x 得到 5 = 2，这是错误的。进阶题目有时会要求你写出解的个数。
8. Equations with Fractions | 含分数的方程
Fractions look intimidating, but the method is always the same: multiply every term by a common denominator, usually the lowest common multiple (LCM) of the denominators, to clear all fractions in one step.
分数看起来吓人，但方法始终一样：把每一项都乘以公分母（通常是各分母的最小公倍数 LCM），一步就能把所有分数消掉。
Example 1: Solve (x + 1) ÷ 3 = 4. Multiply both sides by 3: x + 1 = 12, so x = 11.
例 1：解 (x + 1) ÷ 3 = 4。两边同时乘 3：x + 1 = 12，所以 x = 11。
Example 2: Solve (2x − 1) ÷ 5 + x ÷ 2 = 3. The LCM of 5 and 2 is 10. Multiply every term by 10: 2(2x − 1) + 5x = 30. Expand: 4x − 2 + 5x = 30. Collect like terms: 9x − 2 = 30, so 9x = 32 and x = 32 ÷ 9 ≈ 3.56.
例 2：解 (2x − 1) ÷ 5 + x ÷ 2 = 3。5 与 2 的最小公倍数是 10。把每一项都乘 10：2(2x − 1) + 5x = 30。展开：4x − 2 + 5x = 30。合并同类项：9x − 2 = 30，所以 9x = 32，x = 32 ÷ 9 ≈ 3.56。
Example 3 - cross multiplication | 例 3：交叉相乘： Solve x ÷ 4 = (x + 3) ÷ 6. Cross multiplying gives 6x = 4(x + 3), so 6x = 4x + 12, 2x = 12 and x = 6.
例 3：解 x ÷ 4 = (x + 3) ÷ 6。交叉相乘得 6x = 4(x + 3)，所以 6x = 4x + 12，2x = 12，x = 6。
Top tip | 高分提示： After clearing fractions, always re-read the equation to check that you multiplied every single term, including the whole numbers. A missing term is the most frequent mistake in this type of question.
高分提示：消去分数之后，务必重读方程，检查是否把每一项（包括整数项）都乘了。漏乘某一项是这类题目中最常见的错误。
9. Forming Equations from Word Problems | 根据应用题列方程
Word problems carry many of the marks in IGCSE papers. Follow a fixed routine: define the unknown with a letter, translate each sentence into algebra, solve the equation, then interpret the answer in the context of the question.
应用题在 IGCSE 试卷中占很多分。请遵循固定的流程：用一个字母表示未知数，把每句话翻译成代数式，解方程，最后结合题目情境解释答案。
- Step 1 | 第一步：Choose a letter and state clearly what it represents, e.g. 'let x be the width in cm'。
- Step 2 | 第二步：Write expressions for the other unknown quantities, e.g. length = x + 3。
- Step 3 | 第三步：Use the given total, perimeter or age relationship to write the equation。
- Step 4 | 第四步：Solve and check the answer makes sense (lengths cannot be negative)。
Example 1 - perimeter | 例 1：周长： A rectangle is 3 cm longer than it is wide and its perimeter is 34 cm. Let the width be w, so the length is w + 3. Then 2(w + w + 3) = 34, so 2(2w + 3) = 34, 2w + 3 = 17, 2w = 14 and w = 7. The width is 7 cm and the length is 10 cm.
例 1：一个长方形比宽长 3 cm，周长是 34 cm。设宽为 w，则长为 w + 3。于是 2(w + w + 3) = 34，即 2(2w + 3) = 34，2w + 3 = 17，2w = 14，w = 7。宽为
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