📚 Mastering Linear Equations for Cambridge KS3 Mathematics | 掌握剑桥 KS3 数学中的线性方程
Linear equations are one of the most important building blocks in the Cambridge KS3 mathematics curriculum. They appear in nearly every topic, from solving word problems to understanding graphs and formulas. If you can solve a linear equation confidently, you already have a powerful tool for later work in algebra, geometry and even science. This article takes you through the key ideas step by step, with clear explanations, worked examples, exam-style tips and a short practice set. It is designed to support classroom learning and revision for tasks such as the exercise on page 183, question 2.
线性方程是剑桥 KS3 数学课程中最重要的基础模块之一。它们几乎出现在每个主题中，从解文字题到理解图像和公式都离不开它。如果你能自信地解出一个线性方程，就已经掌握了代数、几何甚至科学后续学习的强大工具。本文将带你一步步梳理核心概念，提供清晰的解释、例题、应试技巧和一组精简练习，旨在帮助你巩固课堂所学并复习类似第 183 页第 2 题这样的练习。
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 unknown is 1. The unknown, usually written x, is never squared, cubed or raised to any higher power, and it never appears under a square-root sign or in a denominator. A linear equation in one variable can always be written in the general form ax + b = c, where a, b and c are numbers and a is not zero.
线性方程是指未知数的最高次数为 1 的方程。未知数（通常写作 x）不会被平方、立方或进行任何更高次运算，也不会出现在根号下或分母中。一元一次方程总是可以写成一般形式 ax + b = c，其中 a、b、c 是数字，且 a 不等于 0。
ax + b = c (a ≠ 0)
7, 2(x - 4) = 10, x ÷ 3 + 2 = 5 and 4x = 12. Expressions such as x² + 1 = 5, 3 ÷ x = 6 and √x = 4 are not linear, because the power of the unknown is 2, -1 and ½ respectively.
常见的例子包括 3x + 5 = 17、2(x - 4) = 10、x ÷ 3 + 2 = 5 和 4x = 12。而 x² + 1 = 5、3 ÷ x = 6 以及 √x = 4 都不是线性方程，因为未知数的次数分别是 2、-1 和 ½。
The word 'linear' means 'line' , all the points lie on a perfectly straight line. Understanding this link between algebra and graphs will help you later with coordinate geometry, gradients and simultaneous equations.
'线性'一词意为'直线'。当把 y = 2x + 1 这样的线性关系画在坐标图上时，所有点都落在一条笔直的直线上。理解代数与图像之间的这种联系，将有助于你后续学习坐标几何、斜率以及联立方程。
2. The Balancing Method | 平衡法
Think of an equation as a set of balance scales. The left-hand side and the right-hand side must always have the same value. Whatever you do to one side of the equation, you must do exactly the same to the other side. This is the golden rule of solving equations, and it is the reason every step is legal.
把方程想象成一台天平。左边和右边的值必须始终相等。你对方程的一边做了什么，就必须对另一边也做完全相同的操作。这是解方程的金科玉律，也是每一步变形都成立的依据。
- Add the same number to both sides. 两边同时加上同一个数。
- Subtract the same number from both sides. 两边同时减去同一个数。
- Multiply both sides by the same non-zero number. 两边同时乘以同一个不为 0 的数。
- Divide both sides by the same non-zero number. 两边同时除以同一个不为 0 的数。
These four operations keep the equation balanced while you gradually isolate the unknown. In KS3 you will usually need two of them, applied in reverse order to the operations in the expression.
这四种操作能在保持方程平衡的同时，帮助你逐步分离出未知数。在 KS3 阶段通常需要用到其中两步，并且顺序要与算式中的运算顺序相反。
3x + 5 = 17 → 3x = 12 → x = 4
Check your answer by substituting back: 3(4) + 5 = 12 + 5 = 17. ✓ king is fast, and in an exam it protects marks that are easily lost to a single sign slip.
把答案代回检验：3(4) + 5 = 12 + 5 = 17。✓ 检验很快，在考试中能帮你保住因一个符号失误而丢掉的分数。
3. One-Step and Two-Step Equations | 一步与两步方程
both sides by 4 to get x = 7. If it is x ÷ 3 = 9, multiply both sides by 3 to get x = 27.
一步方程只需要一个逆运算。若方程是 x + 7 = 12，两边同时减 7 得 x = 5；若是 4x = 28，两边同时除以 4 得 x = 7；若是 x ÷ 3 = 9，两边同时乘以 3 得 x = 27。
Two-step equations need two inverse operations. Work in reverse BIDMAS order: undo addition or subtraction first, then undo multiplication or division.
两步方程需要两个逆运算。按照与运算顺序相反的顺序进行：先消去加减，再消去乘除。
5x - 7 = 23 → 5x = 30 → x = 6
ves -x = 5, so x = -5. Negative and decimal answers are perfectly acceptable and often appear on examination papers.
如果先减 7，数字会变得麻烦，所以顺序很重要。KS3 的解并不总是正整数：4 - x = 9 得到 -x = 5，所以 x = -5。负数解和小数解完全正常，在考试中也经常出现。
4. Equations with Brackets | 含括号的方程
When an equation contains brackets, you have two legal routes. You can expand the brackets first and then solve, or, if the bracket is multiplied by a factor that divides the other side exactly, you can divide both sides by that factor first. Both routes must give the same answer.
当方程中含括号时，有两条合法的路径。你可以先展开括号再求解；或者当括号外因数能整除另一边时，先两边同除这个因数。两条路径必须得到相同答案。
2(x - 4) = 10
= 9. Route 2 is quicker here, but route 1 is safer when the numbers do not divide neatly.
路径一（展开）：2x - 8 = 10，所以 2x = 18，x = 9。路径二（先除）：x - 4 = 5，所以 x = 9。此处路径二更快，但当数字不能整除时，路径一更稳妥。
- x = 7, so x = -6. A frequent higher-tier style question combines brackets on both sides:
括号前的负号会改变括号内每一项的符号：3 - (x + 2) = 7 变为 3 - x - 2 = 7，即 1 - x = 7，所以 x = -6。更常见的进阶题型是两边都含括号：
3(2x + 1) = 4(x + 4) → 6x + 3 = 4x + 16 → 2x = 13 → x = 6.5
The most common error is multiplying only the first term inside the bracket. Every term must be multiplied by the factor outside.
最常见的错误是只乘了括号内的第一项。括号内的每一项都必须乘以括号外的因数。
5. Variables on Both Sides | 两边都有未知数
When the unknown appears on both sides, collect all the x terms on one side and all the numbers on the other. The standard advice is to subtract the smaller coefficient of x, so that the resulting coefficient stays positive and the arithmetic is easier to check.
当未知数出现在两边时，要把所有含 x 的项集中到一边，把所有数字集中到另一边。通常建议减去较小的 x 系数，这样得到的系数保持正数，运算更容易检查。
5x - 3 = 2x + 9 → 3x - 3 = 9 → 3x = 12 → x = 4
oth sides, then divide by 3. Substituting x = 4 gives 17 = 17, so the solution is correct.
两边同时减 2x，再两边同时加 3，最后除以 3。代入 x = 4 得 17 = 17，说明解正确。
ty and is satisfied by every value of x. If the x terms cancel and the statement is false, for example x + 1 = x + 5, there is no solution.
有两种特殊情况值得掌握。若 x 项消去后等式成立，例如 2x + 3 = 2x + 3 或 4(x - 2) = 4x - 8，则该方程是恒等式，任何 x 值都满足它。若 x 项消去后等式不成立，例如 x + 1 = x + 5，则该方程无解。
6. Equations with Fractions and Decimals | 含分数与小数的方程
Fractions look harder than they are. Multiply every term by the lowest common multiple of the denominators to clear the fractions in one step, then solve the resulting linear equation.
分数看起来比实际更难。把每一项都乘以分母的最小公倍数，一步即可消去分数，然后解所得的线性方程。
x ÷ 2 + x ÷ 3 = 5 → 3x + 2x = 30 → 5x = 30 → x = 6
Here the LCM of 2 and 3 is 6, so every term is multiplied by 6. When the unknown appears in a numerator on both sides, cross-multiplication is the fastest method:
这里 2 和 3 的最小公倍数是 6，所以每一项都乘以 6。当两边分子中都含未知数时，交叉相乘最快：
(x + 1) ÷ 3 = (x - 2) ÷ 4 → 4(x + 1) = 3(x - 2) → 4x + 4 = 3x - 6 → x = -10
integer: 0.5x + 1.2 = 3.7 becomes 5x + 12 = 37, so 5x = 25 and x = 5. Remember to multiply every term, including terms that contain no fraction or decimal at all.
对于小数，两边同时乘以 10、100 或 1000，把每个小数都变成整数：0.5x + 1.2 = 3.7 变为 5x + 12 = 37，所以 5x = 25，x = 5。切记要乘以每一项，包括那些本来不含分数或小数的项。
7. Forming Equations from Word Problems | 从文字题列方程
Word problems are where linear equations earn their marks. Follow a fixed routine: define the unknown with a clear statement, translate the words into an equation, solve it, then answer the question in context and check that the answer is sensible.
文字题正是线性方程大显身手的地方。请遵循固定流程：用一句话明确设未知数，把文字翻译成方程，解方程，再结合情境作答，并检查答案是否合理。
- Define the variable. 设未知数，写清它代表什么。
- Write the equation from the given information. 根据已知信息列出方程。
- Solve using the balancing method. 用平衡法求解。
- Answer in words, with units. 用文字作答，并带上单位。
- Check by substituting and by common sense. 代入检验，并用常识判断。
1) + (x + 2) = 72, giving 3x + 3 = 72, 3x = 69 and x = 23. The numbers are 23, 24 and 25.
三个连续整数之和为 72。设最小的为 x，则 x + (x + 1) + (x + 2) = 72，得 3x + 3 = 72，3x = 69，x = 23。三个数是 23、24、25。
+ 2w + 3) = 36, giving 3w + 3 = 18 and w = 5. The width is 5 cm and the length is 13 cm. Always go back and check both dimensions satisfy the original words.
几何题同理。一个矩形的长比宽的两倍多 3 cm，周长是 36 cm。设宽为 w，则长为 2w + 3，于是 2(w + 2w + 3) = 36，得 3w + 3 = 18，w = 5。宽是 5 cm，长是 13 cm
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