📚 Solving Linear Equations: From Balancing to Word Problems | 解一元一次方程：从平衡法到应用题
Linear equations appear throughout the Cambridge KS3 Mathematics curriculum, and they are the first real tool of algebra: once you can solve them, you can unlock sequences, graphs, geometry and later GCSE work. This article covers the essential skills needed to solve one-variable linear equations confidently, including balancing, inverse operations, brackets, unknown on both sides, fractions and word problems. Work through each section in order, and try the practice questions at the end.
线性方程贯穿剑桥 KS3 数学课程，它是代数中第一件真正有用的工具：一旦你学会解方程，就能打开数列、图像、几何以及之后 GCSE 内容的大门。本文涵盖解一元一次方程所需的核心技能，包括平衡法、逆运算、去括号、未知数在两侧、含分数以及应用题。请按顺序学习每个小节，并在最后尝试练习题。
1. What Is a Linear Equation? | 什么是线性方程？
A linear equation is an equation in which the unknown variable appears only to the first power. Terms such as x, 2x and -3x are allowed, but x², x³, 1/x and √x general form of a one-variable linear equation is ax + b = c, where a, b and c are numbers and a ≠ 0. At KS3 you normally work with one unknown, and the solution is usually an integer or a simple rational number such as a fraction or a decimal.
线性方程是未知数只出现一次方的方程。像 x、2x 和 -3x 这样的项是允许的，但 x²、x³、1/x 和 √x 不可以。线性关系的解在图上画出来会形成一条直线，这正是线性方程名字的来源。一元线性方程的一般形式为 ax + b = c，其中 a、b、c 是数且 a ≠ 0。在 KS3 阶段，通常只处理一个未知数，解一般也是整数或简单有理数（分数、小数）。
| Allowed 允许出现 | x, 2x, -3x, x/5, 0.4x |
| Not allowed 不允许出现 | x², x³, 1/x, √x |
A simple example of a linear equation is shown below.
下面是一个最简单的线性方程示例。
x + 5 = 12
2. The Balancing Method | 平衡法
Think of an equation as a set of balance scales. Whatever you do to one side, you must do to the other side to keep the scales balanced. This idea is the foundation of all equation solving, and it is the reason why every step you write down must be applied to both sides at once.
可以把方程想象成一台天平。对等式的一边做什么操作，就必须对另一边做同样的操作，天平才能保持平衡。这一思想是解所有方程的基础，也是为什么你写下的每一步都必须同时作用于两边。
act 5 from both sides because addition and subtraction are opposite operations. The equation stays true because the same amount is removed from each side. Later, when you meet multiplication and division, the same rule applies.
例如，如果 x + 5 = 12，你可以从两边同时减去 5，因为加法和减法是相反的运算。等式依然成立，因为两边去掉的是同一个数。之后遇到乘法和除法时，同样的规则依然适用。
x + 5 - 5 = 12 - 5
x = 7
This principle allows you to add, subtract, multiply or divide, as long as the same operation is applied to both sides. Remember that you may never divide both sides by zero, because division by zero is undefined.
这一原则允许你进行加、减、乘、除，只要两边做的是相同操作即可。请记住，永远不能两边同时除以 0，因为除以 0 没有意义。
3. Inverse Operations and the Order of Unwrapping | 逆运算与"拆解"顺序
Inverse operations undo each other: addition undoes subtraction, and multiplication undoes division. To solve an equation you apply inverse operations to the unknown in the reverse order of BIDMAS, so you strip away the operations in the opposite order to the one in which they were built up.
逆运算是互相抵消的运算：加法抵消减法，乘法抵消除法。解方程时，你要按照 BIDMAS 的相反顺序对未知数施加逆运算，也就是说，按运算"被加上去"的反向顺序把它们一层层剥掉。
5, and then dividing both sides by 3 gives x = 5. Substituting back, 3 × 5 + 4 = 19, so the answer is correct.
以 3x + 4 = 19 为例。未知数 x 先乘以 3，再加上 4，所以要先抵消加法。两边同时减 4 得到 3x = 15，再两边同时除以 3 得到 x = 5。代回检验：3 × 5 + 4 = 19，答案正确。
3x + 4 - 4 = 19 - 4
3x = 15
3x ÷ 3 = 15 ÷ 3
x = 5
4. Two-Step and Multi-Step Equations | 两步与多步方程
5x - 7 = 23: add 7 to both sides to get 5x = 30, then divide both sides by 5 to get x = 6.
KS3 阶段的大多数方程需要两步或更多步。典型的两步方程是 5x - 7 = 23：两边同时加 7 得到 5x = 30，再两边同时除以 5 得到 x = 6。
x and 3x to obtain 5x - 4 = 21, then add 4 to both sides to get 5x = 25, and finally divide by 5 to get x = 5. Simplifying each side first always makes the later steps shorter and reduces the chance of a mistake.
当等式同一侧含有同类项时，先合并再移项。对 2x + 3x - 4 = 21，先把 2x 与 3x 合并得到 5x - 4 = 21，然后两边加 4 得 5x = 25，最后除以 5 得 x = 5。先把每一侧化简，后面的步骤会短很多，出错的概率也更小。
2x + 3x - 4 = 21 → 5x - 4 = 21 → 5x = 25 → x = 5
5. Equations with Brackets | 含括号的方程
t. For 3(x + 4) = 27, expanding gives 3x + 12 = 27, so 3x = 15 and x = 5. Dividing both sides by 3 first gives x + 4 = 9, so x = 5 as well.
遇到括号时有两种常用方法：先展开括号，或者两边同时除以括号外的乘数。对 3(x + 4) = 27，展开得 3x + 12 = 27，所以 3x = 15，x = 5；先两边除以 3 得 x + 4 = 9，同样得 x = 5。
x = 4 and x = -2. With more than one bracket, expand them all before collecting like terms: 2(3x - 1) - (x + 5) = 12 becomes 6x - 2 - x - 5 = 12, then 5x - 7 = 12, so 5x = 19 and x = 19/5 = 3.8.
当括号外的乘数为负数时要特别小心，因为括号内每一项都要变号。对 -2(x - 3) = 10，展开得 -2x + 6 = 10，于是 -2x = 4，x = -2。若有两个以上括号，应全部展开后再合并同类项：2(3x - 1) - (x + 5) = 12 化为 6x - 2 - x - 5 = 12，即 5x - 7 = 12，所以 5x = 19，x = 19/5 = 3.8。
3(x + 4) = 27 → 3x + 12 = 27 → 3x = 15 → x = 5
6. Unknowns on Both Sides | 未知数出现在两侧
2x + 9. Subtracting 2x from both sides gives 3x - 3 = 9, then 3x = 12, and finally x = 4.
当未知数出现在等式两侧时，通过两边同时加减较小的含 x 项，把它从一侧消掉。以 5x - 3 = 2x + 9 为例，两边同时减 2x 得 3x - 3 = 9，于是 3x = 12，最后 x = 4。
quation: the left side is 5 × 4 - 3 = 17 and the right side is 2 × 4 + 9 = 17, so both sides agree.
选择减去较小的系数可以让含 x 的项保持为正，避免负号带来的混乱。务必把答案代回原方程检验：左边 5 × 4 - 3 = 17，右边 2 × 4 + 9 = 17，两边相等。
5x - 3 = 2x + 9 → 3x - 3 = 9 → 3x = 12 → x = 4
7. Equations with Fractions | 含分数的方程
Fractions are easier to handle if you clear them early. One approach is to multiply both sides by the denominator of the fraction; a faster approach is to multiply both sides by the lowest common multiple of all the denominators, so that every fraction disappears in a single step.
分数如果早点清除会好处理得多。一种方法是两边同时乘以分数的分母；更快的方法是把两边同时乘以所有分母的最小公倍数，这样一步就能消掉所有分数。
- 1)/4 = 3, multiply both sides by 4 to get 2x - 1 = 12, then 2x = 13 and x = 13/2 = 6.5.
对 x/2 + x/3 = 5，2 和 3 的最小公倍数是 6。把每一项都乘以 6 得 3x + 2x = 30，于是 5x = 30，x = 6。对 (2x - 1)/4 = 3，两边同时乘以 4 得 2x - 1 = 12，于是 2x = 13，x = 13/2 = 6.5。
Remember that when the numerator of a fraction contains more than one term, it behaves as a single group. That is why the bracket around 2x - 1 must be kept until the fraction has been cleared.
请记住，当分数的分子含有不止一项时，它整体是一个"组"。这也是为什么在清除分数之前，2x - 1 外面的括号必须保留。
x/2 + x/3 = 5 → 3x + 2x = 30 → 5x = 30 → x = 6
8. Special Cases: No Solution and Infinitely Many | 特殊情况：无解与无穷多解
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