📚 Solving Linear Equations | 求解线性方程
Linear equations are a cornerstone of Key Stage 3 mathematics. In the Cambridge Checkpoint syllabus, you first encounter them around page 59, where you start building the skills to find unknown values. Understanding how to solve equations will help you in algebra, geometry, and even science, because almost every later topic - from angle problems to speed-distance-time calculations to rearranging scientific formulae - depends on the same core technique: keep both sides balanced and unwind the operations step by step.
线性方程是 KS3 数学的基石。在剑桥 Checkpoint 大纲中，大约在第 59 页你会首次遇到它们，并开始学习如何求解未知数。理解如何解方程不仅对代数有帮助，对几何甚至科学也很有用，因为后面几乎每一个专题——从角度问题到速度、距离、时间的计算，再到科学公式的变形——都依赖同一个核心技巧：保持两边平衡，并一步步"拆解"运算。
This guide walks through the whole progression, from the simplest one-step equation to equations with brackets, fractions and unknowns on both sides, and finishes with exam-style practice and a revision checklist.
本文按难度递进，从最简单的一步方程，一直讲到含括号、含分数以及未知数出现在两边的方程，最后给出考题风格的练习和复习清单。
1. What is a Linear Equation? | 什么是线性方程？
+ 3 = 7 and 2x = 10 are both linear equations. The goal is to find the value of the unknown (often x) that makes the equation true. That value is called the solution, or the root, of the equation.
线性方程是表明两个表达式相等的数学陈述，其中变量的指数仅为 1。例如，x + 3 = 7 和 2x = 10 都是线性方程。我们的目标是找到使方程成立的未知数（通常用 x 表示）的值。这个值叫作方程的"解"或"根"。
Key features to look for: no variables in the denominator, no x² or higher powers, no roots or powers of the unknown, and only one unknown in most early problems. The two sides of the equation are usually called the left-hand side (LHS) and the right-hand side (RHS).
关键特征：分母中不含变量，没有 x² 或更高次幂，未知数也不出现在根号或指数中，并且在早期问题中通常只有一个未知数。方程的两边通常称为左边（LHS）和右边（RHS）。
expression apart from an equation. 3x + 5 is an expression - it can be simplified but not "solved" for every value of x, whereas an equation is normally true for only one particular value.
区分"表达式"和"方程"很重要。3x + 5 是一个表达式——它可以被化简，但不能被"求解"。3x + 5 = 20 是一个方程——它有等号，因此有解。另外要注意，像 2(x + 3) = 2x + 6 这样的恒等式对任意 x 都成立，而方程通常只对某一个特定的值成立。
2. Balancing the Equation | 平衡方程
Think of an equation as a perfectly balanced scale. Whatever operation you do to one side, you must do exactly the same to the other side to keep the balance. This is the golden rule of solving equations, and every method in this article is just a special case of it.
将方程想象成一个完全平衡的天平。无论你对一边进行什么运算，都必须对另一边进行完全相同的运算以保持平衡。这是解方程的金科玉律，本文后面所有的方法都只是这条规则的特别情形。
ct 5 from both sides to isolate x. The scale stays balanced: x + 5 − 5 = 12 − 5, so x = 7.
如果我们有 x + 5 = 12，我们可以从两边减去 5 来分离 x。天平保持平衡：x + 5 − 5 = 12 − 5，因此 x = 7。
x + 5 = 12 → subtract 5 from both sides → x = 7
x + 5 = 12 → 两边减 5 → x = 7
ring undoes a square root (though the last one rarely appears at this level). Always work backwards through the order of operations - undo the + and − first, then the × and ÷.
用来"拆解"方程的运算叫作逆运算：加法抵消减法，乘法抵消除法（开方抵消平方，虽然在 KS3 阶段很少出现）。始终按照运算顺序的"倒序"来做——先消去 + 和 −，再处理 × 和 ÷。
3. Solving One-Step Equations Using Addition or Subtraction | 用加减法求解一步方程
When a constant is added to or subtracted from the variable, use the inverse operation. Addition is undone by subtracting the same number from both sides, and subtraction is undone by adding the same number to both sides.
当变量上加上或减去一个常数时，就使用逆运算。加法通过在两边同时减去同一个数来消去，减法则通过在两边同时加上同一个数来消去。
7 from both sides to get x = −5, which shows that the answer does not have to be positive.
例题：对于 x + 8 = 15，两边同时减 8，得到 x = 7。对于 y − 4 = 9，两边同时加 4，得到 y = 13。对于 x + 7 = 2，两边同时减 7，得到 x = −5，这说明答案不一定是正数。
| Equation | Inverse operation | Solution |
| x + 12 = 20 | subtract 12 from both sides | x = 8 |
| x − 9 = 3 | add 9 to both sides | x = 12 |
| x + 0.5 = 2.5 | subtract 0.5 from both sides | x = 2 |
| x − 6 = −2 | add 6 to both sides | x = 4 |
Show your working one line at a time, with the equals signs lined up vertically. This keeps your algebra readable and makes it far easier for an examiner to award method marks even if you make a slip at the end.
请一行一行地写清步骤，并让等号纵向对齐。这样既让代数过程清晰易读，也方便阅卷老师在最后一步出错时仍能给出"方法分"。
4. Solving One-Step Equations Using Multiplication or Division | 用乘除法求解一步方程
When the variable is multiplied by a number, divide both sides by that number. When the variable is divided by a number, multiply both sides by that number.
当变量被某个数乘时，两边同时除以这个数；当变量被某个数除时，两边同时乘以这个数。
ltiply both sides by 3 to get x = 12. For −5x = 20, divide both sides by −5 to get x = −4.
例题：对于 2x = 10，两边同时除以 2，得到 x = 5。对于 x ÷ 3 = 4，写成 x/3 = 4，两边同时乘以 3，得到 x = 12。对于 −5x = 20，两边同时除以 −5，得到 x = −4。
= 5 x/3 = 4 → ×3 → x = 12 −5x = 20 → ÷(−5) → x = −4
g a fraction by its reciprocal is a useful shortcut: (2/3)x = 6 becomes x = 6 × (3/2) = 9.
要特别注意正负号。除以负数会让答案变号；而把分数乘以它的倒数是一个很实用的捷径：(2/3)x = 6 可化为 x = 6 × (3/2) = 9。
5. Solving Two-Step Equations | 求解两步方程
ion or subtraction first, then the multiplication or division. This is often remembered as "undo the + or −, then undo the × or ÷".
大多数考题涉及两种运算。要按相反的顺序拆解：先处理加法或减法，再处理乘法或除法。可以记成"先消 + 或 −，再消 × 或 ÷"。
Subtract 4 from both sides: 3x = 15. Divide both sides by 3: x = 5. Check: 3 × 5 + 4 = 19. ✓
例 1：解 3x + 4 = 19。两边减 4：3x = 15。两边除以 3：x = 5。检验：3 × 5 + 4 = 19。✓
5. Add 3 to both sides: x/2 = 8. Multiply both sides by 2: x = 16. Check: 16 ÷ 2 − 3 = 5. ✓
例 2：解 x/2 − 3 = 5。两边加 3：x/2 = 8。两边乘以 2：x = 16。检验：16 ÷ 2 − 3 = 5。✓
ubtract 7 from both sides: −2x = −6. Divide both sides by −2: x = 3. Check: 7 − 2 × 3 = 1. ✓
例 3：解 7 − 2x = 1。两边减 7：−2x = −6。两边除以 −2：x = 3。检验：7 − 2 × 3 = 1。✓
say x + 4 = 19/3; you must remove the +4 first, or divide every single term on both sides.
这里常见的陷阱是：只除含 x 的那一项，而不是整边。在 3x + 4 = 19 中，你不能直接说 x + 4 = 19/3；你必须先消去 +4，或者把两边每一项都除以 3。
6. Equations with the Unknown on Both Sides | 未知数在两边的方程
When x appears on both sides, first collect all the x terms on one side and all the numbers on the other. To keep the equation tidy, many students prefer to move the smaller x term so that the coefficient stays positive.
当 x 出现在两边时，先把所有含 x 的项集中到一边，把所有常数集中到另一边。为了让方程更整洁，很多同学会选择消去较小的那一项，使系数保持为正。
oth sides: 3x = 12. Divide by 3: x = 4. Check: LHS = 5 × 4 − 3 = 17, RHS = 2 × 4 + 9 = 17. ✓
例题：解 5x − 3 = 2x + 9。两边减 2x：3x − 3 = 9。两边加 3：3x = 12。除以 3：x = 4。检验：左边 = 5 × 4 − 3 = 17，右边 = 2 × 4 + 9 = 17。✓
5x − 3 = 2x + 9 → −2x → 3x − 3 = 9 → +3 → 3x = 12 → ÷3 → x = 4
= 8x − 12. Add 12: 16 = 8x. Divide by 8: x = 2. Check: LHS = 4 − 2 = 2, RHS = 14 − 12 = 2. ✓
另一个含负系数的例子：解 4 − x = 7x − 12。两边加 x：4 = 8x − 12。再加 12：16 = 8x。除以 8：x = 2。检验：左边 = 4 − 2 = 2，右边 = 14 − 12 = 2。✓
7. Equations with Brackets | 含括号的方程
There are two reliable approaches to equations with brackets. You can expand the brackets first and then solve, or you can divide both sides by the number outside the bracket when it divides exactly. Both are correct, but choose the one that keeps the numbers simple.
处理含括号的方程有两种可靠方法：先展开括号再解，或者当括号外的数能整除时，两边同时除以这个数。两种方法都对，但要选择让数字更简单的那一种。
+ 4) = 21. Divide both sides by 3: x + 4 = 7. Subtract 4: x = 3. Check: 3 × (3 + 4) = 21. ✓
例 1（先除）：解 3(x + 4) = 21。两边除以 3：x + 4 = 7。再减 4：x = 3。检验：3 × (3 + 4) = 21。✓
2 = 10. Add 2: 2x = 12. Divide by 2: x = 6. Check: LHS = 2 × 17 = 34, RHS = 24 + 10 = 34. ✓
例 2（先展开）：解 2(3x − 1) = 4x + 10。展开左边：6x − 2 = 4x + 10。减 4x：2x − 2 = 10。加 2：2x = 12。除以 2：x = 6。检验：左边 = 2 × 17 = 34，右边 = 24 + 10 = 34。✓
Be careful with a minus sign in front of a bracket: −(2x − 5) becomes −2x + 5, because the negative sign multiplies both terms inside. Forgetting to flip the second sign is one of the most common errors in this topic.
要特别注意括号前的负号：−(2x − 5) 变成 −2x + 5，因为负号要乘到括号里的每一项。忘记改变第二项的符号是这一专题最常见的错误之一。
8. Equations with Fractions | 含分数的方程
Fractions look intimidating but usually disappear as soon as you multiply every term by a common denominator. This converts the equation into a familiar linear form.
分数看起来吓人，但只要把每一项都乘以公分母，它们通常会立刻消失，方程就变成熟悉的一次形式了。
ply every term by 12: 4x + 3x = 84, so 7x = 84 and x = 12. Check: 12/3 + 12/4 = 4 + 3 = 7. ✓
例 1：解 x/3 + x/4 = 7。3 和 4 的最小公分母是 12。把每一项乘以 12：4x + 3x = 84，所以 7x = 84，x = 12。检验：12/3 + 12/4 = 4 + 3 = 7。✓
3: 2x + 1 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Check: (2 × 7 + 1)/3 = 15/3 = 5. ✓
例 2：解 (2x + 1)/3 = 5。两边乘以 3：2x + 1 = 15。减 1：2x = 14。除以 2：x = 7。检验：(2 × 7 + 1)/3 = 15/3 = 5。✓
6x + 12 = 8x − 4. Rearrange: 16 = 2x, so x = 8. Check: LHS = 10/4 = 2.5, RHS = 15/6 = 2.5. ✓
例 3：解 (x + 2)/4 = (2x − 1)/6。交叉相乘：6(x + 2) = 4(2x − 1)。展开：6x + 12 = 8x − 4。整理：16 = 2x，所以 x = 8。检验：左边 = 10/4 = 2.5，右边 = 15/6 = 2.5。✓
When cross-multiplying, always put brackets around each numerator - the numerator has two terms and both must be multiplied. Writing 6x + 2 instead of 6(x + 2) is a frequent slip.
交叉相乘时，一定要给每个分子加括号——分子有两项，两项都必须被乘到。把 6(x + 2) 写成 6x + 2 是常见的疏忽。
9. Forming Equations from Word Problems | 从应用题列方程
Many exam questions ask you to set up your own equation rather than solve one that is given. Follow a consistent routine: read the problem twice, define the unknown with a letter, translate each sentence into algebra, write the equation, solve it, then write the answer in the context of the question with units.
很多考题要求你自己列方程，而不是直接解给定的方程。请遵循固定流程：把题目读两遍、用字母定义未知数、把每句话翻译成代数、写出方程、求解，最后结合题目情境和单位写出答案。
Example 1 (ages): Sara is three times as old as her brother Ben. In five years' time, the sum of their ages will be 34. Let Ben' years: (x + 5) + (3x + 5) = 34, so 4x + 10 = 34, 4x = 24, x = 6. Ben is 6 and Sara is 18.
例 1（年龄）：Sara 的年龄是弟弟 Ben 的三倍。五年后，两人年龄之和为 34。设 Ben 的年龄为 x，则 Sara 为 3x。五年后：(x + 5) + (3x + 5) = 34，即 4x + 10 = 34，4x = 24，x = 6。Ben 6 岁，Sara 18 岁。
3) = 34, so 2(2w + 3) = 34, 4w + 6 = 34, 4w = 28, w = 7. The width is 7 cm and the length is 10 cm.
例 2（周长）：一个长方形的周长是 34 厘米，长比宽多 3 厘米。设宽为 w，则长为 w + 3。于是 2(w + w + 3) = 34，即 2(2w + 3) = 34，4w + 6 = 34，4w = 28，w = 7。宽为 7 厘米，长为 10 厘米。
+ (n + 1) + (n + 2) = 72, so 3n + 3 = 72, 3n = 69, n = 23. The numbers are 23, 24 and 25.
例 3（连续整数）：三个连续整数之和为 72。设最小的为 n，则其余为 n + 1 和 n + 2。于是 n + (n + 1) + (n + 2) = 72，即 3n + 3 = 72，3n = 69，n = 23。三个数是 23、24 和 25。
10. Checking Answers and Common Mistakes | 检验答案与常见错误
Always substitute your solution back into the original equation, using the left-hand side and the right-hand side separately. If both sides give the same value, your answer is almost certainly correct. This habit takes about ten seconds and can rescue several marks in an exam.
一定要把解代回原方程，分别计算左边和右边。如果两边得到相同的值，你的答案几乎肯定正确。这个习惯只需十秒左右，却能在考试中救回好几分。
| Common mistake | Why it is wrong | Correct approach |
| Doing the operation on one side only | Destroys the balance of the equation | Apply it to both sides |
| Dividing only part of a side by a number | Terms are not treated equally | Divide every term |
| Forgetting to expand a negative bracket | Signs inside the bracket are unchanged | Multiply every term by −1 |
| Losing track of a negative sign when moving terms | The sign must change when a term crosses the equals sign | Write the inverse operation explicitly |
| Writing x = 5 without showing working | No method marks if the answer is wrong | Show each balancing step |
ion is x = −4 for a question about the number of sweets, you have made an error somewhere.
还要检查答案在情境中是否合理：长度不能为负，人数必须是整数。如果你算出一包糖果的数量是 x = −4，那说明中间某处出错了。
11. Exam-Style Practice Questions | 考题风格练习
Try these without looking at the answers. They cover every technique in this article and mirror the style of Cambridge Checkpoint and IGCSE
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