📚 Solving Linear Equations | 解线性方程
Linear equations are a fundamental part of Key Stage 3 Mathematics. They allow us to find unknown values by keeping both sides of an equation balanced. Whether you are working with one‑step, two‑step, or equations with variables on both sides, the core idea is always the same: perform the same operation on both sides of the equals sign to isolate the unknown. This article will guide you through the essential types of linear equations you will meet in the Cambridge Lower Secondary curriculum and give you the tools to solve them confidently.
线性方程是 KS3 数学的基础内容。通过保持等式两边平衡，我们可以求出未知数的值。无论你是在处理一步方程、两步方程，还是变量出现在等号两边的方程，核心思想始终不变：在等号两边同时进行相同的运算，把未知数单独留在一边。本文将带你梳理剑桥初中课程中最常见的几类线性方程，并帮助你自信地解题。
1. What Is an Equation? | 什么是方程？
An equation is a mathematical statement that two expressions are equal. It always contains an equals sign (=). The expression on the left‑hand side (LHS) has the same value as the expression on the right‑hand side (RHS). For example, 2x + 3 = 11 is an equation. The letter x is called the variable, or unknown, and solving the equation means finding the value of x that makes the statement true.
方程是一个表示两个表达式相等的数学语句，其中一定含有等号 (=)。左边表达式 (LHS) 的值等于右边表达式 (RHS) 的值。例如，2x + 3 = 11 就是一个方程。字母 x 称为变量或未知数，解方程就是求出使等式成立的 x 的值。
Think of an equation as a balance scale. If you add, subtract, multiply, or divide one side, you must do exactly the same to the other side to keep the scale balanced. This is the golden rule of solving equations.
我们可以把方程想象成一个天平。如果你在一端加上、减去、乘以或除以某个数，就必须在另一端做完全相同的操作，天平才能保持平衡。这就是解方程的金科玉律。
2. One‑Step Equations | 一步方程
One‑step equations are the simplest type. You only need to perform one inverse operation to find the unknown. For instance, if x + 7 = 15, the inverse of adding 7 is subtracting 7. Subtract 7 from both sides: x + 7 − 7 = 15 − 7, giving x = 8.
一步方程是最简单的类型，只需要进行一步逆运算就能求出未知数。例如 x + 7 = 15，加 7 的逆运算是减 7。两边同时减去 7：x + 7 − 7 = 15 − 7，得到 x = 8。
Similarly, if the equation involves multiplication or division, use the opposite operation. For 4x = 28, divide both sides by 4: 4x ÷ 4 = 28 ÷ 4, so x = 7. If x/5 = 9, multiply both sides by 5: (x/5) × 5 = 9 × 5, so x = 45. Always check your answer by substituting it back into the original equation.
类似地，如果方程中出现乘法或除法，就使用相反的运算。对于 4x = 28，两边同时除以 4：4x ÷ 4 = 28 ÷ 4，所以 x = 7。如果 x/5 = 9，两边同时乘以 5：(x/5) × 5 = 9 × 5，所以 x = 45。别忘了把你的答案代入原方程进行检验。
3. Two‑Step Equations | 两步方程
A two‑step equation requires two operations to isolate the variable. A common example is 2x + 3 = 13. The goal is to get x by itself. First, undo the addition or subtraction: subtract 3 from both sides. 2x + 3 − 3 = 13 − 3 gives 2x = 10. Then, undo the multiplication: divide both sides by 2. 2x ÷ 2 = 10 ÷ 2, so x = 5.
两步方程需要两次运算才能把变量单独隔离出来。常见的例子如 2x + 3 = 13。目标是让 x 单独留在等号一边。首先，消去加减法：两边同时减 3，2x + 3 − 3 = 13 − 3 得到 2x = 10。然后消去乘法：两边同时除以 2，2x ÷ 2 = 10 ÷ 2，所以 x = 5。
The order matters: always deal with addition or subtraction before multiplication or division. This is the reverse of the order of operations (BODMAS/BIDMAS). Think of unwrapping a package: you remove the outer layer before the inner one. In 2x + 3, the multiplication (2 × x) is the inner operation, and adding 3 is the outer operation. To ‘unwrap’ x, we reverse the process.
运算顺序很重要：一定要先处理加减，再处理乘除。这与运算顺序 (BODMAS/BIDMAS) 正好相反。就好像拆开一个包裹：你总会先去掉外层，再打开内层。在 2x + 3 中，乘法 (2 × x) 是内层运算，加 3 是外层运算。要“拆出” x，我们就要反向操作。
4. Equations with Negative Coefficients | 系数为负的方程
Sometimes the coefficient of x appears as a negative number. For example, 15 − 2x = 7. There are two ways to approach this. One method is to add 2x to both sides first to make the coefficient positive: 15 − 2x + 2x = 7 + 2x gives 15 = 7 + 2x. Then subtract 7 from both sides: 15 − 7 = 2x, so 8 = 2x, and finally divide by 2: x = 4.
有时候 x 的系数是负数。例如 15 − 2x = 7。有两种思路。一种方法是先在两边同时加上 2x，让系数变正：15 − 2x + 2x = 7 + 2x 得到 15 = 7 + 2x。然后两边减 7：15 − 7 = 2x，所以 8 = 2x，最后除以 2：x = 4。
Alternatively, you can subtract 15 from both sides, giving −2x = 7 − 15, which simplifies to −2x = −8. Then divide both sides by −2, giving x = 4. Both methods are correct; choose the one that feels more comfortable.
另一种做法是两边先减 15，得到 −2x = 7 − 15，化简为 −2x = −8。然后两边除以 −2，得出 x = 4。两种方法都正确，选择你觉得更顺手的那种就好。
5. Equations with Brackets | 带括号的方程
When an equation contains brackets, the first step is usually to expand them. For instance, 3(2x − 1) = 21. Multiply each term inside the bracket by 3: 3 × 2x = 6x and 3 × (−1) = −3. The equation becomes 6x − 3 = 21. Now it is a simple two‑step equation: add 3 to both sides to get 6x = 24, then divide by 6 to find x = 4.
当方程中含有括号时，第一步通常是把括号展开。例如 3(2x − 1) = 21。把括号内的每一项都乘以 3：3 × 2x = 6x，3 × (−1) = −3。方程变为 6x − 3 = 21。这就成了一个简单的两步方程：两边加 3 得到 6x = 24，然后除以 6 得出 x = 4。
Sometimes a number or a negative sign appears directly in front of a bracket without a multiplication sign, such as 2(x + 5) − 3x = 4. Expand 2(x + 5) to 2x + 10. The equation becomes 2x + 10 − 3x = 4. Combine like terms: 2x − 3x = −x, so −x + 10 = 4. Subtract 10 from both sides: −x = −6, so x = 6.
有时一个数或负号直接出现在括号前，没有乘号，比如 2(x + 5) − 3x = 4。把 2(x + 5) 展开成 2x + 10。方程变为 2x + 10 − 3x = 4。合并同类项：2x − 3x = −x，所以 −x + 10 = 4。两边减 10：−x = −6，因此 x = 6。
6. Equations with Variables on Both Sides | 变量在等号两边的方程
Equations such as 5x + 2 = 3x + 10 contain the variable on both sides of the equals sign. The strategy is to collect all variable terms on one side and all constant terms on the other. Choose the side with the smaller coefficient of x to move. Here, subtract 3x from both sides: 5x + 2 − 3x = 3x + 10 − 3x, which gives 2x + 2 = 10. Then solve as a two‑step equation: subtract 2 from both sides to get 2x = 8, then divide by 2 to obtain x = 4.
像 5x + 2 = 3x + 10 这样的方程，变量出现在等号两边。解题策略是把所有含变量的项集中到等号一边，所有常数项集中到另一边。选择移动 x 系数较小的一边。这里，两边同时减去 3x：5x + 2 − 3x = 3x + 10 − 3x，得到 2x + 2 = 10。然后按两步方程来解：两边减 2 得 2x = 8，再除以 2 得 x = 4。
If the equation is 2x − 3 = 4x + 7, you might subtract 2x from both sides: −3 = 2x + 7. Then subtract 7: −10 = 2x, and divide by 2: x = −5. Alternatively, add 3 to both sides and subtract 4x, but the result is the same. Always aim to keep the x‑term positive if possible.
如果方程是 2x − 3 = 4x + 7，你可以在两边同时减 2x：−3 = 2x + 7。然后减 7：−10 = 2x，除以 2：x = −5。也可以先把两边的常数项合并，但结果一样。尽量让含 x 的项保持正数，这样计算会更顺手。
7. Checking Your Answer by Substitution | 代入检验答案
No solution is complete without a check. Substitute your answer back into the original equation. For 5x + 2 = 3x + 10 with x = 4, LHS = 5(4) + 2 = 20 + 2 = 22. RHS = 3(4) + 10 = 12 + 10 = 22. Both sides match, so the solution is correct. This habit catches arithmetic mistakes and builds confidence.
写完答案不等于解题结束，代入原方程检验是不可缺少的一步。对于 5x + 2 = 3x + 10，当 x = 4 时，左边 = 5(4) + 2 = 20 + 2 = 22。右边 = 3(4) + 10 = 12 + 10 = 22。两边相等，说明解答正确。这个习惯能帮你发现计算错误，也让你对自己更有信心。
Substitution is also a key skill used in many other topics, such as graphs and sequences. Practise it every time you solve an equation, and it will soon become automatic.
代入检验也是图像、数列等许多其他课题中的关键技能。每解一个方程都练习代入，你很快就能做到不假思索。
8. Equations with Fractions | 含有分数的方程
When fractions make an equation look messy, eliminate them by multiplying every term by the common denominator. For example, x/3 + 1 = 5/6. The LCM of 3 and 6 is 6. Multiply every term by 6: 6·(x/3) + 6·1 = 6·(5/6). This gives 2x + 6 = 5. Subtract 6 from both sides: 2x = −1, so x = −½.
如果方程中含有分数，看起来比较杂乱，可以先找出所有分母的最小公倍数，给每一项都乘上这个数。例如 x/3 + 1 = 5/6。3 和 6 的最小公倍数是 6。每一项都乘以 6：6·(x/3) + 6·1 = 6·(5/6)，得到 2x + 6 = 5。两边减 6：2x = −1，所以 x = −½。
Another common example: (2x + 1)/4 = 3. Multiply both sides by 4: 2x + 1 = 12. Then subtract 1 and divide by 2: 2x = 11, x = 5.5 or 11/2. The clearing‑fractions technique turns a fraction equation into a familiar integer equation.
另一个常见例子：(2x + 1)/4 = 3。两边同时乘以 4：2x + 1 = 12。然后减 1，除以 2：2x = 11，x = 5.5 或 11/2。这种“去分母”的技巧可以把分数方程变成我们熟悉的整数方程。
9. Forming Equations from Word Problems | 根据应用题列出方程
Many real‑life situations can be modelled with linear equations. Read the problem carefully, identify the unknown, and represent it with a letter. Then build an equation that matches the relationships described. For example: “Three times a number added to five gives twenty‑six. Find the number.” Let the number be n. Then 3n + 5 = 26. Solve: 3n = 21, n = 7.
很多现实情境都可以用线性方程来建立数学模型。仔细读题，找出未知量，用一个字母表示它。然后根据题意写出一个方程。例如：“一个数的 3 倍加上 5 等于 26，求这个数。”设这个数为 n，那么 3n + 5 = 26。解方程：3n = 21，n = 7。
Word problems often involve ages, geometry, or money. For example, “A rectangle has a length that is 3 cm more than twice its width. The perimeter is 36 cm. Find its width.” Let the width be w cm. Length = 2w + 3 cm. Perimeter = 2(length + width) = 2(2w + 3 + w) = 2(3w + 3) = 6w + 6. Set equal to 36: 6w + 6 = 36, so 6w = 30, w = 5. Width = 5 cm, length = 13 cm. Always check that the answer fits the original conditions.
应用题常常涉及年龄、几何或金钱。例如：“一个长方形的长比宽的 2 倍还多 3 cm。周长是 36 cm。求宽。”设宽为 w cm，那么长 = 2w + 3 cm。周长 = 2(长 + 宽) = 2(2w + 3 + w) = 2(3w + 3) = 6w + 6。让周长等于 36：6w + 6 = 36，所以 6w = 30，w = 5。宽 = 5 cm，长 = 13 cm。记得检查答案是否符合题意。
10. Common Mistakes and How to Avoid Them | 常见错误及避免方法
One frequent mistake is doing operations in the wrong order. Students often divide before adding or subtracting. Remember: reverse BODMAS. First undo addition/subtraction, then multiplication/division. For 2x + 3 = 11, dividing by 2 first gives x + 1.5 = 5.5, which then requires subtracting 1.5, a messy route that often leads to errors. Always subtract 3 first to keep integers.
一个常见错误是运算顺序错误。学生经常在加减之前先做乘除。要记住：与 BODMAS 顺序相反。首先消去加减法，再处理乘除法。对于 2x + 3 = 11，如果先除以 2 会得到 x + 1.5 = 5.5，然后需要减 1.5，这样不仅绕了弯路，还容易算错。应该先减 3，保持整洁的整数运算。
Another mistake is forgetting to perform an operation on both sides. When moving a term, many learners simply change its sign mentally without writing the step. Write out each step clearly: “add 5 to both sides” or “subtract 2x from both sides”. This prevents sign errors.
Also, when dealing with brackets, ensure you multiply every term inside by the factor outside, especially if a negative sign precedes the bracket.
另一个错误是忘记对等号两边同时进行运算。移项时，很多学生只是心里想一下“变号”，却不把步骤写出来。要清楚地写出每一步：“两边加 5”或“两边减 2x”。这能防止符号错误。另外，处理括号时，一定要把括号外的因数乘以括号内的每一项，尤其当括号前是负号时更要注意。
11. Practice Patterns for Fluency | 流利解题的典型练习
To become fluent in solving linear equations, practise a variety of patterns every day. Start with simple forms: x + a = b, ax = b, x/a = b. Move to two‑step: ax + b = c, ax − b = c. Then try with negatives, such as a − bx = c. Incorporate brackets and expand. Finally, tackle equations with variables on both sides and fractions. Aim to solve each equation in under a minute.
想在解线性方程上做到流利，每天要练习不同类型的题目。从最简单的形式开始：x + a = b，ax = b，x/a = b。再到两步方程：ax + b = c，ax − b = c。然后尝试含有负系数的方程，如 a − bx = c。接着练习带括号的方程并进行展开。最后挑战两边都有变量以及含分数的方程。争取在一分钟内解出一道题。
Below is a quick reference table of equation types and the first step to solve them:
下面是一个快速参考表，列出了不同方程类型以及解题的第一步：
| Equation type | 方程类型 | Example | 例子 | First step | 第一步 |
| One‑step addition | x + 4 = 10 | Subtract 4 | 减 4 |
| One‑step multiplication | 5x = 35 | Divide by 5 | 除以 5 |
| Two‑step (ax + b) | 4x + 3 = 19 | Subtract 3 | 减 3 |
| Negative coefficient | 12 − 3x = 3 | Add 3x (or subtract 12) | 加 3x（或减 12） |
| Brackets | 2(x − 5) = 14 | Expand | 展开 |
| Variables both sides | 6x + 1 = 2x + 13 | Subtract 2x | 减 2x |
| Fractions | x/2 + 3 = 7 | Multiply by 2 | 乘以 2 |
12. Connecting to Further Topics | 联系更高阶的课题
Mastering linear equations at KS3 provides a strong foundation for many areas of mathematics. In KS4 and beyond, you will use these skills to solve simultaneous equations, quadratic equations by factorising, and inequalities. Even in coordinate geometry, finding the equation of a straight line relies on the same balancing principles. The idea of isolating an unknown appears again in rearranging formulae, such as changing the subject of v = u + at to t = (v − u)/a.
在 KS3 阶段掌握线性方程，能为许多数学领域打下坚实基础。到了 KS4 乃至更高层次，你将运用这些技能去解联立方程、用因式分解解二次方程以及求解不等式。即使在坐标几何中，求直线方程也同样依赖保持等号平衡的原则。隔离未知数这个思想还会在公式变形中再次出现，比如把 v = u + at 改写为 t = (v − u)/a。
The confidence you gain from solving linear equations now will make later topics feel much more approachable. Treat every equation as a puzzle to be unravelled logically – the more you practise, the sharper your problem‑solving abilities become.
现在通过解线性方程培养起来的自信心，会让你后续的课题变得容易许多。把每一个方程都看作一个需要逻辑推理解开的谜题——练习得越多，你的解题能力就会越强。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导