Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Linear equations are a cornerstone of algebra at Key Stage 3. They appear in everything from simple number puzzles to real‑world problems involving distance, cost and measurement. Mastering the techniques to solve them gives you a toolkit you will use again and again in secondary maths and beyond. This article walks you through the core concepts, step‑by‑step methods and common pitfalls, with plenty of worked examples to build your confidence.

一元一次方程是KS3代数的基石。从简单的数字谜题到涉及距离、成本和测量的实际问题，方程随处可见。掌握解方程的方法就等于获得了一个你在中学数学乃至更高阶段都会反复使用的工具包。本文带你梳理核心概念、分步方法和常见误区，并提供大量例题来帮助你建立信心。

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1. Equations vs Expressions | 方程与表达式的区别

An expression is a combination of numbers, letters (variables) and operation symbols, like 3x + 5 or 7y – 2. An equation is a statement that two expressions are equal, signalled by an equals sign (=). For example, 2x + 3 = 11 is an equation; it says ‘2 times a number plus 3 equals 11’. In KS3 we work with linear equations, where the variable appears only to the power of 1 (no squares, cubes or higher powers).

表达式是由数字、字母（变量）和运算符号组合而成的式子，如 3x + 5 或 7y – 2。方程则是表明两个表达式相等的陈述，用等号（=）来表示。例如，2x + 3 = 11 就是一个方程，它表达了“一个数的2倍加3等于11”。在KS3阶段，我们学习线性方程，即变量只出现一次方（没有平方、立方或更高次幂）的方程。

Thinking of an equation as a balance will help you understand why we perform the same operation on both sides. The equals sign is the pivot of the scales – whatever you do to one side, you must do to the other to keep the scales level.

把方程想象成一个天平，这有助于你理解为什么我们要在等号两边同时进行相同的运算。等号就是天平的支点——你对一边做什么，对另一边也必须做什么，才能保持平衡。

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2. The Balance Method | 天平法原理

Imagine an old‑fashioned pan balance. If both pans hold the same weight, the beam is horizontal. To keep it level, any change you make on the left must be mirrored exactly on the right. This is the golden rule for solving equations: perform the same operation on both sides of the equals sign. The aim is to isolate the variable on one side, leaving a number on the other side that gives the solution.

想象一个老式的托盘天平。如果两个托盘里的重量相等，横梁就是水平的。为了保持水平，你在左边做的任何改变，都必须在右边完全复制。这就是解方程的黄金法则：在等号两边同时进行相同的运算。我们的目标是把变量单独留在等式的一边，让另一边只剩一个数，这个数就是方程的解。

Consider x + 5 = 12. The variable x has 5 added to it. To ‘undo’ the addition, subtract 5 from both sides:

考虑方程 x + 5 = 12。变量 x 被加上了5。为了“撤销”这个加法，我们在两边同时减去5：

x + 5 – 5 = 12 – 5 → x = 7

The same logic applies to subtraction, multiplication and division – always do the inverse operation.

同样的逻辑也适用于减法、乘法和除法——总是进行逆运算。

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3. Solving One‑Step Equations | 解一步方程

One‑step equations need only a single inverse operation to reveal the solution. The four basic types are:

一步方程只需进行一次逆运算就能求出解。四种基本类型如下：

Addition: x + a = b → subtract a from both sides. For instance, x + 7 = 15 → x = 15 – 7 → x = 8.

加法型： x + a = b → 两边同时减去 a。例如，x + 7 = 15 → x = 15 – 7 → x = 8。

Subtraction: x – a = b → add a to both sides. For instance, y – 4 = 11 → y = 11 + 4 → y = 15.

减法型： x – a = b → 两边同时加上 a。例如，y – 4 = 11 → y = 11 + 4 → y = 15。

Multiplication: a x = b → divide both sides by a (provided a ≠ 0). For instance, 4p = 20 → p = 20 ÷ 4 → p = 5.

乘法型： a x = b → 两边同时除以 a（a ≠ 0）。例如，4p = 20 → p = 20 ÷ 4 → p = 5。

Division: x / a = b → multiply both sides by a. For instance, m / 3 = 6 → m = 6 × 3 → m = 18.

除法型： x / a = b → 两边同时乘以 a。例如，m / 3 = 6 → m = 6 × 3 → m = 18。

Practising these basic moves until they become automatic makes more complicated equations much easier.

反复练习这些基本步骤，直到它们变成自然而然的操作，会让更复杂的方程容易得多。

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4. Solving Two‑Step Equations | 解两步方程

Two‑step equations involve two operations, such as multiplication plus addition. The rule of thumb is to undo the addition or subtraction first, then the multiplication or division. This order follows the reverse of the normal order of operations (BIDMAS/BODMAS).

两步方程包含两种运算，例如先乘再加。经验法则是先处理加减，再处理乘除。这个顺序与正常的运算顺序（BIDMAS/BODMAS）相反。

Take 2x + 3 = 11. Step one: subtract 3 from both sides.

以 2x + 3 = 11 为例。第一步：两边同时减去3。

2x + 3 – 3 = 11 – 3 → 2x = 8

Step two: divide both sides by 2.

第二步：两边同时除以2。

2x ÷ 2 = 8 ÷ 2 → x = 4

Another example: 5y – 7 = 18. Add 7 to both sides: 5y = 25. Then divide by 5: y = 5.

再举一个例子：5y – 7 = 18。两边加7：5y = 25。然后两边除以5：y = 5。

Always write the intermediate step clearly. Showing your working helps prevent mistakes and earns partial marks in assessments.

要清晰地写出中间步骤。展示过程有助于避免错误，在测评中也能拿到过程分。

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5. Equations with Brackets | 含括号的方程

When an equation contains brackets, your first job is to expand them using the distributive law. After expanding, you reduce the equation to a familiar two‑step or sometimes multi‑step form.

如果方程中含有括号，你的第一项任务就是利用分配律将其展开。展开之后，方程就会变成你熟悉的两步，甚至多步形式。

Example: 3(x – 2) = 9. Expand the left side: 3 × x – 3 × 2 = 3x – 6. The equation becomes 3x – 6 = 9. Now add 6 to both sides:

例如：3(x – 2) = 9。展开左边：3 × x – 3 × 2 = 3x – 6。方程变为 3x – 6 = 9。然后两边加6：

3x = 15 → x = 5

If a negative number is multiplying the bracket, be careful with signs. For –2(4 – y) = 10, expand to –8 + 2y = 10. Then add 8 to both sides: 2y = 18, so y = 9. Always double‑check your sign work.

如果括号外是负数，一定要注意符号。比如 –2(4 – y) = 10，展开得 –8 + 2y = 10。然后两边加8：2y = 18，所以 y = 9。要始终仔细检查符号。

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6. Variables on Both Sides | 变量在等式两边

Equations often have the unknown on both the left‑hand and right‑hand sides, such as 5x + 2 = 3x + 10. The strategy is to collect all variable terms on one side and all constant terms on the other. This is done by adding or subtracting terms to both sides.

方程中经常会出现未知数同时在左边和右边的情况，比如 5x + 2 = 3x + 10。解题策略是把所有含变量的项移到一边，所有常数项移到另一边。这可以通过在两边加减相同的项来实现。

Step 1: remove the smaller variable term, in this case 3x, by subtracting 3x from both sides.

第一步：消去较小的变量项，这里就是 3x，在两边同时减去 3x。

5x – 3x + 2 = 3x – 3x + 10 → 2x + 2 = 10

Step 2: subtract 2 from both sides.

第二步：两边减2。

2x = 8 → x = 4

You can check your answer by substituting back into the original equation: 5(4) + 2 = 22 and 3(4) + 10 = 22. Both sides match.

你可以把答案代入原方程进行检验：5(4) + 2 = 22，3(4) + 10 = 22。两边相等。

It does not matter whether you move the variable terms to the left or to the right, as long as you do it correctly. Many students find it easiest to keep the variable on the side where its coefficient is larger and positive.

把变量项移到左边还是右边都没关系，只要操作正确即可。很多学生发现，把变量保留在系数较大且为正的一边最为简便。

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7. Equations Involving Fractions | 涉及分数的方程

Fractions in equations can look intimidating, but you can eliminate them by multiplying every term on both sides by the lowest common denominator (LCD). This transforms the equation into an integer form you already know how to solve.

方程中含有分数可能会令人望而生畏，但你可以通过在等号两边每一项都乘以最简公分母（LCD）来消除分数。这样方程就会变成你已经会解的整数形式。

Example: x/3 + 2 = 5. The denominator is 3, so multiply every term by 3:

例如：x/3 + 2 = 5。分母是3，所以每一项都乘以3：

3 × (x/3) + 3 × 2 = 3 × 5 → x + 6 = 15

Then subtract 6: x = 9.

然后减去6：x = 9。

When a fraction is a single rational expression, like (2x + 1)/5 = 3, multiply both sides by 5 directly:

如果分数是一个单独的有理式，比如 (2x + 1)/5 = 3，可以直接两边乘以5：

(2x + 1) = 15 → 2x = 14 → x = 7

Always remember to multiply every term, not just the fraction terms. A common error is to forget to multiply the whole‑number terms by the LCD.

务必记住要乘以每一项，而不仅仅是分数项。一个常见的错误是忘记把整数项也乘以最简公分母。

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8. Checking Your Solution | 检验你的解

Substituting your answer back into the original equation is one of the most powerful habits you can develop. It verifies that your solution is correct and helps catch arithmetic mistakes. Even a simple slip in signs can turn a right answer into a wrong one.

把得到的答案代回原方程，这是你可以培养的最强有力的习惯之一。它能证实你的解是正确的，并帮助发现计算错误。哪怕只是一个简单的符号错误，都可能让本应正确的答案出错。

Take the equation 4x – 5 = 2x + 7, with the claimed solution x = 6. Left side: 4(6) – 5 = 24 – 5 = 19. Right side: 2(6) + 7 = 12 + 7 = 19. The two sides match, so x = 6 is correct. If they had not matched, you would re‑examine your working.

以方程 4x – 5 = 2x + 7 为例，假设得出的解是 x = 6。左边：4(6) – 5 = 24 – 5 = 19。右边：2(6) + 7 = 12 + 7 = 19。两边相等，所以 x = 6 正确。如果不相等，你就需要重新检查解题过程。

Checking is especially valuable in tests because it gives you immediate feedback without needing an answer key.

检验在考试中尤其有价值，因为它能在没有答案的情况下立即给你反馈。

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9. Forming Equations from Word Problems | 从文字题建立方程

Real‑world problems are often described in words, and your task is to translate them into algebraic equations. Read the question carefully, identify the unknown quantity (assign a letter to it), and then build an equation statement using the relationships described.

现实世界的问题通常用文字描述，你的任务是把它们翻译成代数方程。仔细读题，找出未知量（用一个字母表示），然后根据所描述的关系建立方程。

Example: ‘When a number is multiplied by 5 and then 12 is subtracted, the result is 28.’ Let the number be n. The equation is 5n – 12 = 28. Solve: 5n = 40 → n = 8.

例如：“一个数乘以5再减去12，结果是28。” 设这个数为 n。方程为 5n – 12 = 28。解方程：5n = 40 → n = 8。

For a two‑stage problem: ‘The sum of twice a number and 7 is the same as three times the number minus 2.’ This gives 2x + 7 = 3x – 2. Collect terms: 7 + 2 = 3x – 2x → 9 = x. Always re‑read the sentence to check your equation matches the wording.

再来看一个两步的问题：“一个数的2倍加7等于这个数的3倍减2。” 由此得到 2x + 7 = 3x – 2。合并同类项：7 + 2 = 3x – 2x → 9 = x。一定要反复读原句，确保你的方程与表述一致。

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10. Common Mistakes to Avoid | 常见错误规避

Many errors in solving equations come from rushing or forgetting the balance rule. Watch out for these typical slips:

解方程中的很多错误都源于匆忙或忘记了天平法则。请留意以下这些典型的“小失误”：

Forgetting to operate on both sides. If you subtract 4 from the left, you must subtract 4 from the right. Writing ‘x + 4 = 10 → x = 10’ is incorrect.

忘记在两边同时运算。 如果你在左边减去4，右边也必须减去4。写成 “x + 4 = 10 → x = 10” 是错误的。

Misapplying the order of inverse operations. In 2x + 3 = 11, dividing by 2 first gives x + 1.5 = 5.5, which is messier and can lead to errors. Always undo addition/subtraction before multiplication/division.

用错逆运算的顺序。 在 2x + 3 = 11 中，如果先除以2就会得到 x + 1.5 = 5.5，这个步骤更繁琐且容易出错。务必先还原加减，再还原乘除。

Sign errors when expanding brackets. –2(x – 3) should become –2x + 6, not –2x – 6. A missing negative sign can throw off the whole solution.

展开括号时的符号错误。 –2(x – 3) 应该得到 –2x + 6，而不是 –2x – 6。遗漏一个负号就可能导致整个解错误。

Not clearing fractions correctly. Multiply every term by the denominator. In x/2 + 3 = 8, the correct step is x + 6 = 16, not x + 3 = 16.

去分母时出错。 每一项都要乘以分母。在 x/2 + 3 = 8 中，正确的步骤是 x + 6 = 16，而不是 x + 3 = 16。

Losing a solution or misreading the answer. If the equation reduces to something like 0 = 0, it is true for all values of x (identity). If it reduces to a false statement like 0 = 5, there is no solution. At KS3 these cases are less common, but it is good to be aware.

丢失解或者误读答案。 如果方程化简成类似 0 = 0 的结果，说明它对 x 的所有值都成立（恒等式）。如果化简成一个矛盾的式子比如 0 = 5，则无解。在KS3阶段这些情况较少见，但有所了解是好的。

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11. Practice Makes Permanent | 熟能生巧

The best way to become fluent in solving linear equations is to practise regularly, starting with simple one‑step equations and gradually increasing the complexity. Keep a notebook where you write each step clearly, and always check your answer by substitution. Over time, these methods will become second nature.

熟练解一元一次方程的最佳途径就是经常练习，从简单的一步方程入手，逐步增加难度。准备一个笔记本，清晰地写下每一步，并始终用代入法检验答案。假以时日，这些方法就会成为你的第二天性。

Try these on your own:

不妨自己做做看：

• 4a + 2 = 18
• 7b – 3 = 4b + 9
• 2(c – 5) = 3c + 4
• (x + 4)/3 = 2x – 1

(Solutions: a = 4; b = 4; c = –14; x = 1.4 or 7/5. Cover the answers with a piece of paper before you start!)

（答案：a = 4；b = 4；c = –14；x = 1.4 或 7/5。开始前先用一张纸盖住答案！）

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12. Linking Equations to Graphs | 方程与图像的联系

As you progress through KS3, you will see that linear equations are closely related to straight‑line graphs. When you solve 2x + 3 = 9, you are actually finding the x‑coordinate of the point where the line y = 2x + 3 meets the horizontal line y = 9. This visual link reinforces the balance idea and prepares you for topics like simultaneous equations and inequalities later on.

随着你在KS3阶段不断进步，你会发现线性方程与直线图像密切相关。当你解 2x + 3 = 9 时，实际上是在寻找直线 y = 2x + 3 与水平线 y = 9 的交点的 x 坐标。这一图像联系进一步强化了平衡概念，也为以后学习联立方程和不等式等主题做好了准备。

Keep practising, stay curious, and remember that every equation is just a balance waiting to be solved.

坚持练习，保持好奇心，并记住：每一个方程都不过是等待被解开的一个天平。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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