Mastering Linear Equations and Formulae | 掌握线性方程与公式

📚 Mastering Linear Equations and Formulae | 掌握线性方程与公式

Linear equations are one of the most important building blocks of KS3 mathematics. They give us a formal way to represent relationships, find missing values, and solve practical problems in science, finance, and everyday life. In this article, we will work through the core methods for solving linear equations and using formulae, with a focus on Cambridge-style questions.

线性方程是 KS3 数学最重要的基础模块之一。它们为我们提供了一种正式的方法来表示关系、找出缺失值，并解决科学、金融和日常生活中的实际问题。在本文中，我们将讲解解线性方程和使用公式的核心方法，重点关注剑桥风格题型。

1. What is a Linear Equation? | 什么是线性方程？

A linear equation is an equation in which the unknown appears only to the power of 1, and its graph is a straight line. Examples include x + 5 = 12, 3y − 2 = 10, and 2(n + 1) = 8.

线性方程是指未知数的次数只为 1 的方程，其图像是一条直线。例如 x + 5 = 12、3y − 2 = 10 和 2(n + 1) = 8。

An equation always has two sides separated by an equals sign. The equals sign tells us that both sides have exactly the same value.

方程总是由等号连接左右两边。等号表示两边具有完全相同的值。

ax + b = c

2. The Balance Method | 平衡法

Think of an equation as a balance scale. Whatever you do to one side, you must do to the other side to keep it balanced. This is why we add, subtract, multiply, or divide both sides by the same amount.

把方程想象成一架天平。你对一侧做了什么，就必须对另一侧做同样的操作以保持平衡。这就是为什么我们要对两边同时加、减、乘或除以同一个数。

Add the same value to both sides | 两边加上相同的值
Subtract the same value from both sides | 两边减去相同的值
Multiply both sides by the same non-zero value | 两边乘以相同的非零数
Divide both sides by the same non-zero value | 两边除以相同的非零数

3. Solving Simple One-Step Equations | 解一步方程

For x + 7 = 15, subtract 7 from both sides. You get x = 15 − 7, so x = 8.

对于 x + 7 = 15，两边同时减去 7。得到 x = 15 − 7，所以 x = 8。

x + 7 = 15 → x = 8

For 5x = 35, divide both sides by 5. You get x = 35 ÷ 5, so x = 7.

对于 5x = 35，两边同时除以 5。得到 x = 35 ÷ 5，所以 x = 7。

5x = 35 → x = 7

4. Solving Two-Step Equations | 解两步方程

Solve 3x − 4 = 11. First add 4 to both sides: 3x = 15. Then divide both sides by 3: x = 5.

解 3x − 4 = 11。首先两边加 4：3x = 15。然后两边除以 3：x = 5。

3x − 4 = 11 → 3x = 15 → x = 5

Always reverse the order of operations: undo addition or subtraction before multiplication or division.

总是按照运算顺序的逆序：先处理加减，再处理乘除。

5. Equations with Brackets | 含括号的方程

If an equation contains brackets, expand them first using the distributive law, then solve as normal.

如果方程含有括号，先用分配律展开，然后按常规方法解。

a(b + c) = ab + ac

Solve 2(x + 3) = 14. Expand: 2x + 6 = 14. Subtract 6: 2x = 8. Divide by 2: x = 4.

解 2(x + 3) = 14。展开：2x + 6 = 14。减 6：2x = 8。除以 2：x = 4。

6. Equations with Variables on Both Sides | 变量在两侧的方程

Solve 5x + 2 = 3x + 10. Collect the x terms on one side. Subtract 3x from both sides: 2x + 2 = 10. Then subtract 2: 2x = 8. Divide by 2: x = 4.

解 5x + 2 = 3x + 10。把含 x 的项集中到一边。两边同时减 3x：2x + 2 = 10。再减 2：2x = 8。除以 2：x = 4。

You can collect variables on the left or right; just keep the balance.

你可以把变量移到左边或右边，只要保持平衡即可。

7. Substituting into Formulae | 代入公式

A formula is a rule that connects two or more variables. To evaluate a formula, replace each letter with its given number.

公式是连接两个或多个变量的规则。求公式的值时，把每个字母替换成给定的数。

For v = u + at, with u = 2, a = 3 and t = 4: v = 2 + 3 × 4 = 2 + 12 = 14.

对于 v = u + at，当 u = 2、a = 3、t = 4 时：v = 2 + 3 × 4 = 2 + 12 = 14。

After substituting, follow BIDMAS: Brackets, Indices, Division and Multiplication, Addition and Subtraction.

代入后按照 BIDMAS 顺序：括号、指数、除法和乘法、加法和减法。

8. Rearranging Simple Formulae | 简单公式变形

Sometimes you need to make a different letter the subject. Use the same balancing method to isolate the required variable.

有时你需要把另一个字母变成主项。使用相同的平衡法来分离所需变量。

Make x the subject of y = 3x + 5. Subtract 5: y − 5 = 3x. Divide by 3: x = (y − 5) ÷ 3.

把 y = 3x + 5 中的 x 变成主项。减 5：y − 5 = 3x。除以 3：x = (y − 5) ÷ 3。

y = 3x + 5 → x = (y − 5) ÷ 3

9. Word Problems and Real-Life Contexts | 文字题与实际情境

Many Cambridge questions ask you to form an equation from a written scenario. Read carefully, define the unknown with a letter, then translate the words into algebra.

许多剑桥题目要求你从文字情境中建立方程。仔细阅读，用字母设定未知数，然后把文字翻译成代数式。

A rectangle length is 3 cm more than its width. The perimeter is 22 cm. Let width be w. Length is w + 3. Perimeter: 2(w + w + 3) = 22. Simplify: 2(2w + 3) = 22. Expand: 4w + 6 = 22. Subtract 6: 4w = 16. Divide by 4: w = 4. So width is 4 cm.

一个矩形的长比宽多 3 cm，周长是 22 cm。设宽为 w。长为 w + 3。周长：2(w + w + 3) = 22。化简：2(2w + 3) = 22。展开：4w + 6 = 22。减 6：4w = 16。除以 4：w = 4。所以宽是 4 cm。

10. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Watch out for these common errors when solving equations. They can change your answer completely if you are not careful.

解方程时要注意这些常见错误。如果不小心，它们会完全改变你的答案。

Forgetting to do the same to both sides | 忘记对两边做相同运算
Mixing up inverse operations | 混淆逆运算
Not expanding brackets correctly | 括号展开不正确
Losing negative signs when moving terms | 移项时丢失负号
Substituting values without following BIDMAS | 代入数值时没有遵循 BIDMAS

Always write each step clearly. Check your final answer by putting it back into the original equation.

始终清楚地写出每一步。将最终答案代回原方程进行检验。

11. Quick Practice Questions | 快速练习题

Try these questions to build confidence. The answers are shown after each arrow, but cover them first if you can.

尝试以下问题来建立信心。答案显示在箭头之后，但尽量先盖住答案。

Solve x + 9 = 17 | 解 x + 9 = 17 → x = 8
Solve 4x − 5 = 19 | 解 4x − 5 = 19 → x = 6
Solve 3(y + 2) = 21 | 解 3(y + 2) = 21 → y = 5
Solve 7x − 3 = 3x + 13 | 解 7x − 3 = 3x + 13 → x = 4
If s = ut + ½at², find s when u = 0, t = 3, a = 10 | 若 s = ut + ½at²，当 u = 0、t = 3、a = 10 时，求 s → s = 45

12. Key Summary | 要点总结

Keep these rules in mind when working through any linear equation or formula question.

在处理任何线性方程或公式问题时，请牢记这些规则。

Do the same operation to both sides | 对方程两边做相同运算
Undo operations in reverse order | 按逆序撤销运算
Expand brackets before solving | 解方程前先展开括号
Substitute carefully and follow BIDMAS | 代入时仔细并遵循 BIDMAS
Check your answer in the original equation | 将答案代回原方程检验

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Mastering Linear Equations and Formulae | 掌握线性方程与公式