KS3 Cambridge Mathematics: Solving Linear Equations | 解一元一次方程

📚 KS3 Cambridge Mathematics: Solving Linear Equations | 解一元一次方程

Linear equations are the foundation of algebra at Key Stage 3. In Cambridge Lower Secondary Mathematics, you are expected to solve equations using inverse operations and the balancing method, even when brackets or variables on both sides are involved. This article is suitable for KS3 students preparing for Cambridge Checkpoint or school assessments.

线性方程是 KS3 代数的基础。在剑桥初中数学中，你需要掌握使用逆运算和平衡法解方程，即使方程中含有括号或两边都有变量。本文适合准备剑桥 Checkpoint 或校内评估的 KS3 学生。

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

A linear equation is an algebraic statement in which the highest power of the unknown is 1. It usually contains an equals sign and exactly one unknown, such as x or y. The solution is the value of the unknown that makes the statement true.

线性方程是指未知数的最高次数为 1 的代数等式。它通常包含一个等号和一个未知数，例如 x 或 y。方程的解就是使等式成立的未知数的值。

Examples of linear equations:

线性方程的例子：

• 2x + 3 = 11
• 5y – 7 = 2y + 8
• 3(x – 4) = 9
• x/4 + 1 = 6

Expressions like x² + 2 = 6 are not linear because the unknown is squared. Equations such as xy = 12 are also not linear because there are two unknowns multiplied together.

像 x² + 2 = 6 这样的表达式不是线性方程，因为未知数被平方了。像 xy = 12 这样的方程也不是线性方程，因为两个未知数相乘。

In Cambridge KS3, you will mainly work with one unknown and a variable power of 1. Recognising this helps you choose the correct solving method.

在剑桥 KS3 阶段，你主要会接触一个未知数且变量次数为 1 的方程。识别这一点有助于你选择正确的求解方法。

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2. The Balancing Method | 平衡法

An equation is like a balance scale. Whatever you do to one side, you must do to the other to keep it balanced. This is the core idea behind solving any linear equation and is heavily assessed in Cambridge Checkpoint.

方程就像一个天平。你对一边做的任何操作，都必须对另一边做同样的操作，以保持平衡。这是解任何线性方程的核心思想，也是剑桥 Checkpoint 中重点考查的内容。

For example, to solve x + 3 = 7, subtract 3 from both sides:

例如，要解 x + 3 = 7，两边同时减去 3：

x + 3 – 3 = 7 – 3 ⇒ x = 4

The symbol ⇒ means ‘leads to’ and shows the next step in your working. You should write one operation per line to make your method clear.

符号 ⇒ 表示“推出”，显示解题过程中的下一步。你应该每行只写一步运算，使解题过程更清晰。

You can check the balance by substituting x = 4 into x + 3 = 7. The left side becomes 4 + 3 = 7, which equals the right side.

你可以通过把 x = 4 代入 x + 3 = 7 来检验平衡。左边变成 4 + 3 = 7，与右边相等。

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3. Solving One-Step Equations | 解一步方程

One-step equations require only one inverse operation to isolate the unknown. Addition and subtraction are inverse operations, as are multiplication and division. Choosing the correct inverse is the key skill here.

一步方程只需要一次逆运算就能求出未知数。加法和减法是互逆运算，乘法和除法也是互逆运算。选择正确的逆运算这里的关键技能。

Type 1: Addition equation. Solve x + 5 = 12. Subtract 5 from both sides to get x = 7.

类型 1：加法方程。解 x + 5 = 12。两边同时减去 5，得到 x = 7。

x = 12 – 5 = 7

Type 2: Subtraction equation. Solve x – 6 = 9. Add 6 to both sides to get x = 15.

类型 2：减法方程。解 x – 6 = 9。两边同时加上 6，得到 x = 15。

x = 9 + 6 = 15

Type 3: Multiplication equation. Solve 4x = 20. Divide both sides by 4 to get x = 5.

类型 3：乘法方程。解 4x = 20。两边同时除以 4，得到 x = 5。

x = 20 ÷ 4 = 5

Type 4: Division equation. Solve y ÷ 3 = 9. Multiply both sides by 3 to get y = 27.

类型 4：除法方程。解 y ÷ 3 = 9。两边同时乘以 3，得到 y = 27。

y = 9 × 3 = 27

When the coefficient of x is negative, divide by the negative coefficient. For example, -2x = 8 gives x = 8 ÷ (-2) = -4.

当 x 的系数为负数时，要除以这个负系数。例如 -2x = 8 得到 x = 8 ÷ (-2) = -4。

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4. Solving Two-Step Equations | 解两步方程

A two-step equation has two operations acting on the unknown. You must undo them in reverse order, following the reverse of the order of operations. This means you handle addition or subtraction before multiplication or division.

两步方程中有两种运算作用于未知数。你必须按照运算顺序的逆序来逐步消去它们。也就是说，先处理加减法，再处理乘除法。

Example: Solve 3x + 2 = 14.

例子：解方程 3x + 2 = 14。

3x + 2 = 14

3x = 14 – 2 = 12

x = 12 ÷ 3 = 4

First subtract 2, then divide by 3. Always check by substituting x = 4 back into the original equation.

先减去 2，然后除以 3。始终要把 x = 4 代回原方程进行检验。

Another example: Solve x/5 – 2 = 3. First add 2 to both sides, then multiply by 5.

另一个例子：解方程 x/5 – 2 = 3。首先两边加 2，然后两边乘以 5。

x/5 = 3 + 2 = 5

x = 5 × 5 = 25

The reverse order is essential. If you divide before subtracting, you will often make a mistake with the constant term.

逆序处理非常重要。如果你先除后减，常常会在常数项上出错。

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5. Equations with Brackets | 带括号的方程

When an equation contains brackets, expand them first using the distributive law. Then collect like terms and solve using the balancing method. This topic appears frequently in Cambridge Checkpoint tests.

当方程中含有括号时，先用分配律展开括号。然后合并同类项，再用平衡法求解。这个主题经常出现在剑桥 Checkpoint 测试中。

Example: Solve 2(x + 3) = 16.

例子：解方程 2(x + 3) = 16。

2(x + 3) = 16

2x + 6 = 16

2x = 10

x = 5

Remember that a minus sign before a bracket changes every sign inside, for example -2(x – 3) = -2x + 6. This is a very common source of errors.

记住括号前有负号会改变括号内每一项的符号，例如 -2(x – 3) = -2x + 6。这是非常常见的错误来源。

Example with a negative sign: Solve -3(2x – 1) = 15.

带负号的例子：解方程 -3(2x – 1) = 15。

-6x + 3 = 15

-6x = 12

x = 12 ÷ (-6) = -2

If the equation has brackets on both sides, expand them all before collecting like terms.

如果方程两边都有括号，要先展开所有括号，再合并同类项。

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6. Equations with Variables on Both Sides | 两边都含未知数的方程

If the unknown appears on both sides of the equation, first collect all unknown terms on one side and all numbers on the other side. Then simplify and solve.

如果未知数出现在方程两边，首先把所有的未知项移到一边，把所有数字移到另一边。然后化简并求解。

Example: Solve 5x – 3 = 2x + 9.

例子：解方程 5x – 3 = 2x + 9。

5x – 2x = 9 + 3

3x = 12

x = 4

To decide which side to move the variable to, choose the side with the larger coefficient. This keeps the variable positive.

为了决定把变量移到哪一边，选择系数较大的一边。这样能让未知数保持为正。

• 5x – 2x = 3x, which is positive.
• If you moved 5x to the right, you would get -3x = 12, leading to x = -4 but with an extra negative step.

可补充：

5x – 2x = 3x，为正。如果把 5x 移到右边，会得到 -3x = 12，虽然也能得到 x = -4，但多了一个负号步骤。

Another example: Solve 7x + 2 = 3x – 6.

另一个例子：解方程 7x + 2 =

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