Solving Equations with Brackets (Cambridge KS3 p92 q1) | 求解带括号的方程（剑桥KS3数学第92页第1题）

📚 Solving Equations with Brackets (Cambridge KS3 p92 q1) | 求解带括号的方程（剑桥KS3数学第92页第1题）

Welcome to this KS3 Cambridge Mathematics revision article. Today we focus on solving linear equations that include brackets, a key skill tested in the Cambridge Checkpoint curriculum. We will work through a specific example from your practice book: Page 92, Question 1. This question asks us to solve 2(x+3)=14. Following the steps carefully will help you build confidence in algebra and ace your assessments.

欢迎阅读这篇KS3剑桥数学复习文章。今天我们重点学习求解带括号的一元一次方程，这是剑桥Checkpoint课程中考查的重要技能。我们将通过练习册中的一个具体问题进行讲解：第92页第1题。题目要求解方程 2(x+3)=14。认真跟随解题步骤将有助于你建立代数的信心，并在考试中取得好成绩。

1. The Problem from Page 92 | 第92页的问题

Let’s take a close look at the exact problem from your Cambridge Checkpoint Mathematics Practice Book 8. On page 92, the first question reads: “Solve the equation 2(x + 3) = 14.” This is a linear equation containing a bracket, so we must first deal with the bracket before isolating the variable x. The goal is to find the value of x that makes both sides of the equation equal.

让我们仔细看看剑桥Checkpoint数学练习册8第92页的具体问题。第92页的第一题是：“解方程 2(x + 3) = 14。”这是一个含有括号的线性方程，因此我们必须先处理括号，然后再分离出变量x。我们的目标是找到使等式两边相等的x值。

2. What is an Equation? | 什么是方程？

An equation is a mathematical statement that shows two expressions are equal, using the equals sign ‘=’. For example, 2(x + 3) = 14 is an equation. The left-hand side is 2(x + 3) and the right-hand side is 14. Solving an equation means finding the value of the unknown (usually x) that makes the statement true. In KS3, we focus on linear equations where the variable has an exponent of 1, meaning no x² or higher powers.

方程是使用等号’=’表示两个表达式相等的数学陈述。例如，2(x + 3) = 14 就是一个方程。左边是2(x + 3)，右边是14。解方程意味着找出使该陈述成立的未知数（通常是x）的值。在KS3阶段，我们重点学习线性方程，其中变量的指数为1，意味着没有x²或更高次幂。

3. The Balancing Method | 平衡法

To solve any equation, you must always keep it balanced. Imagine an old-fashioned balance scale. Whatever operation you do to one side of the equation, you must do exactly the same to the other side. This maintains the equality. If you add, subtract, multiply or divide on one side, do the identical operation on the other side. This principle is the foundation of solving all linear equations.

求解任何方程，你必须始终保持其平衡。想象一台旧式天平秤。无论你对等式的一边做什么操作，都必须对另一边做完全相同的操作。这保持了等式的成立。如果你在一边加、减、乘或除，就要在另一边执行相同的运算。这条原则是解决所有线性方程的基础。

4. Expanding Brackets | 去括号

When an equation contains brackets, the first step is usually to expand them using the distributive property. The distributive property states that a(b + c) = ab + ac. In our problem, the bracket is 2(x + 3). We multiply the 2 by every term inside the bracket: 2 × x gives 2x, and 2 × 3 gives 6. Therefore, 2(x + 3) expands to 2x + 6. The original equation 2(x + 3) = 14 now becomes 2x + 6 = 14. This is a simpler equation without brackets.

当方程含有括号时，第一步通常是用分配律去括号。分配律指出 a(b + c) = ab + ac。在我们的问题中，括号是2(x + 3)。我们将2乘以括号内的每一项：2 × x 得到 2x，2 × 3 得到 6。因此，2(x + 3) 展开为 2x + 6。原方程 2(x + 3) = 14 现在变为 2x + 6 = 14。这是一个没有括号的更简单方程。

A common mistake is to multiply only the first term. Always ensure the outside number multiplies both the x-term and the constant term inside the brackets. For practice, try expanding 3(2x – 5) on your own – you should get 6x – 15.

一个常见的错误是只乘第一项。务必确保括号外的数同时乘以括号内的x项和常数项。自己尝试展开 3(2x – 5)，应该得到 6x – 15。

5. Step-by-Step Solution: p92 Question 1 | 逐步求解：第92页第1题

Now let’s solve the bracket-free equation 2x + 6 = 14 using the balancing method. We want to isolate x on one side. Observe the operations attached to x: it is multiplied by 2, and then 6 is added. We reverse these operations in the opposite order – first deal with the addition, then the multiplication. Subtract 6 from both sides to remove the constant term. Then divide both sides by 2 to undo the multiplication.

现在让我们用平衡法求解无括号的方程 2x + 6 = 14。我们希望将x单独留在等式一边。观察与x相关的运算：x先乘以2，然后加上6。我们按相反顺序逆操作——先处理加法，再处理乘法。两边减去6以消去常数项，然后两边除以2以撤销乘法。

Step	Equation	Operation
After expanding	2x + 6 = 14	Given: 2(x+3)=14
Subtract 6 from both sides	2x = 8	2x + 6 − 6 = 14 − 6
Divide both sides by 2	x = 4	2x ÷ 2 = 8 ÷ 2

The solution to the equation is x = 4. This means that if you substitute 4 back into the original bracket form, you will get a true statement. The step-by-step clearing of operations turns a seemingly tricky equation into two simple arithmetic moves.

方程的解是 x = 4。这意味着如果你把4代回原来的括号形式，你会得到一个成立的等式。逐步消去运算将看似复杂的方程变成了两个简单的算术步骤。

2(x + 3) = 14 → x = 4

6. Checking Your Answer | 检验答案

Always verify your solution by substituting it into the original equation. This habit not only catches mistakes but deepens your understanding. For x = 4, the left-hand side becomes 2(4 + 3) = 2 × 7 = 14. The right-hand side is 14. Since both sides are equal, our solution is correct. If they did not match, you would know to retrace your steps.

一定要将解代回原方程进行验证。这个习惯不仅能捕捉错误，还能加深理解。当x=4时，左边变为 2(4 + 3) = 2 × 7 = 14。右边是14。因为两边相等，所以我们的解正确。如果不相等，你就能知道需要重新检查步骤。

You can also check mentally: ‘What number plus 3 gives 7 when multiplied by 2?’ Working backwards from 14: half of 14 is 7, and 7 minus 3 is 4. That matches our solution.

你也可以心算检验：“什么数加3后再乘以2等于14？”从14倒推：14的一半是7，7减3等于4。这与我们的解相符。

7. Common Mistakes to Avoid | 常见错误

Even careful students slip up. Here are the most frequent errors when solving bracket equations, and how to steer clear of them.

即使是细心的学生也会犯错误。以下是求解括号方程时最常见的错误，以及如何避免它们。

Forgetting to multiply both terms inside the bracket: Some write 2(x+3) = 2x+3 instead of 2x+6. Always apply the outside factor to every term inside. 忘记乘以括号内的所有项：有些人会把 2(x+3) 写成 2x+3，而不是 2x+6。一定要将外面的因数乘以括号内的每一项。
Incorrect balancing: Subtracting 6 from only one side breaks the equality. Whatever you do to one side, you must do to the other. 平衡步骤错误：仅从一边减去6会破坏等式。对一边做什么，就必须对另一边做同样的操作。
Division mishandling: When dividing 2x = 8, students might divide only the 8 by 2 and write x = 8, or miscalculate 8 ÷ 2. Double-check your arithmetic. 除法操作失误：当对 2x = 8 除以2时，学生可能只把8除以2而写成 x=8，或者算错 8÷2。再次检查你的算术。
Not checking the answer: Skipping verification allows simple errors to go undetected. Always plug your answer back in. 不检验答案：跳过验证会让简单错误无法被发现。始终将答案代入检验。

8. Equations with Variables on Both Sides | 两边含有变量的方程

In KS3, you will soon encounter equations that have x on both sides, such as 3(x − 1) = 2x + 4. The method is an extension of what we have just practised. First, expand the bracket: 3(x − 1) becomes 3x − 3. Now the equation reads 3x − 3 = 2x + 4. The next step is to collect the variable terms on one side. Subtract 2x from both sides to move all x-terms to the left: 3x − 2x − 3 = 4, which simplifies to x − 3 = 4. Then add 3 to both sides, giving x = 7. Always keep the balance and check: 3(7 − 1) = 3×6 = 18, and 2×7+4 = 18. It works!

在KS3阶段，你很快会遇到两边都含有x的方程，例如 3(x − 1) = 2x + 4。解题方法是我们刚才练习的延伸。首先去括号：3(x − 1) 变为 3x − 3。现在方程读作 3x − 3 = 2x + 4。下一步是将含有变量的项移到一边。两边减去2x，把所有x项移到左边：3x − 2x − 3 = 4，简化为 x − 3 = 4。然后两边加3，得到 x = 7。始终维持平衡并检验：3(7 − 1) = 3×6 = 18，而 2×7+4 = 18。完美解出！

9. Real-World Application | 现实世界应用

Equations with brackets model many real-life situations. Imagine a rectangular garden where the length is 3 metres more than the width. If the perimeter is 14 metres, we can set up an equation. Let the width be x, then length is x + 3. The perimeter formula is 2(length + width) = 2(x + (x+3)) = 2(2x+3). Setting this equal to 14 gives 2(2x+3)=14. Divide both sides by 2 to get 2x+3=7, subtract 3, then divide by 2: x=2. So the width is 2 m and the length is 5 m. This is exactly the same structure as our page 92 problem.

带括号的方程可以模拟许多现实情形。想象一个矩形花园，其长度比宽度多3米。如果周长为14米，我们可以建立方程。设宽为x，则长为 x+3。周长公式为 2(长 + 宽) = 2(x + (x+3)) = 2(2x+3)。令其等于14，得到 2(2x+3)=14。两边除以2得 2x+3=7，减3，再除以2：x=2。所以宽为2米，长为5米。这与我们第92页的问题结构完全相同。

Seeing algebra in context helps you appreciate why these skills matter. Whether you are designing a floor plan or splitting a restaurant bill, linear equations are powerful tools.

在具体情境中

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Solving Equations with Brackets (Cambridge KS3 p92 q1) | 求解带括号的方程（剑桥KS3数学第92页第1题）