Solving Equations with Brackets (p229_1) | 解带括号的方程 (p229_1)

📚 Solving Equations with Brackets (p229_1) | 解带括号的方程 (p229_1)

Solving equations is a fundamental skill in KS3 Mathematics. In particular, equations containing brackets appear frequently in Cambridge exam questions, such as the one found on page 229, question 1. This article will guide you through solving the equation 3(2x − 4) = 2(x + 5) step by step, helping you build confidence in algebraic manipulation.

解方程是KS3数学中的一项基本技能。尤其是带有括号的方程经常出现在剑桥考试题中，例如第229页第1题。本文将逐步指导你解方程3(2x − 4) = 2(x + 5)，帮助你在代数运算中建立信心。

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1. The Problem Statement | 题目陈述

The equation we need to solve is presented exactly as it appears in the exercise: 3(2x − 4) = 2(x + 5). Both sides contain a bracket, which means we must first expand them to simplify the equation. Even though brackets can make an equation look more complicated, applying the distributive law will reveal a simple linear equation that is easy to solve.

我们需要解的方程按照练习中的呈现方式是这样的：3(2x − 4) = 2(x + 5)。两边各有一个括号，这意味着我们必须首先展开它们以简化方程。尽管括号会让方程看起来更复杂，但应用分配律后就会呈现一个容易求解的简单线性方程。

3(2x − 4) = 2(x + 5)

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2. Expanding Brackets | 展开括号

Expanding brackets means multiplying the term outside the bracket by every term inside the bracket. This relies on the distributive law, which states a(b + c) = a×b + a×c. The same rule applies when there is a minus sign: a(b − c) = a×b − a×c. We must carefully apply this to both sides of the equation.

展开括号意味着用括号外的项去乘以括号内的每一项。这依赖于分配律，即 a(b + c) = a×b + a×c。当括号内有减号时同样的规则也适用：a(b − c) = a×b − a×c。我们必须小心地将它应用于方程的两边。

For the left side, multiply 3 by 2x to get 6x, and multiply 3 by −4 to get −12. Pay special attention to the sign: the term inside is ‘minus 4’, so the product is negative.

对于左边，用 3 乘以 2x 得到 6x，再用 3 乘以 −4 得到 −12。特别注意符号：括号内是“减4”，所以乘积为负。

Left side: 3(2x − 4) = 6x − 12

On the right side, multiply 2 by x to obtain 2x, and 2 by +5 to obtain +10. The right-hand bracket does not have a minus sign, so both products stay positive.

在右边，用 2 乘以 x 得到 2x，再用 2 乘以 +5 得到 +10。右边的括号没有减号，因此两个乘积都是正的。

Right side: 2(x + 5) = 2x + 10

After expanding both brackets, the equation becomes much simpler: 6x − 12 = 2x + 10. The brackets have done their job and can now be removed from our working.

展开两个括号之后，方程就变得简单多了：6x − 12 = 2x + 10。括号的作用已经完成，现在可以从我们的计算过程中去掉了。

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3. Setting up the Simplified Equation | 建立简化方程

Now that the brackets are gone, we are left with a standard linear equation: 6x − 12 = 2x + 10. This is the form we are most comfortable solving. We can think of it as a balance scale where the left and right sides must stay equal while we rearrange terms.

既然括号已经去掉，我们面对的是一个标准的一次方程：6x − 12 = 2x + 10。这是我们最熟悉的求解形式。我们可以把它想象成一个天平，当我们重新排列各项时，左右两边必须保持相等。

6x − 12 = 2x + 10

The goal from here is to gather all terms containing x on one side of the equation and all constant numbers on the other side. This is often called ‘collecting like terms’ or ‘isolating the variable’.

从这里开始的目标是把所有含 x 的项聚集在方程的一边，把所有常数项移到另一边。这通常被称为“合并同类项”或“分离变量”。

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4. Collecting Like Terms | 合并同类项

To move terms from one side to the other, we perform inverse operations. Since we have +2x on the right, we subtract 2x from both sides. Similarly, the left side has −12, so we add 12 to both sides. Performing the same operation on both sides keeps the equation balanced.

为了将项从一边移到另一边，我们要进行逆运算。因为右边有 +2x，我们把两边同时减去 2x。同样地，左边有 −12，我们把两边同时加上 12。在两边进行相同的运算能保持方程平衡。

Subtracting 2x: 6x − 12 − 2x = 2x + 10 − 2x, which simplifies to 4x − 12 = 10. Then adding 12: 4x − 12 + 12 = 10 + 12, giving 4x = 22.

减去 2x：6x − 12 − 2x = 2x + 10 − 2x，简化后得到 4x − 12 = 10。然后加上 12：4x − 12 + 12 = 10 + 12，得到 4x = 22。

4x = 22

Always check that you have changed the sign of any term that you move across the equals sign. Moving a term changes its sign: +2x became −2x on the left, and −12 became +12 on the right.

始终要检查你移动穿过等号的项是否改变了符号。移项会改变符号：+2x 移到左边变成了 −2x，而 −12 移到右边变成了 +12。

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5. Solving for x | 求解 x

Now we have 4x = 22. To find the value of one x, we divide both sides by 4. This is because division is the inverse of multiplication, and we need to undo the multiplication of x by 4.

现在我们得到 4x = 22。为了求出一个 x 的值，我们把两边同时除以 4。这是因为除法是乘法的逆运算，而我们需要取消 x 被乘以 4 的效果。

x = 22 ÷ 4

22 divided by 4 simplifies to 5.5 as a decimal, or 11/2 as a fraction. Both forms are acceptable in KS3 answers, but it is good practice to leave the answer as a simplified fraction unless the question specifies otherwise.

22 除以 4 化简后得到小数 5.5，或者分数 11/2。在 KS3 的答案中两种形式都可以接受，但除非题目另有要求，否则将答案保留为化简后的分数是一个好习惯。

x = 5.5 or x = 11/2

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6. Checking the Solution | 检验解

It is always wise to verify your answer by substituting the value back into the original equation. This confirms that no mistake was made during the rearranging and calculating steps.

将求出的值代回原方程来检验答案总是明智的。这能确认在移项和计算过程中没有发生任何错误。

Substitute x = 5.5 into the left-hand side: 3(2 × 5.5 − 4) = 3(11 − 4) = 3 × 7 = 21. Now substitute into the right-hand side: 2(5.5 + 5) = 2 × 10.5 = 21. Both sides equal 21, so the solution is correct.

把 x = 5.5 代入左边：3(2 × 5.5 − 4) = 3(11 − 4) = 3 × 7 = 21。再代入右边：2(5.5 + 5) = 2 × 10.5 = 21。两边都等于 21，因此解是正确的。

If the two sides had produced different numbers, we would know that an error had occurred and we should retrace our steps. Checking not only builds confidence but also develops careful mathematical habits.

如果两边算出的数字不同，我们就知道出现了错误，应重新检查计算步骤。检验不仅能建立信心，还能培养严谨的数学习惯。

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7. Common Mistakes | 常见错误

When solving equations with brackets, students often make similar errors. Recognising these can help you avoid them in your own work. The table below outlines some of the most frequent mistakes and how to prevent them.

在解带括号的方程时，学生经常会犯一些相似的错误。认识这些错误能帮助你在自己的作业中避开它们。下表列出了一些最常见的错误以及如何防止犯错。

Common Mistake	常见错误
Forgetting to multiply both terms inside the bracket when expanding	展开时忘记把括号内的两项都进行乘法运算
Mishandling negative signs, e.g. writing 3(−4) as +12 instead of −12	处理负号不正确，例如将 3(−4) 误写成 +12 而不是 −12
Moving terms across the equals sign without changing their signs	移项时不改变符号就将其移到等号另一边
Dividing incorrectly, such as dividing only one part of a term	除法运算错误，例如只对项的某一部分作除法
Not simplifying the final fraction, leaving 22/4 instead of 11/2	未化简最终分数，将 22/4 保留而没化成 11/2

By checking each step carefully—especially when dealing with negatives and division—you can consistently arrive at the correct answer.

通过仔细检查每一步——尤其是在处理负数和除法时——你就能始终如一地得到正确答案。

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8. Real-world Application | 实际应用

Equations with brackets are not just abstract exercises; they often represent real situations. For instance, consider a shop offering a special deal: the price of a jacket after a discount of 4 pounds is multiplied by 3 for a bundle, while another store sells two jackets with a 5-pound voucher each. If the final cost is the same in both shops, the equation 3(2x − 4) = 2(x + 5) can be used, where x is the original price of one jacket.

带括号的方程不仅仅是抽象的练习；它们常常代表真实情境。例如，假设一家商店提供特价：一件夹克减去 4 英镑折扣后的价格乘以 3 作为捆绑销售价，而另一家商店以每件附赠 5 英镑代金券的形式销售两件夹克。如果最终花费相同，就可以用方程 3(2x − 4) = 2(x + 5) 来求解，其中 x 是一件夹克的原价。

In this scenario, the solution x = 5.5 means the original price is £5.50. This is a typical application of brackets in financial literacy problems, which are common in KS3 Cambridge Mathematics.

在这个情境中，解 x = 5.5 意味着原价是 5.50 英镑。这是括号在金融素养问题中的典型应用，这种问题在 KS3 剑桥数学中很常见。

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9. Alternative Methods | 替代解法

While expanding brackets first is the standard and most reliable approach, it is useful to see that another method exists. Instead of balancing terms on both sides, you can bring everything to one side and set the expression equal to zero. This method is particularly powerful when solving quadratic equations later on.

尽管先展开括号是标准且最可靠的方法，但了解存在另一种方法也很有用。你可以把所有项都移到一边，并令表达式等于零，而不是在两边整理各项。这个方法在以后解二次方程时会特别有用。

Write the equation as 3(2x − 4) − 2(x + 5) = 0. Expand to get 6x − 12 − 2x − 10 = 0. Notice that subtracting 2(x+5) gives −2x −10. Combining like terms yields 4x − 22 = 0, then adding 22 to both sides gives 4x = 22, and finally x = 5.5. The result is identical, but this technique emphasises that equations can be rearranged into the

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