Solving Linear Equations with Brackets | 含括号一元一次方程的解法

📚 Solving Linear Equations with Brackets | 含括号一元一次方程的解法

This article walks you through the types of algebraic equations found in Exercise 1 on page 106 of the Cambridge Checkpoint Maths workbook for Key Stage 3. The focus is on solving linear equations that contain one or more pairs of brackets, a skill that is fundamental for building confidence in algebra. You will learn how to expand brackets correctly, use inverse operations to isolate the unknown, and check your solutions.

本文带你一步步解析剑桥初中数学教材第106页练习1中出现的一元一次方程题。核心能力是求解带括号的线性方程，这是建立代数自信的基础。你将学会如何正确展开括号、利用逆运算分离未知数，以及如何检验你的解。

1. Understanding the Equation Structure | 认识方程结构

A typical equation from the exercise looks like 3(x + 2) = 15. The number outside the bracket, 3, multiplies everything inside the bracket. The equals sign tells us that the expression on the left balances the number on the right. Our job is to find the value of x that makes this true.

练习中常见的方程形式如 3(x + 2) = 15。括号外的数字 3 会乘以括号内的每一项。等号表示左边的式子与右边的数值相等。我们的任务就是求出使等式成立的 x 的值。

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2. Expanding Brackets First | 第一步：展开括号

Always begin by multiplying the term outside the brackets by each term inside. For 3(x + 2), we compute 3 × x = 3x and 3 × 2 = 6. The expanded form is 3x + 6. Keep the right‑hand side unchanged: 3x + 6 = 15.

一定要先把括号外面的项乘以括号内的每一项。以 3(x + 2) 为例，计算 3 × x = 3x，3 × 2 = 6。展开后得到 3x + 6。等号右边保持不动：3x + 6 = 15。

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3. Using Inverse Operations to Isolate the Variable | 利用逆运算隔离未知数

Think of the equation as a balance. To isolate x, we undo the addition first. Subtract 6 from both sides: 3x + 6 − 6 = 15 − 6, giving 3x = 9. Then we undo the multiplication by dividing both sides by 3: 3x ÷ 3 = 9 ÷ 3, so x = 3.

把方程想象成一个天平。要隔离 x，我们首先处理加法：从两边同时减去 6：3x + 6 − 6 = 15 − 6，得到 3x = 9。然后处理乘法，两边同时除以 3：3x ÷ 3 = 9 ÷ 3，所以 x = 3。

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4. Solving a Two‑Step Example | 两步求解示例

Take 2(3x − 4) = 10. Expand: 2 × 3x = 6x, 2 × (−4) = −8, so 6x − 8 = 10. Add 8 to both sides: 6x = 18. Divide by 6: x = 3. Always write each step clearly; skipping steps often leads to sign errors.

以 2(3x − 4) = 10 为例。展开：2 × 3x = 6x，2 × (−4) = −8，得到 6x − 8 = 10。两边加 8：6x = 18。再除以 6：x = 3。每一步都应该写清楚；跳过步骤常常导致符号出错。

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5. Equations with a Negative Sign Outside the Brackets | 括号外为负号的情况

When you see −(2x + 1), it means −1 times the bracket. Expand carefully: −1 × 2x = −2x, −1 × 1 = −1. For example, solve 7 − 2(x + 1) = 3. First rewrite as 7 + (−2)(x + 1) = 3. Expand: (−2)(x) = −2x, (−2)(1) = −2, so 7 − 2x − 2 = 3. Simplify the left: 5 − 2x = 3. Subtract 5: −2x = −2. Divide by −2: x = 1.

当看到 −(2x + 1) 时，它代表 −1 乘以整个括号。小心展开：−1 × 2x = −2x，−1 × 1 = −1。例如，解 7 − 2(x + 1) = 3。先把方程看成 7 + (−2)(x + 1) = 3。展开：(−2)(x) = −2x，(−2)(1) = −2，得到 7 − 2x − 2 = 3。左边化简：5 − 2x = 3。两边减 5：−2x = −2。除以 −2：x = 1。

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6. When the Variable Appears on Both Sides | 未知数出现在两边时

Some problems extend the skill to equations like 4(x + 1) = 2(x + 5). Expand both sides: 4x + 4 = 2x + 10. Collect x‑terms on one side by subtracting 2x from both sides: 2x + 4 = 10. Subtract 4: 2x = 6, so x = 3. Always aim to have variables on one side and constants on the other.

有些题目会把技能延伸到类似 4(x + 1) = 2(x + 5) 的方程。两边同时展开：4x + 4 = 2x + 10。把含 x 的项移到同一边：两边同时减去 2x，得 2x + 4 = 10。再减 4：2x = 6，因此 x = 3。始终要把未知项和常数项分别集中到等号两边。

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7. Dealing with Fractions Inside or Outside the Brackets | 处理括号内外的分数

Consider (1/2)(4x + 6) = 5. Multiply both sides by 2 to clear the fraction: 4x + 6 = 10. Then subtract 6: 4x = 4, x = 1. Alternatively, expand first: (1/2)(4x) + (1/2)(6) = 2x + 3 = 5, then 2x = 2, x = 1. Both paths are correct.

考虑 (1/2)(4x + 6) = 5。可以先在两边乘以 2 消去分数：4x + 6 = 10。然后减 6：4x = 4，x = 1。也可以先展开：(1/2)(4x) + (1/2)(6) = 2x + 3 = 5，再得 2x = 2，x = 1。两种途径都是正确的。

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8. Checking Your Solution | 检验你的解

Always substitute your answer back into the original equation. For x = 3 in 2(3x − 4) = 10: left side becomes 2(3×3 − 4) = 2(9 − 4) = 2×5 = 10, which matches the right side. Checking catches arithmetic slips and helps you understand the structure of the equation.

一定要把求出的解代回原方程。对 x = 3 代入 2(3x − 4) = 10：左边变成 2(3×3 − 4) = 2(9 − 4) = 2×5 = 10，与右边一致。检验能发现计算失误，并帮助你理解方程的结构。

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

Mistake: Forgetting to multiply both terms inside the bracket. 错误：忘记乘括号内的每一项。	Fix: Write an intermediate line showing each product. 纠正：写出中间步骤，展示每一项的乘积。
Mistake: Mishandling negative signs before brackets. 错误：处理括号前的负号不当。	Fix: Rewrite with −1 and bracket the expansion. 纠正：把负号写成 −1，再展开。
Mistake: Adding instead of subtracting to isolate the variable. 错误：为隔离变量时应减却误加。	Fix: Think of the equation as a balanced scale; perform the same operation on both sides. 纠正：把方程看作天平，两边同时进行相同的运算。

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10. Practice Problems from p106_1 | 来自 p106_1 的练习题

Try these on your own, then check the answers below. (a) 5(x + 3) = 30, (b) 2(4x − 1) = 14, (c) 10 − 3(x − 2) = 4, (d) (1/3)(6x + 9) = 7, (e) 4(x + 1) = 2x + 10.

自己动手试试下面这些题，然后对照下面的答案。 (a) 5(x + 3) = 30，(b) 2(4x − 1) = 14，(c) 10 − 3(x − 2) = 4，(d) (1/3)(6x + 9) = 7，(e) 4(x + 1) = 2x + 10。

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11. Solutions with Step‑by‑Step Reasoning | 带分步推理的解答

(a) 5x + 15 = 30 → 5x = 15 → x = 3.
(b) 8x − 2 = 14 → 8x = 16 → x = 2.
(c) Expand −3(x − 2) → 10 − 3x + 6 = 4 → 16 − 3x = 4 → −3x = −12 → x = 4.
(d) Multiply by 3: 6x + 9 = 21 → 6x = 12 → x = 2.
(e) Expand: 4x + 4 = 2x + 10 → 2x = 6 → x = 3.

(a) 5x + 15 = 30 → 5x = 15 → x = 3。
(b) 8x − 2 = 14 → 8x = 16 → x = 2。
(c) 展开 −3(x − 2) → 10 − 3x + 6 = 4 → 16 − 3x = 4 → −3x = −12 → x = 4。
(d) 两边乘 3：6x + 9 = 21 → 6x = 12 → x = 2。
(e) 展开：4x + 4 = 2x + 10 → 2x = 6 → x = 3。

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12. Why This Skill Matters for KS3 and Beyond | 这项技能对 KS3 及以后的重要性

Mastering equations with brackets prepares you for multi‑step problems in algebra, such as solving simultaneous equations, expanding quadratics, and rearranging formulae. The careful, logical thinking developed here is exactly what Cambridge assessments reward. Return to this guide whenever you need a quick recap, and always remember: expand first, then use inverse operations step by step.

掌握带括号的方程为你解决更复杂的代数问题打下基础，比如解联立方程、展开二次式、变形公式。这里培养的严谨、有逻辑的思维正是剑桥评估体系所看重的。任何时候需要快速复习，都可以回到这篇指南。记住：先展开，再一步步用逆运算求解。

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