📚 Solving Linear Equations with Brackets | 解带括号的一元一次方程
Equations with brackets appear frequently in KS3 Mathematics. Learning to solve them accurately builds a strong foundation for algebra at IGCSE level and beyond. This article breaks down the process into clear steps, with worked examples and common pitfalls to avoid.
含有括号的方程在KS3数学中经常出现。学会准确求解这类方程，为IGCSE及更高层次的代数学习打下坚实基础。本文将解题过程分解为清晰的步骤，并配有例题和常见错误分析。
1. What is a Linear Equation? | 什么是一元一次方程？
he power of 1. It forms a straight line when graphed. Examples: 2x + 5 = 11, 3(y − 2) = 9.
一元一次方程是指未知数（通常为x）的最高次数为1的方程。它的图像是一条直线。例如：2x + 5 = 11，3(y − 2) = 9。
Linear equations can involve brackets, fractions, and negative numbers. The key is to keep the equation balanced by performing the same operation on both sides.
线性方程可能包含括号、分数和负数。关键是通过在等号两边进行相同的运算来保持方程平衡。
2. Expanding Brackets Review | 括号展开回顾
× x + 3 × 4 = 3x + 12. Always multiply each term inside the bracket by the factor outside.
在解含括号的方程前，先回顾如何用括号外的因数乘以括号内的每一项。例如，3(x + 4) 表示 3 × x + 3 × 4 = 3x + 12。务必用外面的因数乘以括号里的每一项。
| Expression | 表达式 | Expanded Form | 展开式 |
|---|---|
| 2(a + 5) | 2a + 10 |
| −3(m − 2) | −3m + 6 |
| 4(2y − 1) | 8y − 4 |
Note the sign rules: a negative factor reverses the signs inside the bracket. This is a common place where mistakes occur.
注意符号规则：负因数会改变括号内各项的符号。这是常见的出错点。
3. The Goal: Isolate the Variable | 目标：分离变量
The aim when solving any linear equation is to get the variable on its own on one side of the equals sign. This usually involves three stages: expand brackets if any, collect like terms, then use inverse operations to undo addition, subtraction, multiplication or division.
解任何线性方程的目标都是将变量单独留在等号的一边。这通常包括三个阶段：如果有括号先展开，合并同类项，然后使用逆运算（加/减、乘/除）逐步求解。
Remember: whatever you do to one side, you must do to the other. This is the golden rule of equation solving.
记住：对等式一边做什么，对另一边也要做什么。这是解方程的黄金法则。
4. Solving Equations with Brackets: Step-by-Step | 解带括号方程：逐步法
Follow this systematic approach to solve equations containing brackets:
按照以下系统方法求解含有括号的方程：
- Step 1: Expand – Use the distributive law to remove all brackets.
- 第1步：展开 – 使用分配律去掉所有括号。
- Step 2: Simplify – Combine like terms on each side if necessary.
- 第2步：化简 – 如有必要，在等号每一边合并同类项。
- Step 3: Rearrange – Move variable terms to one side and constant terms to the other using inverse operations.
- 第3步：移项 – 使用逆运算将含变量项移到一边，常数项移到另一边。
- Step 4: Solve – Isolate the variable by multiplying or dividing.
- 第4步：求解 – 通过乘除运算使变量的系数变为1。
- Step 5: Check – Substitute your solution back into the original equation.
- 第5步：检验 – 将解代入原方程验证。
5. Worked Example 1 | 例题1
Solve 3(y + 4) = 27.
3(y + 4) = 27
Step 1 – Expand: 3×y + 3×4 = 3y + 12. So the equation becomes 3y + 12 = 27.
第1步 – 展开：3y + 12 = 27。
Step 2 – Subtract 12 from both sides: 3y = 15.
第2步 – 两边减去12：3y = 15。
Step 3 – Divide both sides by 3: y = 5.
第3步 – 两边除以3：y = 5。
Check: 3(5+4) = 3×9 = 27 ✔.
检验：3(5+4) = 3×9 = 27 ✔。
6. Worked Example 2 (with negative numbers) | 例题2（含负数）
Solve −2(x − 5) = 8.
−2(x − 5) = 8
Expand carefully: −2 × x = −2x, −2 × (−5) = +10. Equation: −2x + 10 = 8.
小心展开：−2x + 10 = 8。
Subtract 10 from both sides: −2x = −2.
两边减去10：−2x = −2。
Divide by −2: x = 1.
除以−2：x = 1。
Check: −2(1 − 5) = −2 × (−4) = 8 ✔.
检验：−2(1 − 5) = −2 × (−4) = 8 ✔。
7. When the Variable Appears on Both Sides | 当变量出现在等号两边
d: 2x + 2 = x + 5. Then subtract x from both sides: x + 2 = 5. Finally, subtract 2: x = 3.
方程可能既有括号又带两边变量，如2(x + 1) = x + 5。先展开：2x + 2 = x + 5，然后两边减去x得x + 2 = 5，最后减去2得x = 3。
When moving terms, it is often easier to subtract the smaller variable term first to keep the coefficient positive.
移项时，通常先减去较小的含变量项，尽量让变量的系数保持正数。
8. Common Mistakes to Avoid | 常见错误避免
| Mistake | 错误 | Why It's Wrong | 为什么错 | Correction | 正确做法 |
|---|---|---|
| 2(x + 3) = 2x + 3 | The 2 was not multiplied by 3. / 2没有乘以3。 | 2x + 6 |
| −3(a − 2) = −3a − 6 | −3 × (−2) = +6, not −6. / 负数乘负数得正。 | −3a + 6 |
| 4 − (2x + 1) = 4 − 2x + 1 | The minus sign flips both signs. / 减号要变号。 | 4 − 2x − 1 |
| Forgetting to balance both sides. / 忘记等号两边平衡。 | If you add 5 to one side, you must add 5 to the other. / 一边加5，另一边也必须加5。 | Always do the same to both sides. / 始终两边同操作。 |
Writing each step clearly helps you spot these mistakes before they affect your final answer.
把每一步写清楚，能让你在最终答案受影响之前发现这些错误。
9. Checking Your Solution | 检验你的解
Always substitute your answer back into the original equation, including all brackets. Calculate the left-hand side and right-hand side separately. If they match, you are correct. This habit is especially valuable in exams to catch sign errors.
始终将答案代入原方程（保留括号）检验。分别计算左边和右边的值。如果两边相等，你的答案就是对的。这个习惯在考试中特别有用，能发现符号错误。
3x − 1) = 10, found x = 2. Check: 2(3×2 − 1) = 2(6 − 1) = 2×5 = 10. Left = Right. Correct.
例如：解2(3x − 1) = 10，得x = 2。检验：2(3×2 − 1) = 2(6 − 1) = 2×5 = 10。左边等于右边，正确。
10. Practice Questions (with Solutions) | 练习题（附解答）
Try these before looking at the solutions. Write out every step.
在查看解答之前，先试做这些题。写出每一步。
1. 5(m + 2) = 35
2. −4(y − 3) = 12
3. 2(3x − 1) = x + 8
4. 7 − 2(t + 1) = 3t − 5
Solutions | 解答：
1. Expand: 5m + 10 = 35 → 5m = 25 → m = 5.
展开：5m + 10 = 35 → m = 5。
2. Expand: −4y + 12 = 12 → −4y = 0 → y = 0.
展开：−4y + 12 = 12 → y = 0。
3. 6x − 2 = x + 8 → 5x = 10 → x = 2.
6x − 2 = x + 8 → x = 2。
2 = 3t − 5 → 5 − 2t = 3t − 5 → add 2t to both sides: 5 = 5t − 5 → add 5: 10 = 5t → t = 2.
展开：7 − 2t − 2 = 3t − 5 → 5 − 2t = 3t − 5 → t = 2。
11. Real-World Application: Perimeter Problems | 实际应用：周长问题
= perimeter. If perimeter = 34 cm, solve: 2(3x + 2) = 34 → 6x + 4 = 34 → 6x = 30 → x = 5.
含括号的方程经常出现在几何问题中。一个长方形的长为(x + 3) cm，宽为(2x − 1) cm，其周长可表示为2[(x+3) + (2x−1)] = 周长。若周长 = 34 cm，解方程：2(3x + 2) = 34 → 6x + 4 = 34 → 6x = 30 → x = 5。
This shows how algebraic equations model real situations, making the skill highly practical.
这表明代数方程能模拟实际情境，说明这项技能非常实用。
12. Summary and Next Steps | 总结与下一步
To solve equations with brackets confidently, always expand first, then simplify, and apply inverse operations step by step. Check your answer in the original equation. Practise with a mix of positive and negative coefficients until the process becomes automatic.
要自信地解带括号的方程，总是先展开再化简，然后一步步使用逆运算。把答案代回原方程检验。混合练习正负系数题目，直到过程变得自动化。
Once you master this, you are ready to tackle equations with fractions, double brackets, and unknowns on both sides – all of which appear later in the KS3 curriculum.
掌握这一点后，就可以应对含有分数、双重括号以及两边都有未知数的方程，这些都会在KS3后期课程中出现。
Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
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