📚 Solving Linear Equations with Brackets | 解含括号的一元一次方程
Linear equations with brackets appear throughout the Cambridge KS3 Mathematics course. They test your ability to apply the distributive law, collect like terms, and isolate the unknown correctly. This article builds the skill step by step, with clear examples and common mistakes to avoid.
含括号的一元一次方程贯穿剑桥 KS3 数学课程。这类题目考查分配律、合并同类项以及正确分离未知数的能力。本文逐步建立解题技能，并给出清晰示例和需要避免的常见错误。
1. Understanding the Equation Structure | 理解方程结构
An equation is a balance: whatever you do to one side, you must do to the other. When brackets are involved, the expression inside the brackets must be treated as a single group until it is expanded.
方程就像一个天平：你对一边做什么，就必须对另一边做同样的事情。当式子中含有括号时，括号内的表达式在展开之前必须作为一个整体来处理。
3(x + 2) = 18
Here, the whole quantity x + 2 is multiplied by 3. You cannot simply subtract 2 first; the bracket must be removed using multiplication.
这里，x + 2 整个量都乘以 3。你不能先直接减去 2；必须先通过乘法去掉括号。
2. The Distributive Law: Removing Brackets | 分配律：去括号
The distributive law states that a(b + c) = ab + ac. Multiply the term outside the bracket by every term inside the bracket, keeping the signs correct.
分配律指出 a(b + c) = ab + ac。把括号外的项与括号内的每一项相乘，并保持符号正确。
3(x + 2) = 3×x + 3×2 = 3x + 6
So the equation becomes:
因此方程变为：
3x + 6 = 18
If the bracket is preceded by a minus sign, for example 5 − 2(x − 1), expand 2(x − 1) first, then apply the minus to the whole result: 5 − 2x + 2.
如果括号前是减号，例如 5 − 2(x − 1)，先展开 2(x − 1)，再把减号作用于整个结果：5 − 2x + 2。
3. Collecting Like Terms | 合并同类项
After removing brackets, simplify each side by collecting terms that contain the same variable or are just numbers. This reduces the equation to a simpler form.
去括号后，把含有相同变量的项或纯数字项合并，从而化简方程。
4(x − 3) + 2x = 5(x + 1)
Expand first: 4x − 12 + 2x = 5x + 5. Then collect like terms: 6x − 12 = 5x + 5.
先展开：4x − 12 + 2x = 5x + 5。然后合并同类项：6x − 12 = 5x + 5。
This step makes the next stage - moving terms - much easier and less error-prone.
这一步让下一步的移项变得更加容易，也不容易出错。
4. Moving Terms Across the Equal Sign | 移项
To solve an equation, move variable terms to one side and number terms to the other. Remember that moving a term across the equal sign changes its sign.
解方程时，把含变量的项移到一边，把数字项移到另一边。记住，移项过等号要变号。
6x − 12 = 5x + 5
Subtract 5x from both sides: 6x − 5x − 12 = 5, giving x − 12 = 5.
两边同时减去 5x：6x − 5x − 12 = 5，得到 x − 12 = 5。
Then add 12 to both sides: x = 5 + 12, so x = 17.
然后两边同时加上 12：x = 5 + 12，所以 x = 17。
5. Isolating the Variable | 孤立变量
After moving terms, you often have something like 3x = 12. To isolate x, divide both sides by the coefficient of x.
移项后，你通常会得到类似 3x = 12 的形式。为了分离出 x，把两边同时除以 x 的系数。
3x = 12 → x = 12 ÷ 3 = 4
Always apply the same operation to the whole of each side. Never divide only part of the left-hand side or only part of the right-hand side.
始终对等式的整个两边施加相同的运算。绝不能只除左边的一部分或右边的一部分。
If the coefficient is negative, divide by the negative number and keep the sign rules in mind.
如果系数是负数，就除以这个负数，并注意符号规则。
6. Dealing with Negative Coefficients | 处理负系数
Negative coefficients are a common source of error. Suppose you have −4x = 20. To find x, divide both sides by −4.
负系数是常见的错误来源。假设你有 −4x = 20。要求出 x，两边同时除以 −4。
−4x = 20 → x = 20 ÷ (−4) = −5
Another tricky case is when a bracket has a negative multiplier, such as −2(x − 3) = 14. Expand to −2x + 6 = 14, not −2x − 6.
另一个棘手的情况是括号前有负数乘数，例如 −2(x − 3) = 14。展开得到 −2x + 6 = 14，而不是 −2x − 6。
Then solve: −2x = 8, so x = 8 ÷ (−2) = −4.
然后求解：−2x = 8，所以 x = 8 ÷ (−2) = −4。
7. Variables on Both Sides | 变量在等式两边
When variables appear on both sides, move the smaller variable term first. This keeps the coefficient of x positive, which reduces sign errors.
当变量出现在等式两边时，先移动较小的变量项。这样可以让 x 的系数保持为正，减少符号错误。
2(x + 3) = 3(x − 1)
Expand: 2x + 6 = 3x − 3. Move the smaller variable term 2x across: 6 = 3x − 2x − 3, so 6 = x − 3. Add 3: x = 9.
展开：2x + 6 = 3x − 3。把较小的变量项 2x 移过去：6 = 3x − 2x − 3，得到 6 = x − 3。加 3：x = 9。
This method is more reliable than moving the larger term and ending with a negative coefficient, although both paths are valid.
这个方法比移动较大项并最终得到负系数更可靠，虽然两条路径都是正确的。
8. Equations with Fractions | 含分数的方程
If brackets and fractions appear together, remove the fraction first by multiplying every term by the lowest common denominator.
如果括号和分数同时出现，先通过将每一项乘以最小公分母来去掉分数。
(x + 2) / 3 = 5
Multiply both sides by 3: x + 2 = 15, so x = 13.
两边同时乘以 3：x + 2 = 15，所以 x = 13。
For a more complex case such as (2(x − 1)) / 5 = 4, multiply by 5 first: 2(x − 1) = 20, then expand and solve.
对于更复杂的情况，如 (2(x − 1)) / 5 = 4，先乘以 5：2(x − 1) = 20，再展开求解。
2x − 2 = 20 → 2x = 22 → x = 11
9. Checking Your Solution | 检验解
Always substitute your answer back into the original equation to make sure both sides are equal. This catches arithmetic mistakes and sign errors.
一定要把你的答案代回原方程，检查两边是否相等。这可以发现算术错误和符号错误。
For 3(x + 2) = 18, we found x = 4. Substitute: 3(4 + 2) = 3 × 6 = 18, which is correct.
对于 3(x + 2) = 18，我们得到 x = 4。代回：3(4 + 2) = 3 × 6 = 18，正确。
If the two sides do not match, retrace your steps from the beginning rather than trying to fix only the final line.
如果两边不相等，请从头重新检查步骤，而不是只修改最后一行。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Many students lose marks on bracket equations because of a few repeated errors. The table below summarises them.
许多学生在括号方程上丢分，是因为一些反复出现的错误。下表总结了这些错误。
| Mistake | Correct approach |
|---|---|
| Forgetting to multiply every term inside the bracket | Multiply the outside term by each term, one by one |
| Incorrect sign when a minus is before the bracket | Expand the bracket first, then apply the minus to all terms |
| Moving a term without changing its sign | Treat moving across the equal sign as reversing the sign |
| Dividing only part of a side by the coefficient | Divide every term on both sides by the same number |
Writing each step on a new line and checking signs at every line can eliminate most of these problems.
把每一步写在新的一行，并在每一行检查符号，可以消除大部分问题。
11. Real-Life Applications | 实际应用
Linear equations with brackets model many real situations, such as calculating costs, areas, or unknown lengths. Being able to set up and solve them is an essential KS3 skill.
含括号的一元一次方程可以模拟许多实际情况，例如计算费用、面积或未知长度。能够建立并求解这类方程是 KS3 的基本技能。
For example, a rectangle has length (x + 4) cm and width 3 cm. If its area is 27 cm², the equation is 3(x + 4) = 27.
例如，一个长方形的长为 (x + 4) cm，宽为 3 cm。如果它的面积是 27 cm²，方程就是 3(x + 4) = 27。
Solving gives x + 4 = 9, so x = 5 cm. This shows how abstract algebra connects directly to measurement and everyday problems.
求解得到 x + 4 = 9，所以 x = 5 cm。这表明抽象代数与测量和日常问题直接相关。
12. Practice and Summary | 练习与总结
To master bracket equations, practise in this order: expand brackets, collect like terms, move terms, isolate x, and check. Always work neatly and keep the equation balanced.
要掌握括号方程，请按以下顺序练习：去括号、合并同类项、移项、分离 x、检验。始终书写工整并保持等式平衡。
Try solving this equation: 2(x − 5) + 3x = 4(x + 1). The solution is x = 14. If you got a different answer, review the steps in sections 3 and 4.
尝试解这个方程：2(x − 5) + 3x = 4(x + 1)。答案是 x = 14。如果你得到不同的答案，请复习第 3 和第 4 节的步骤。
With regular practice, solving linear equations with brackets becomes automatic, giving you a strong foundation for graphs, inequalities, and more advanced algebra.
通过定期练习，解含括号的一元一次方程会变得自动化，为你学习图像、不等式和更高级的代数打下坚实基础。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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