Solving Equations with Brackets | 解含括号的一元一次方程

📚 Solving Equations with Brackets | 解含括号的一元一次方程

In KS3 Cambridge Mathematics, you often meet equations that contain brackets. To solve them correctly, you need to expand the brackets first using the distributive law, then simplify and isolate the unknown. This article guides you through the essential steps, common pitfalls and practice examples.

在剑桥初中数学中，你经常会遇到含有括号的方程。要正确求解，必须先运用分配律展开括号，然后化简并求出未知数。本文将带你掌握关键步骤、常见易错点以及配套练习。

1. Why brackets appear in equations | 为什么方程中会出现括号

Brackets group terms together and show that a multiplier applies to everything inside. For example, 3(x + 2) means three times the entire quantity x + 2. Equations become more powerful when quantities are collected this way.

括号将各项组合在一起，表明乘数作用于括号内的整体。例如 3(x + 2) 表示三倍的 (x + 2)。采用这种分组方式使方程的表达能力更强。

If we leave brackets unexpanded, the equation cannot be simplified directly. We must remove them first.

如果不展开括号，方程无法直接化简，必须先去括号。

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2. The distributive law | 分配律

The distributive law states that a(b + c) = ab + ac. The term outside multiplies each term inside. This is the foundation of all bracket expansions in algebra.

分配律指出 a(b + c) = ab + ac。外面的项要与里面每一项相乘。这是代数中所有括号展开的基础。

5(y − 4) = 5 × y − 5 × 4 = 5y − 20

5(y − 4) = 5 × y − 5 × 4 = 5y − 20

Notice how the sign inside the brackets stays with the term. When a minus sign is present, multiplication still follows the same rule.

注意括号内的符号跟随该项移动。遇到减号时，乘法规则同样适用。

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3. Expanding single brackets step by step | 逐步展开单个括号

Take a simple expression: 4(2x + 3). Multiply 4 by 2x to get 8x, then 4 by 3 to get 12. The expanded form is 8x + 12.

考虑简单表达式 4(2x + 3)。先用 4 乘 2x 得 8x，再用 4 乘 3 得 12。展开后为 8x + 12。

What about 2(5 − 3x)? Multiply 2 by 5 gives 10, and 2 by (−3x) gives −6x. Result: 10 − 6x.

那么 2(5 − 3x) 呢？2 乘以 5 得 10，2 乘以 (−3x) 得 −6x。结果为 10 − 6x。

Always re-check the signs: positive × negative = negative.

务必再次检查符号：正数乘负数得负数。

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4. Equations with one set of brackets | 含一组括号的方程

Consider the equation 3(x + 4) = 21. First expand: 3x + 12 = 21. Then subtract 12 from both sides: 3x = 9. Finally divide by 3: x = 3.

考虑方程 3(x + 4) = 21。先展开：3x + 12 = 21。然后两边同时减 12：3x = 9。最后除以 3：x = 3。

Some students forget to multiply the second term inside the bracket. Remember: the multiplier applies to every term.

有些同学会漏乘括号里的第二项。记住：乘数作用于每一项。

Common error: 3(x + 4) → 3x + 4 (wrong)

常见错误：3(x + 4) → 3x + 4（错误）

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5. Equations with multiple brackets | 含多组括号的方程

When an equation has brackets on both sides, expand each set independently. Example: 2(x + 5) = 3(2x − 1).

当方程两侧都有括号时，独立展开每一组。例如：2(x + 5) = 3(2x − 1)。

Left side: 2x + 10. Right side: 6x − 3. Equation becomes 2x + 10 = 6x − 3. Then solve.

左边：2x + 10。右边：6x − 3。方程变为 2x + 10 = 6x − 3。然后求解。

Collect variable terms on one side: subtract 2x from both sides to get 10 = 4x − 3. Add 3: 13 = 4x, so x = 13/4 or 3.25.

把含变量项移到同一边：两边减 2x 得 10 = 4x − 3。加 3 得 13 = 4x，因此 x = 13/4 即 3.25。

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6. Dealing with negative signs before brackets | 处理括号前的负号

A minus sign in front of a bracket means multiply every term inside by −1. For instance, −(2x − 5) becomes −2x + 5.

括号前面的负号表示括号内每一项都要乘以 −1。例如 −(2x − 5) 变为 −2x + 5。

If the equation is 7 − (y + 3) = 2, expand: 7 − y − 3 = 2, which simplifies to 4 − y = 2. So −y = −2 and y = 2.

若方程为 7 − (y + 3) = 2，展开得 7 − y − 3 = 2，化简为 4 − y = 2。因此 −y = −2，y = 2。

Avoid the common trap of writing −(a + b) = −a + b. The minus must change both signs.

避免常见陷阱：误写为 −(a + b) = −a + b。负号必须同时改变两项的符号。

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7. Combining like terms after expanding | 展开后合并同类项

After removing brackets, you often have to simplify by collecting like terms. Like terms share exactly the same variable part and exponent.

去括号后，经常需要通过合并同类项来化简。同类项具有完全相同的变量部分和指数。

Example: 4(x + 2) + 3(x − 5). First expand: 4x + 8 + 3x − 15. Combine x terms: 7x. Combine constant terms: −7. Final simplified expression: 7x − 7.

例如：4(x + 2) + 3(x − 5)。先展开：4x + 8 + 3x − 15。合并 x 项：7x。合并常数项：−7。最终化简为 7x − 7。

• 4x and 3x are like terms — they both contain x to the power 1.
• 4x 和 3x 是同类项——它们都包含 x 的一次方。

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8. Solving step-by-step strategy | 逐步求解策略

Follow a clear order when tackling any equation with brackets:

解含括号方程时，遵循清晰的顺序：

• Expand all brackets first. 首先展开所有括号。
• Simplify each side by collecting like terms. 合并每一边的同类项，化简。
• Use addition or subtraction to get variable terms on one side and constants on the other. 利用加减法将含变量项移到一边，常数项移到另一边。
• Divide by the coefficient to find the unknown. 除以系数求出未知数。

This four-step method works for almost all linear equations at KS3 level.

这种四步法几乎适用于所有初中阶段的一元一次方程。

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9. Checking your solution | 检验答案

Always substitute your answer back into the original equation to verify. If the left-hand side equals the right-hand side, your solution is correct.

务必将答案代回原方程检验。若左边等于右边，则答案正确。

Take the equation 5(2x − 1) = 25. Our solution: x = 3. Check: 5(2×3 − 1) = 5(6 − 1) = 5×5 = 25. Works.

以方程 5(2x − 1) = 25 为例。解得 x = 3。检验：5(2×3 − 1) = 5(6 − 1) = 5×5 = 25。正确。

Checking not only confirms accuracy but also helps you spot expansion or sign errors.

检验不仅能确认准确性，还能帮你发现展开或符号错误。

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10. Common mistakes to avoid | 常见错误与规避

Mistake 1: Forgetting to multiply the constant term inside the brackets. E.g. 2(x + 6) → 2x + 6. Correct: 2x + 12.

错误一：漏乘括号里的常数项。例如 2(x + 6) → 2x + 6。正确为 2x + 12。

Mistake 2: Mishandling negative multipliers. −3(x − 2) should become −3x + 6, not −3x − 2.

错误二：处理负乘数失误。−3(x − 2) 应得到 −3x + 6，而非 −3x − 2。

Mistake 3: Losing the equality when moving terms. Always perform the same operation on both sides.

错误三：移项时忘记保持等号平衡。务必对等号两边进行相同操作。

Mistake 4: Dropping the minus sign when subtracting brackets, such as 4 − (x + 3) becoming 4 − x + 3. It should be 4 − x − 3.

错误四：减去括号时丢掉负号，如 4 − (x + 3) 错写成 4 − x + 3。正确的应为 4 − x − 3。

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11. Worked examples for practice | 实战例题

Example 1: Solve 2(3x + 4) = 20. Expand: 6x + 8 = 20. Subtract 8: 6x = 12. Divide: x = 2.

例题 1：解 2(3x + 4) = 20。展开：6x + 8 = 20。减 8：6x = 12。除以 6：x = 2。

Example 2: Solve 5(x − 2) = 3(x + 4). Expand: 5x − 10 = 3x + 12. Subtract 3x: 2x − 10 = 12. Add 10: 2x = 22, x = 11.

例题 2：解 5(x − 2) = 3(x + 4)。展开：5x − 10 = 3x + 12。减 3x：2x − 10 = 12。加 10：2x = 22，x = 11。

Example 3: Solve 4 − 2(y + 3) = y − 1. Expand carefully: 4 − 2y − 6 = y − 1. Simplify left: −2y − 2 = y − 1. Add 2y to both sides: −2 = 3y − 1. Add 1: −1 = 3y, so y = −1/3.

例题 3：解 4 − 2(y + 3) = y − 1。仔细展开：4 − 2y − 6 = y − 1。左边化简：−2y − 2 = y − 1。两边加 2y：−2 = 3y − 1。加 1：−1 = 3y，y = −1/3。

These examples mirror typical exam-style questions at the Cambridge Checkpoint level.

这些例题模拟了剑桥 Checkpoint 考试中的典型题型。

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12. Summary and key takeaways | 总结与要点

Success with bracket equations relies on three habits: apply the distributive law accurately, treat negative signs with care, and always simplify before solving. Once brackets are gone, the equation behaves like any standard linear equation.

解好含括号方程依赖于三个习惯：正确运用分配律、谨慎处理负号、求解前一定先化简。一旦去掉括号，方程就变成了普通的一元一次方程。

Regular practice of expanding and simplifying will build confidence. Remember to check your answers by substitution to catch any sign errors early.

通过反复练习展开与化简，你会逐渐建立信心。记得用代入法检验答案，以便及早发现符号错误。

Step	Action	步骤	操作
1	Expand all brackets	1	展开所有括号
2	Collect like terms	2	合并同类项
3	Move variable terms to one side	3	将含变量项移到一边
4	Divide by the coefficient	4	除以系数

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