📚 Solving Linear Equations with Brackets | 解含括号的一元一次方程
Linear equations are the foundation of algebra, and when brackets appear, they can seem tricky at first. Understanding how to expand brackets and systematically isolate the unknown variable is a crucial skill in the Cambridge KS3 Mathematics curriculum. This article will guide you through the methods step by step, with clear examples and common pitfalls to avoid.
一元一次方程是代数的基础，当方程中出现括号时，起初可能会让人觉得棘手。掌握展开括号并系统地解出未知变量，是剑桥 KS3 数学课程中的重要技能。本文将逐步引导你掌握解题方法，配以清晰的例子和需要避免的常见错误。
1. What are Linear Equations with Brackets? | 什么是含括号的一元一次方程？
A linear equation with brackets contains one or more expressions inside brackets, such as 3(x + 2) = 15 or 5(2y – 4) = 10. The unknown variable appears only to the first power, making the equation linear. The brackets indicate multiplication of the term outside by each term inside.
含括号的一元一次方程是指方程中括号内含有一个或多个表达式，例如 3(x + 2) = 15 或 5(2y – 4) = 10。未知变量只出现一次方，因此方程是线性的。括号表示外面的项与括号内每一项相乘。
These equations often arise from real-life problems, such as calculating total costs with discounts, or finding unknown lengths in geometry. The key to solving them is to remove the brackets correctly and then apply inverse operations.
这类方程常出现在现实问题中，例如计算折扣后的总价，或在几何中求未知长度。解题的关键是正确地去掉括号，然后运用逆运算。
2. The Distributive Law | 分配律的运用
To expand brackets, we use the distributive law: a(b + c) = ab + ac, and a(b – c) = ab – ac. The multiplier ‘a’ is distributed to every term inside the brackets. If the multiplier is negative, the signs of the terms inside will change.
要展开括号，我们使用分配律：a(b + c) = ab + ac，以及 a(b – c) = ab – ac。乘数 ‘a’ 要分配给括号内的每一项。如果乘数是负数，括号内各项的符号会改变。
For example, 4(m + 3) becomes 4 × m + 4 × 3 = 4m + 12. And -2(3n – 5) becomes -6n + 10. Always pay attention to the sign in front of the bracket.
例如，4(m + 3) 展开为 4 × m + 4 × 3 = 4m + 12。而 -2(3n – 5) 展开为 -6n + 10。要始终注意括号前面的符号。
| Expression 表达式 | Expanded Form 展开式 |
|---|---|
| 2(x + 5) | 2x + 10 |
| -3(y – 4) | -3y + 12 |
| -(2a + 3) | -2a – 3 |
3. Step-by-Step Method to Solve | 逐步解题法
Once the brackets are expanded, the equation becomes a simpler linear equation without brackets. The general steps are: expand the brackets, collect like terms on each side, move variable terms to one side using inverse operations, move constants to the other side, and finally divide by the coefficient of the variable.
一旦括号被展开，方程就变成了不含括号的简单一次方程。一般步骤为：展开括号，合并每一边的同类项，利用逆运算把含变量项移到一边，把常数项移到另一边，最后除以变量的系数。
Let’s outline the process with the equation 2(3x – 1) = 10. First expand: 6x – 2 = 10. Then add 2 to both sides: 6x = 12. Finally divide by 6: x = 2. Always check your answer by substituting it back into the original equation.
我们以方程 2(3x – 1) = 10 为例进行说明。首先展开：6x – 2 = 10。然后两边加 2：6x = 12。最后两边除以 6：x = 2。始终要将答案代回原方程来进行检验。
4. Moving Terms and Simplifying | 移项与化简
Moving terms means adding or subtracting terms on both sides of the equation to isolate the variable. If you have 3x + 5 = 2x + 9, subtract 2x from both sides to get x + 5 = 9, then subtract 5 to obtain x = 4.
移项指的是在方程两边同时加上或减去项，以隔离变量。若方程为 3x + 5 = 2x + 9，两边减 2x 得到 x + 5 = 9，再两边减 5 得到 x = 4。
Brackets often hide like terms. For example, 4p + 3(2 – p) = 11. Expand first: 4p + 6 – 3p = 11. Combine like terms: p + 6 = 11. Then p = 5.
括号往往会隐藏同类项。例如，4p + 3(2 – p) = 11。先展开：4p + 6 – 3p = 11。合并同类项：p + 6 = 11。于是 p = 5。
5. Worked Example 1 | 示例详解 1
Solve 5(2y + 3) – 4y = 27. Step 1: Expand the bracket: 10y + 15 – 4y = 27. Step 2: Combine the y-terms: 6y + 15 = 27. Step 3: Subtract 15 from both sides: 6y = 12. Step 4: Divide by 6: y = 2.
解方程 5(2y + 3) – 4y = 27。第一步：展开括号：10y + 15 – 4y = 27。第二步：合并含 y 的项：6y + 15 = 27。第三步：两边减去 15：6y = 12。第四步：除以 6：y = 2。
Check the solution: Left-hand side = 5(2×2 + 3) – 4×2 = 5(4+3) – 8 = 5×7 – 8 = 35 – 8 = 27. Right-hand side = 27. So it is correct.
检验解：左边 = 5(2×2 + 3) – 4×2 = 5(4+3) – 8 = 5×7 – 8 = 35 – 8 = 27。右边 = 27。因此正确。
5(2y + 3) – 4y = 27 → y = 2
6. Dealing with Negative Signs | 处理负号
A negative sign before a bracket changes the sign of every term inside. For example, 8 – 2(x + 1) = 4. Here, the ‘-2’ multiplies both terms: 8 – 2x – 2 = 4. Combine constants: 6 – 2x = 4. Then subtract 6: -2x = -2, so x = 1.
括号前的负号会改变括号内每一项的符号。例如，8 – 2(x + 1) = 4。这里的 ‘-2’ 与两项相乘：8 – 2x – 2 = 4。合并常数项：6 – 2x = 4。然后两边减 6：-2x = -2，所以 x = 1。
Be extra careful when the expression involves subtraction: 3(x – 4) – 2(x + 5) = 0. Expand: 3x – 12 – 2x – 10 = 0. Combine: x – 22 = 0, hence x = 22.
当表达式中带有减法时要格外小心：3(x – 4) – 2(x + 5) = 0。展开：3x – 12 – 2x – 10 = 0。合并：x – 22 = 0，因此 x = 22。
7. Brackets on Both Sides | 两边都有括号的方程
If an equation has brackets on both sides, expand each side separately first. For example, 2(3a + 1) = 4(a – 2). Expanding gives: 6a + 2 = 4a – 8. Then bring variables together: subtract 4a from both sides to get 2a + 2 = -8. Subtract 2: 2a = -10, and a = -5.
如果方程两边都有括号，先分别展开每一边。例如， 2(3a + 1) = 4(a – 2)。展开得：6a + 2 = 4a – 8。然后合并变量：两边减 4a 得到 2a + 2 = -8。两边减 2：2a = -10，解得 a = -5。
Remember that the equal sign acts like a balance: whatever you do to one side, you must do to the other. When expanding, treat the multiplier and its sign as a whole.
记住等号就像一个天平：你对一边做了什么，另一边也必须做同样的事情。展开时，要把乘数及其符号视为一个整体。
8. Equations Involving Fractions | 含分数的方程
Sometimes brackets interact with fractions, like 1/2 (4x + 8) = 7. The fraction outside the bracket multiplies each term inside. So, (1/2)×4x = 2x, and (1/2)×8 = 4, giving 2x + 4 = 7, thus x = 1.5.
有时括号会与分数结合，例如 1/2 (4x + 8) = 7。括号外的分数要与括号内每一项相乘。于是 (1/2)×4x = 2x，(1/2)×8 = 4，得到 2x + 4 = 7，因此 x = 1.5。
For a more complex case: (2/3)(6y – 9) = 10. Expand: (2/3)×6y = 4y, (2/3)×(-9) = -6. So 4y – 6 = 10, 4y = 16, y = 4. Alternatively, you can multiply both sides by the denominator first to clear the fraction.
对于更复杂的情况：(2/3)(6y – 9) = 10。展开：(2/3)×6y = 4y，(2/3)×(-9) = -6。因此 4y – 6 = 10，4y = 16，y = 4。另一种方法是先两边乘以分母以消去分数。
1/2 (4x + 8) = 7 → 2x + 4 = 7 → x = 3/2
9. Checking Your Solution | 检验解
Substitution is the most reliable way to verify your answer. Take the original equation and replace the variable with the obtained value. If both sides evaluate to the same number, the solution is correct. This also helps catch sign errors or expansion mistakes.
代入法是最可靠的验算方式。取原方程，将变量替换为所得数值。如果两边计算出的结果相等，则解是正确的。这也有助于发现符号错误或展开错误。
Example: For 3(2x – 1) + 4 = 19, we found x = 3. Left side: 3(2×3 – 1) + 4 = 3(6 – 1) + 4 = 3×5 + 4 = 19. Right side: 19. It holds.
例子：对于 3(2x – 1) + 4 = 19，我们解得 x = 3。左边：3(2×3 – 1) + 4 = 3(6 – 1) + 4 = 3×5 + 4 = 19。右边：19。等式成立。
10. Common Pitfalls and How to Avoid Them | 常见错误与避免方法
- Forgetting to multiply all terms: In 2(x + 5) + 3, students often write 2x + 5 + 3. The correct expansion is 2x + 10 + 3. Always apply the multiplier to every term inside the bracket.
- 遗漏乘法项： 在 2(x + 5) + 3 中，学生常写成 2x + 5 + 3。正确的展开是 2x + 10 + 3。始终将乘数作用于括号内的每一项。
- Sign errors: With -3(2n – 4), the negative sign must be distributed: -6n + 12, not -6n – 12.
- 符号错误： 对于 -3(2n – 4)，负号必须分配进去：-6n + 12，而不是 -6n – 12。
- Incorrect moving of terms: When moving a term across the equals sign, students sometimes forget to change its sign. Always use inverse operations.
- 移项错误： 当把一项移到等号另一边时，学生有时忘记改变符号。永远使用逆运算。
11. Practice Problems | 练习题
Try solving these equations step by step. Expand the brackets first, then solve using inverse operations. Check your answers afterward.
试着一题一题地解下列方程。先展开括号，然后用逆运算求解。之后检验答案。
| Equation 方程 | Solution 解 |
|---|---|
| 4(x + 3) = 32 | x = 5 |
| 2(3y – 5) + 7 = 23 | y = 4 |
| 5(2a + 1) = 3(a + 9) | a = 22/7 |
| 6 – 3(2m – 1) = 9 | m = 0 |
| 1/3(9t + 6) = 5 | t = 1 |
12. Summary and Key Points | 总结与要点
Solving linear equations with brackets is all about applying the distributive law accurately and then following the logical steps of algebra. Always expand brackets first, collecting like terms to simplify. Use inverse operations to isolate the variable, and never skip the verification step.
解含括号的一元一次方程，关键在于准确运用分配律，然后遵循代数的逻辑步骤。一定要先展开括号，合并同类项以化简。利用逆运算隔离变量，并且绝不能跳过验算这一步。
Remember that practice is essential. The more equations you solve, the more confident you become in manipulating brackets and signs correctly. These skills form a strong foundation for more advanced algebra and problem solving in KS3 and beyond.
记住，练习至关重要。你解的方程越多，在正确处理括号和符号方面就会越自信。这些技能将为 KS3 及往后更高阶的代数和问题解决打下坚实的基础。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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