P117.1: Solving Equations with Brackets | 解带括号的方程

📚 P117.1: Solving Equations with Brackets | 解带括号的方程

This revision guide is based on the concepts from page 117, Exercise 1 in the Cambridge Lower Secondary Mathematics course. You will learn how to solve linear equations that contain brackets, a key skill for KS3 learners. We break down the method step by step, with clear examples and tips to help you avoid common errors.

本复习指南基于剑桥初中数学课程第117页练习1的内容。你将学会如何解含有括号的线性方程，这是KS3学生必须掌握的关键技能。我们将分步详细讲解方法，并提供清晰的示例和建议，帮助你避免常见错误。

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1. What Are Linear Equations? | 什么是线性方程？

A linear equation is an equation where the unknown variable (often x) is not raised to any power higher than 1. For example, 2x + 3 = 7 is a linear equation. Equations with brackets like 2(x + 1) = 8 are also linear because after expanding the brackets, the highest power of x remains 1.

线性方程是指未知数（通常为x）的最高次数为1的方程。例如，2x + 3 = 7 就是一个线性方程。像 2(x + 1) = 8 这样含有括号的方程也是线性方程，因为去掉括号后，x 的最高次幂仍然为1。

In KS3, you are expected to solve equations where the variable appears on one or both sides, and where brackets must be expanded first. Being able to recognise a linear equation helps you choose the right solving strategy.

在KS3阶段，你需要能够求解未知数出现在一边或两边的方程，以及需要先去括号的方程。学会识别线性方程能够帮助你选择正确的求解策略。

The general form of a linear equation in one variable is ax + b = c, where a, b and c are constants. Solving means finding the value of x that makes the equation true. When brackets appear, the equation can always be transformed into this general form.

一元一次线性方程的一般形式是 ax + b = c，其中 a、b 和 c 是常数。求解就是要找出能使等式成立的 x 值。当出现括号时，总可以先把方程转化成这种一般形式。

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2. The Distributive Law: Expanding Brackets | 分配律：去括号

The distributive law states that a(b + c) = ab + ac. This rule is the foundation for expanding brackets in algebra. For example, 3(y − 2) becomes 3 × y − 3 × 2, which simplifies to 3y − 6.

分配律指出 a(b + c) = ab + ac。这一规则是代数中去括号的基础。例如，3(y − 2) 变成 3 × y − 3 × 2，化简后就是 3y − 6。

When you solve an equation like 4(x + 3) = 28, you must first apply the distributive law to remove the bracket: 4x + 12 = 28. Only after expanding can you begin to isolate x.

当你解方程 4(x + 3) = 28 时，必须先用分配律去掉括号：4x + 12 = 28。只有展开之后，才能开始分离变量 x。

A common mistake is to only multiply the first term inside the bracket. Remember, the term outside multiplies every term inside. So 2(3x + 5) is 6x + 10, not 6x + 5. Be especially careful when there is a minus sign, such as −2(x − 4) = −2x + 8.

一个常见错误是只乘括号里的第一项。请记住，括号外的项要乘以括号内的每一项。因此 2(3x + 5) 等于 6x + 10，而不是 6x + 5。当有负号时要格外小心，例如 −2(x − 4) = −2x + 8。

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3. Simple Equations with One Bracket | 含单个括号的简单方程

Let’s solve the equation 5(x + 2) = 35. Step 1: Expand the left side to get 5x + 10 = 35. Step 2: Subtract 10 from both sides: 5x = 25. Step 3: Divide both sides by 5: x = 5. Always write each step on a new line to keep your work clear.

我们来解方程 5(x + 2) = 35。第1步：展开左边得到 5x + 10 = 35。第2步：两边同时减去10：5x = 25。第3步：两边同时除以5：x = 5。每一步都要换行书写，保持卷面整洁。

For the equation 3(2x − 4) = 18, first expand: 6x − 12 = 18. Then add 12 to both sides: 6x = 30. Finally divide by 6: x = 5. Notice that adding 12 cancels the −12 on the left, moving it to the right as +12.

对于方程 3(2x − 4) = 18，首先展开：6x − 12 = 18。然后两边加12：6x = 30。最后除以6：x = 5。注意，加12抵消了左边的−12，将其以+12的形式移到了右边。

You can check your solution by substituting x = 5 back into the original equation: 3(2×5 − 4) = 3(10 − 4) = 3×6 = 18, which matches the right side. This confirms the answer is correct.

你可以通过将 x = 5 代回原方程来检验：3(2×5 − 4) = 3(10 − 4) = 3×6 = 18，与右边相等。这就确认了答案是正确的。

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4. Equations with Multiple Brackets | 含多个括号的方程

Some equations have brackets on both sides, such as 2(x + 3) = 4(x − 1). Begin by expanding both sides: 2x + 6 = 4x − 4. Next, collect variable terms on one side and numbers on the other. Subtract 2x from both sides: 6 = 2x − 4. Then add 4: 10 = 2x, so x = 5.

有些方程两边都含有括号，例如 2(x + 3) = 4(x − 1)。首先两边同时展开：2x + 6 = 4x − 4。接着，将含变量的项移到一边，数字移到另一边。两边同时减去 2x：6 = 2x − 4。然后加4：10 = 2x，因此 x = 5。

When an equation has a term like 3(2 − x), expand carefully: 6 − 3x. The order of terms matters; it is fine to write −3x + 6. For example, solve 3(2 − x) = 9. Expanding gives 6 − 3x = 9. Subtract 6: −3x = 3. Divide by −3: x = −1.

当方程中出现像 3(2 − x) 这样的项时，展开要仔细：6 − 3x。项的顺序没有关系，写成 −3x + 6 也可以。例如，解方程 3(2 − x) = 9。展开得 6 − 3x = 9。减去6：−3x = 3。除以−3：x = −1。

If you have a negative sign in front of a bracket, like −(x + 4), treat it as −1 times the bracket. Thus −(x + 4) = −x − 4. Similarly, 5 − 2(x − 3) expands to 5 − 2x + 6, which simplifies to 11 − 2x. Always rewrite subtraction before expanding when mixing signs.

如果括号前面是一个负号，例如 −(x + 4)，可以把它当作 −1 乘以括号。因此 −(x + 4) = −x − 4。类似地，5 − 2(x − 3) 展开为 5 − 2x + 6，化简得到 11 − 2x。当正负号混合时，最好先把减法改写成加负数再去括号。

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5. Combining Like Terms | 合并同类项

After expanding brackets, you often need to simplify by combining like terms. Like terms are those that contain exactly the same variable raised to the same power. For instance, 4x and −2x are like terms, while 4x and 3 are not.

去括号后，你通常需要合并同类项来进行化简。同类项是指所含变量相同且次数也相同的项。例如，4x 和 −2x 是同类项，而 4x 和 3 不是。

Consider the equation 2(3x + 1) + 4x = 26. First expand: 6x + 2 + 4x = 26. Now combine the x-terms: 6x + 4x = 10x. The equation becomes 10x + 2 = 26. Subtract 2: 10x = 24, so x = 2.4 or 12/5.

考虑方程 2(3x + 1) + 4x = 26。首先展开：6x + 2 + 4x = 26。现在合并 x 项：6x + 4x = 10x。方程变为 10x + 2 = 26。减去2：10x = 24，因此 x = 2.4 或 12/5。

Always combine like terms on each side of the equation before moving terms across the equals sign. This reduces mistakes and makes the equation easier to handle.

在将各项移到等号另一边之前，一定要把等号两边的同类项分别合并。这样可以减少错误，并使方程更容易处理。

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6. Isolating the Variable | 分离变量

The goal of solving any linear equation is to get the variable alone on one side. This usually involves performing inverse operations: addition ‘undoes’ subtraction, multiplication ‘undoes’ division, and vice versa.

解任何线性方程的目标都是将变量单独留在等号的一边。这通常需要用到逆运算：加法“消除”减法，乘法“消除”除法，反之亦然。

Work in the reverse order of BIDMAS (or BODMAS). When isolating x, first look at any added or subtracted numbers, then deal with coefficients multiplying x. For example, in 7x + 3 = 38, subtract 3 first, then divide by 7.

按照运算顺序（BIDMAS/BODMAS）的逆序来操作。分离 x 时，先处理加减的常数，再处理与 x 相乘的系数。例如在 7x + 3 = 38 中，先减3，再除以7。

When the coefficient of x is a fraction, such as (2/3)x = 10, multiply both sides by the denominator and then divide by the numerator, or simply multiply by the reciprocal: x = 10 × (3/2) = 15. The same principle applies after brackets have been removed.

当 x 的系数是分数时，例如 (2/3)x = 10，可以两边同时乘以分母，再除以分子，或者直接乘以倒数：x = 10 × (3/2) = 15。去掉括号后，同样可以应用这一原则。

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7. Equations Involving Fractions | 涉及分数的方程

Some equations contain brackets and fractional terms together, like (x + 1)/2 = 5. This can be seen as ‘x + 1 divided by 2 equals 5’. To remove the fraction, multiply both sides by 2: x + 1 = 10, so x = 9.

有些方程同时含有括号和分数，例如 (x + 1)/2 = 5。可以理解为“x + 1 除以 2 等于 5”。为了去掉分数，两边同时乘以 2：x + 1 = 10，所以 x = 9。

For a harder example: (2x − 3)/5 = 4. Multiply both sides by 5: 2x − 3 = 20. Add 3: 2x = 23, so x = 11.5. Treat the numerator as if it were inside a bracket until you clear the denominator.

再看一个难一点的例子：(2x − 3)/5 = 4。两边乘以5：2x − 3 = 20。加3：2x = 23，所以 x = 11.5。在去掉分母之前，要把分子当作括号内的整体来对待。

When an equation has multiple fractions, such as x/3 + 1 = (x − 2)/2, find a common denominator, 6, and multiply every term by 6: 2x + 6 = 3(x − 2). Then expand and solve: 2x + 6 = 3x − 6, so 12 = x. Checking always confirms accuracy.

当方程含有多个分数时，例如 x/3 + 1 = (x − 2)/2，先找到公分母6，再将每一项乘以6：2x + 6 = 3(x − 2)。接着展开求解：2x + 6 = 3x − 6，因此 12 = x。验算总能确认准确性。

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8. Checking Your Solution | 检查所得的解

Substituting your answer back into the original equation is the most reliable way to verify your solution. If you get the same value on both sides, your solution is correct. This step is worth making a habit, especially under exam conditions.

将你得到的答案代回原方程，是验证解的最可靠方法。如果两边得出相同的值，那么你的解就是正确的。养成验算的习惯很有价值，尤其是在考试环境下。

For the equation 4(2x − 1) = 20, we found x = 3. Check: left side = 4(2×3 − 1) = 4(6 − 1) = 4×5 = 20. Left equals right, so x = 3 is indeed correct. This quick check can often catch simple arithmetic errors.

对于方程 4(2x − 1) = 20，我们解得 x = 3。验算：左边 = 4(2×3 − 1) = 4(6 − 1) = 4×5 = 20。左边等于右边，所以 x = 3 确实正确。这个快速检查常常能发现简单的计算错误。

If your check gives a false statement, like 10 = 12, retrace your steps. Look for sign mistakes, expansion errors, or incorrect inverse operations. Common fault points include forgetting to multiply both terms in a bracket or mishandling negative signs.

如果验算得出一个错误的等式，比如 10 = 12，就要往回检查步骤。看看有没有符号错误、展开错误或逆运算用错。常见的失分点包括忘记乘以括号里的每一项，或者负号处理不当。

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9. Common Mistakes to Avoid | 应避免的常见错误

Mistake 1: Only multiplying the first term inside the bracket. For 5(x + 3) = 20, writing 5x + 3 = 20 leads to wrong answers. The correct expansion is 5x + 15 = 20. Always use the distributive law fully.

错误1：只乘括号内的第一项。对于 5(x + 3) = 20，写成 5x + 3 = 20 会导致错误答案。正确的展开是 5x + 15 = 20。一定要完整应用分配律。

Mistake 2: Mistreating the minus sign. In the equation 8 − 3(x + 1) = 2, many students write 8 − 3x + 3 = 2, but it is 8 − 3x − 3 = 2. Remember, the minus sign applies to all terms inside the bracket.

错误2：负号处理不当。在方程 8 − 3(x + 1) = 2 中，很多学生写成 8 − 3x + 3 = 2，但正确的是 8 − 3x − 3 = 2。记住，负号作用于括号内的每一项。

Mistake 3: Not performing the same operation on both sides. When you add 3 to one side, you must add 3 to the other. Equations represent balance; any step must maintain that balance. Drawing a line down the centre of your working can help.

错误3：等号两边没有进行相同的运算。当你在一边加3时，另一边也必须加3。方程表示平衡，任何步骤都必须保持这种平衡。在书写中间画一条竖线可能有助于提醒自己。

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10. Real-Life Applications | 实际应用

Solving equations with brackets is not just an abstract exercise; it models many real-world situations. For instance, if a mobile phone plan costs a £10 fixed fee plus £5 per gigabyte, the total cost for g gigabytes is 10 + 5g. Setting this expression equal to £35 gives the equation 10 + 5g = 35, easily solved as g = 5.

解带括号的方程不只是一个抽象的练习，它可以模拟许多现实情况。例如，如果一个手机套餐每月固定费用为10英镑，每用1GB额外支付5英镑，那么g GB的总费用为10 + 5g。让这个式子等于35英镑，就得到方程 10 + 5g = 35，容易解得 g = 5。

When a problem involves brackets, like sharing costs equally: three friends equally pay for a meal costing £(2x + 18) and each contributes £12. Then (2x + 18)/3 = 12. Multiply by 3: 2x + 18 = 36, so 2x = 18 and x = 9. This shows how brackets arise naturally.

当问题涉及括号时，比如平摊费用：三位朋友共同支付一顿 £(2x + 18) 的餐费，每人出12英镑。于是有 (2x + 18)/3 = 12。乘以3：2x + 18 = 36，所以 2x = 18，x = 9。这展示了括号是如何自然出现的。

Geometry also uses equations with brackets frequently. The perimeter of a rectangle with length (x + 5) cm and width 3 cm is 2[(x + 5) + 3]. If the perimeter is 28 cm, the equation 2(x + 8) = 28 leads to x + 8 = 14, so x = 6 cm. This reinforces both algebra and measurement skills.

几何中也经常用到带括号的方程。一个长为 (x + 5) cm、宽为3 cm的矩形，其周长为 2[(x + 5) + 3]。如果周长是28 cm，那么方程 2(x + 8) = 28 就变成 x + 8 = 14，因此 x = 6 cm。这同时强化了代数与度量的技能。

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11. More Advanced Examples | 进阶示例

Equations can have nested brackets, such as 2[3x + (x − 1)] = 18. Always work from the innermost brackets outward: 2[3x + x − 1] = 18 becomes 2[4x − 1] = 18. Then expand: 8x − 2 = 18. Add 2: 8x = 20, so x = 2.5 or 5/2. Patience is key when simplifying in stages.

方程中也可能出现嵌套括号，例如 2[3x + (x − 1)] = 18。一定要从最内层的括号开始向外处理：2[3x + x − 1] = 18 变为 2[4x − 1] = 18。展开：8x − 2 = 18。加2：8x = 20，所以 x = 2.5 或 5/2。分步化简时，耐心是关键。

Another challenging type involves variables on both sides after expansion, e.g., 3(2x − 4) = 5x + 1. Expand to get 6x − 12 = 5x + 1. Subtract 5x: x − 12 = 1. Add 12: x = 13. Always aim to have only x terms on one side at the end of your rearrangement.

另一种较难的类型是展开后变量出现在两边，例如 3(2x − 4) = 5x + 1。展开得到 6x − 12 = 5x + 1。减去5x：x − 12 = 1。加12：x = 13。在移项结束时，始终要确保只有一边含有 x 项。

When the equation contains a subtraction of a bracket inside another bracket, use care: 5 − [2(x + 1) − 4] = x + 3. First simplify inside the square brackets: 2(x + 1) − 4 = 2x + 2 − 4 = 2x − 2. Then the equation becomes 5 − (2x − 2) = x + 3, which gives 5 − 2x + 2 = x + 3 → 7 − 2x = x + 3 → 4 = 3x → x = 4/3. Writing small intermediate steps prevents errors.

当方程中的括号外层是减法时，要格外小心：5 − [2(x + 1) − 4] = x + 3。先化简方括号内部：2(x + 1) − 4 = 2x + 2 − 4 = 2x − 2。于是方程变为 5 − (2x − 2) = x + 3，即 5 − 2x + 2 = x + 3 → 7 − 2x = x + 3 → 4 = 3x → x = 4/3。写下简短的中间步骤可以防止错误。

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12. Key Tips Summary | 要点总结

Always expand brackets first using the distributive law. After that, simplify each side by combining like terms. Use inverse operations to isolate the variable, doing the same operation on both sides of the equation at every step.

始终先用分配律去括号。之后，合并同类项以化简方程两边。利用逆运算分离变量，并在每一步都对等号两边进行相同的运算。

Write each step clearly on a new line. If an equation contains fractions, multiply through by a common denominator to work with integers. And never skip the checking step — substituting the answer back into the original equation builds confidence and catches errors.

每一步都要清晰地换行书写。如果方程含有分数，可将每一项乘以公分母，转化为整数运算。绝对不要跳过验算步骤——将答案代回原方程能够建立信心，并揪出错误。

Practice with a variety of equations, from simple one-bracket cases to those with multiple brackets and fractions. With consistent practice, solving equations with brackets becomes a smooth and almost automatic process.

要练习各种类型的方程，从简单的单括号方程到含有多个括号和分数的复杂方程。通过持续练习，解带括号的方程将变得流畅自然，几乎成为一种自动化的过程。

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