📚 Solving Linear Equations: KS3 Cambridge Maths p36 Worked Example | 解线性方程：剑桥KS3数学第36页例题解析
Linear equations are a fundamental part of the Cambridge KS3 mathematics curriculum. Being able to solve equations like the one on page 36, question 1, builds the foundation for algebra, problem solving, and later topics such as graphs and inequalities. In this article, we will work through the step-by-step solution of the equation 3(2x − 1) = 4x + 5, check the answer, discuss common mistakes, and explore related problems that appear in the same exercise.
线性方程是剑桥初中数学课程的核心内容。能够解出类似第36页第1题的方程，为代数、问题解决以及后续的图像与不等式等课题打下坚实基础。本文将逐步解析方程 3(2x − 1) = 4x + 5，验证答案，讨论常见错误，并探讨同一练习中的相关问题。
1. The Problem | 问题呈现
We are asked to solve the equation 3(2x − 1) = 4x + 5 from page 36, exercise 1. The target is to find the value of x that makes both sides equal. This is called a linear equation because the unknown x appears only to the first power, there are no x² terms, and no x appears in a denominator or under a square root sign.
The equation has two important features that decide our method. First, the left-hand side contains a bracket, which must be expanded before anything else can be combined. Second, the unknown x appears on both sides of the equals sign, so we will need to collect all the x terms on one side and all the plain numbers on the other. Solving the equation means finding the single value of x for which the left-hand side and the right-hand side give exactly the same number.
题目要求解第36页练习1中的方程 3(2x − 1) = 4x + 5。目标是找出使等式两边相等的 x 值。它被称为线性方程，是因为未知数 x 只出现一次幂，没有 x² 项，x 也不出现在分母或根号之下。
这个方程有两个关键特征，决定了我们的解题方法。第一，左边含有括号，必须先展开，之后才能合并同类项。第二，未知数 x 同时出现在等号两侧，因此我们需要把所有含 x 的项集中到一边，把所有常数项集中到另一边。解方程的意义，就是找出那唯一的 x 值，使左边与右边恰好得到相同的数值。
2. Expanding Brackets | 展开括号
The first step is to expand the bracket on the left-hand side. Multiply 3 by each term inside: 3 × 2x = 6x and 3 × (−1) = −3. This gives 6x − 3. So the equation becomes 6x − 3 = 4x + 5.
Students often lose marks here by multiplying only the first term inside the bracket. Remember that the number outside the bracket multiplies every term inside it, not just the one nearest to it. A useful habit is to draw a small arrow from the 3 to each term before writing anything down.
3(2x − 1) = 4x + 5
6x − 3 = 4x + 5
第一步是展开左侧的括号。将 3 乘以括号内的每一项：3 × 2x = 6x，3 × (−1) = −3，得到 6x − 3。于是方程变为 6x − 3 = 4x + 5。
很多同学在这里失分，是因为只乘了括号里的第一项。要记住：括号外的数字要乘以括号内的每一项，而不只是最靠近它的那一项。一个实用的习惯是先用小箭头把 3 分别指向括号内的每一项，再动笔书写。
3. Collecting the x Terms on One Side | 把含 x 的项集中到一边
Now we have 6x − 3 = 4x + 5. The unknown appears on both sides, so we choose to remove the smaller x term, which is 4x on the right. Subtract 4x from both sides:
6x − 3 − 4x = 4x + 5 − 4x
2x − 3 = 5
Notice that the right-hand side is now free of x, and the left-hand side has been simplified from 6x − 4x to 2x. Whatever operation we perform, it must be applied to both sides; that is what keeps the equation balanced and the solution unchanged.
现在我们得到 6x − 3 = 4x + 5。未知数出现在两边，因此我们选择消去较小的 x 项，也就是右边的 4x。两边同时减去 4x：得到 2x − 3 = 5。
可以看到，右边已经不含 x，而左边由 6x − 4x 化简为 2x。无论进行什么运算，都必须同时作用于等号两边；这正是保持等式平衡、解不发生改变的关键。
4. Isolating x and Finding the Solution | 求出 x 的值
We now have the much simpler equation 2x − 3 = 5. Add 3 to both sides to remove the −3:
2x − 3 + 3 = 5 + 3
2x = 8
Finally, divide both sides by 2, the coefficient of x:
x = 8 ÷ 2 = 4
So the solution is x = 4 and remember that the answer to a linear equation is normally a single value, not a range.
现在方程已经简化为 2x − 3 = 5。两边同时加 3，消去 −3，得到 2x = 8。最后两边同时除以 x 的系数 2，得到 x = 8 ÷ 2 = 4。
因此方程的解是 x = 4。请务必把答案写成明确的结论 x = 4，而不要停留在 2x = 8 这一步；同时要记住，线性方程的解通常是一个具体的数值，而不是一个范围。
5. Checking Your Answer | 验证答案
ck into the original equation, not the simplified one, so that any earlier error will be exposed.
Left-hand side: 3(2 × 4 − 1) = 3(8 − 1) = 3 × 7 = 21.
Right-hand side: 4 × 4 + 5 = 16 + 5 = 21.
. In an examination, a substitution line like this costs only a few seconds but can save several marks, and it also demonstrates to the examiner that you understand what a solution actually means.
只有经过检验的解才值得信任。请把 x = 4 代回原方程，而不是代回化简后的方程，这样一旦前面出错就能立刻暴露出来。
左边：3(2 × 4 − 1) = 3(8 − 1) = 3 × 7 = 21；右边：4 × 4 + 5 = 16 + 5 = 21。两边都等于 21，说明 x = 4 是正确的。
在考试中，这样一行代入检验只需几秒钟，却可能挽回好几分，同时也向阅卷老师表明你理解"解"的真正含义。
6. The Balance Method Explained | 天平法原理
Every step above is an application of the balance method. Imagine the equation as a pair of scales with the two sides in perfect balance. Adding the same number to both pans, subtracting the same number from both pans, multiplying both pans by the same non-zero number, or dividing both pans by the same non-zero number keeps the scales level.
That is why we may subtract 4x from both sides, add 3 to both sides, and divide both sides by 2 without changing the value of x. The one forbidden operation is dividing by zero or by an expression that could be zero; fortunately, in KS3 linear equations this situation does not arise.
以上每一步都是"天平法"的应用。把方程想象成一架天平，两边完全平衡。在两个托盘中同时加上相同的数、同时减去相同的数、同时乘以相同的非零数，或同时除以相同的非零数，天平都会保持平衡。
这就是为什么我们可以两边同时减 4x、两边同时加 3、两边同时除以 2，而 x 的值始终不变。唯一被禁止的运算是除以 0，或除以一个可能为 0 的式子；幸运的是，在 KS3 的线性方程中不会遇到这种情形。
7. Common Mistakes to Avoid | 常见错误与规避
Most lost marks in this topic come from a small number of repeated slips rather than from genuine misunderstanding. The table below lists the ones that appear most often in Cambridge KS3 exercise books and examination scripts.
本专题失分大多来自少数几种反复出现的疏忽，而不是真正的不理解。下表列出了剑桥 KS3 练习册与考卷中最常见的错误。
| Mistake | 错误 | Why it fails | 错在哪里 | Correct approach | 正确做法 |
| Writing 3(2x − 1) = 6x − 1 | The −1 inside the bracket was never multiplied by 3 | Multiply every term: 6x − 3 |
| Changing only one side | The balance of the equation is destroyed | Apply every operation to both sides |
| Sign error when moving terms | Ignoring that 4x becomes −4x when it crosses the equals sign | Think of it as subtracting 4x from both sides |
| Stopping at 2x = 8 | x has not been isolated, so the equation is unsolved | Divide by the coefficient: x = 4 |
8. An Alternative Method: Working Backwards | 另一种方法：逆运算倒推
A second valid strategy is to treat the equation as a chain of operations applied to x and then undo them in reverse order. Starting from x, the left-hand side does: multiply by 2, subtract 1, multiply by 3; the right-hand side does: multiply by 4, add 5. Because both descriptions must produce the same number, we can compare them directly.
Set both expressions equal and reason in reverse: 3(2x − 1) = 4x + 5 behaves exactly like 6x − 3 = 4x + 5 ely 2x, must equal the difference between 5 and −3, namely 8. Hence 2x = 8 and x = 4. This "difference" n work you should always show the formal balance-method steps to secure full method marks.
另一种可行策略，是把方程看作施加在 x 上的一串运算，再按相反顺序逐一还原。从 x 出发，左边依次是：乘 2、减 1、乘 3；右边是：乘 4、加 5。由于两种描述必须得到同一个数，我们可以直接比较它们。
让两个表达式相等并倒推：3(2x − 1) = 4x + 5 与 6x − 3 = 4x + 5 完全等价，所以 6x 与 4x 的差 2x，必须等于 5 与 −3 的差 8。于是 2x = 8，x = 4。这种"差值"思路适合心算快速检验，但书面作答时仍应写出规范的天平法步骤，以拿到完整的方法分。
9. Related Practice Problems | 相关练习
Page 36 exercise 1 continues with equations built on exactly the same three skills: expanding a bracket, collecting like terms, and dividing by a coefficient. Try the following, then check your answers underneath.
第36页练习1的后续题目，考察的正是同样三项技能：展开括号、合并同类项、除以系数。请先自行练习下列题目，再对照下方的答案。
- Solve 2(x + 3) = 14 | 解 2(x + 3) = 14
- Solve 5(2x − 1) = 3x + 21 | 解 5(2x − 1) = 3x + 21
- Solve 4(x − 2) = 2(x + 5) | 解 4(x − 2) = 2(x + 5)
- Solve 3(2x + 1) − 4 = 2(3x − 2) + 9 | 解 3(2x + 1) − 4 = 2(3x − 2) + 9
quation simplifies to 6x − 1 = 6x + 5, which has no solution because −1 can never equal 5.
答案依次为：x = 4；x = 2；x = 9；第四题化简后得到 6x − 1 = 6x + 5，因为 −1 永远不可能等于 5，所以该方程无解。
10. Equations with Brackets on Both Sides | 两边都有括号的方程
As soon as a bracket appears on each side, the method simply repeats itself. Expand both brackets, collect the x terms on the side where the coefficient is larger, and collect the numbers on the other side. Take 4(x − 2) = 2(x + 5) from the list above:
4x − 8 = 2x + 10
2x = 18
x = 9
ror. If an equation leads to something like 0x = 7, it has no solution; if it leads to 0x = 0, every value of x works and the equation is called an identity.
一旦两边都出现括号，方法只是重复一遍：展开两个括号，把含 x 的项集中到系数较大的那一边，把数字集中到另一边。以 4(x − 2) = 2(x + 5) 为例，展开得 4x − 8 = 2x + 10，整理得 2x = 18，故 x = 9。
检验：4(9 − 2) = 28，2(9 + 5) = 28。因为 4x 大于 2x，减去 2x 可以让所有系数保持为正，从而降低符号出错的可能。若方程化为 0x = 7，则无解；若化为 0x = 0，则任何 x 都成立，这样的方程称为恒等式。
11. Why This Matters for Later Topics | 与后续内容的联系
The techniques practised on page 36 are not isolated skills. The same expansion and balancing steps are used to rearrange formulas such as v = u + at, to solve simultaneous equations by substitution, to find where two straight-line graphs y = 6x − 3 and y = 4x + 5 intersect, and to handle inequalities, where the only extra rule is that multiplying or dividing by a negative number reverses the inequality sign.
In fact, solving 3(2x − 1) = 4x + 5 oint has x-coordinate 4, with y-coordinate 21. Understanding this link makes the algebra feel purposeful rather than mechanical.
第36页所练的技巧并非孤立的知识点。同样的展开与平衡步骤，也用于变形公式（如 v = u + at）、用代入法解联立方程、求两条直线 y = 6x − 3 与 y = 4x + 5 的交点，以及处理不等式——那里唯一的额外规则是：乘或除以负数时不等号要反向。
事实上，解 3(2x − 1) = 4x + 5 就等同于问两条直线 y = 6x − 3 与 y = 4x + 5 在哪里相交；答案 x = 4 说明交点横坐标为 4，纵坐标为 21。理解了这一联系，代数就不再是机械的运算，而有了明确的意义。
12. Exam Tips and Summary | 考试技巧与小结
In summary, the equation 3(2x − 1) = 4x + 5 es by 2 to get x = 4. Substituting x = 4 into the original equation gives 21 on both sides, confirming the answer.
For examination technique, show every line of working, keep the equals signs aligned vertically, and write the final answer as a separate statement. Where a question carries more than one mark, the method marks are usually awarded for the expansion, the collecting of terms and the correct division, even if an arithmetic slip creeps in. Finally, always substitute your answer back into the original equation: it takes seconds and it is the single most reliable way to protect your marks.
综上，方程 3(2x − 1) = 4x + 5 分四步解决：展开括号得 6x − 3 = 4x + 5；两边减 4x 得 2x − 3 = 5；两边加 3 得 2x = 8；两边除以 2 得 x = 4。把 x = 4 代入原方程，两边都等于 21，答案得到确认。
考试技巧方面：每一步过程都要写出，等号上下对齐，最后把答案单独写成一行结论。当一道题分值超过一分时，方法分通常分别给在展开括号、合并同类项和正确除法上，即使中间出现计算疏忽也仍可得分。最后，一定要把答案代回原方程检验：只需几秒钟，却是保住分数最可靠的办法。
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