📚 Solving Linear Equations: Cambridge KS3 Maths p124_2 | 解一元一次方程：剑桥 KS3 数学 p124_2
Cambridge KS3 Maths page 124, exercise 2 is a core algebra drill: solve linear equations that contain brackets, unknowns on both sides, and small fractional coefficients. This article explains the complete method step by step, using the style of question 2 as a fully worked model, and finishes with a checklist of the mistakes that cost marks most often.
剑桥 KS3 数学第 124 页的练习 2 是初中代数的基础题型：解含有括号、两边都含未知数以及简单分数系数的一元一次方程。本文以第 2 题的题型为例，完整讲解解题方法，并在最后总结最容易丢分的错误清单。
1. Understanding the Question | 理解题意
g every value of x that makes the left-hand side equal to the right-hand side. The word linear tells you that x appears only to the first power: you will never meet x², √x or 1/x in this exercise.
在剑桥 KS3 数学第 124 页的练习 2 中，题目要求解含一个未知数 x 的方程。典型例子是 2(3x − 1) = 4(x + 2)。解方程意味着找出所有使左边等于右边的 x 值。Linear（线性）这个词告诉你：未知数 x 只出现一次幂，本练习中绝不会出现 x²、√x 或 1/x 这类形式。
Read the question twice before you write anything. Underline the brackets and any like terms. This helps you choose the correct first step and prevents the two classic errors: dropping a bracket, or moving a term without changing its sign.
动笔之前请把题目读两遍。用下划线标出括号和同类项。这样能帮你判断正确的第一步，并避免两个经典错误：漏掉括号，或者移项时忘记变号。
Every equation in this exercise can be solved by the same five-step routine:
本练习中的每一道方程都可以用同一套五步流程解决：
- Expand all brackets | 展开所有括号
- Collect like terms on each side separately | 分别在等号两边合并同类项
- Move all x terms to one side, all numbers to the other | 把含 x 的项移到一边，常数项移到另一边
- Divide by the coefficient of x | 除以 x 的系数
- Substitute back to check | 代回原方程检验
2. The Balance Model | 天平模型
Think of an equation as a balance scale. The equals sign is the pivot. Whatever you do to the left-hand side you must also do to the right-hand side, otherwise the scale tips and the statement is no longer true.
把方程想象成一架天平。等号就是支点。你对左边做的任何运算，都必须在右边做同样的运算，否则天平就会倾斜，等式不再成立。
This single idea justifies every legal move in algebra: adding the same number to both sides, subtracting the same term from both sides, multiplying both sides by the same non-zero number, and dividing both sides by the same non-zero number. It is also why you may swap the two sides of an equation freely, and why the answer can be negative or a fraction — no rule forbids it.
这一个观念解释了代数中所有合法的变形：两边同时加上同一个数、同时减去同一个项、同时乘以同一个非零数、同时除以同一个非零数。它也解释了为什么可以随意交换等号左右两边，以及为什么答案可以是负数或分数——没有任何规则禁止它们。
If a = b, then a + c = b + c and a − c = b − c
If a = b and c ≠ 0, then ac = bc and a ÷ c = b ÷ c
Never divide both sides by an expression that could be zero, and never multiply both sides by an unknown. In this exercise the coefficient of x is always a definite non-zero number, so both operations are safe.
不要用可能为零的式子去除两边，也不要用未知数去乘两边。在本练习中，x 的系数始终是一个确定的非零数，所以这两种运算都是安全的。
3. Expanding Brackets | 展开括号
The first step in almost every question on this page is to remove the brackets. Multiply the number in front of the bracket by every term inside it, one at a time, keeping the signs attached to the terms.
这一页上几乎所有题目的第一步都是去括号。用括号前的数分别乘以括号内的每一项，一次乘一项，并且把符号跟着项一起处理。
a(b + c) = ab + ac
−a(b − c) = −ab + ac
So 2(3x − 1) becomes 6x − 2, and 4(x + 2) becomes 4x + 8. Notice the minus sign inside the first bracket: it multiplies by 2 as well, giving −2, not +2.
因此 2(3x − 1) 展开为 6x − 2，4(x + 2) 展开为 4x + 8。注意第一个括号内的负号：它同样要乘以 2，得到 −2，而不是 +2。
When a bracket is preceded by a minus sign, treat that as multiplying by −1. The expression 3(x − 4) − 2(x + 1) expands to 3x − 12 − 2x − 2. Every term inside the second bracket changes sign, which is the single most frequent source of lost marks at KS3.
当括号前是负号时，把它看作乘以 −1。式子 3(x − 4) − 2(x + 1) 展开为 3x − 12 − 2x − 2。第二个括号内的每一项都要变号，这是 KS3 阶段最常见的失分点。
4. Collecting Like Terms | 合并同类项
Like terms have exactly the same letter part. 6x and 4x are like terms; 6x and 6x² are not; 6x and 6 are not. After expanding, scan each side of the equation and combine the x terms together and the numbers together.
同类项是指字母部分完全相同的项。6x 与 4x 是同类项；6x 与 6x² 不是；6x 与 6 也不是。展开之后，请分别扫描等号两边，把 x 项合并在一起，把常数合并在一起。
| Expression | 表达式 | Like terms? | 是否同类项 | Combined form | 合并结果 |
| 6x + 4x | yes 是 | 10x |
| 6x − 2 + 3x | yes 是 | 9x − 2 |
| 5x + 5x² | no 不是 | 5x + 5x² |
| −3x + 7 − x | partly 部分 | −4x + 7 |
Remember that x on its own counts as 1x, and −x counts as −1x. Writing the invisible coefficient 1 on top of the term makes collecting far more reliable.
记住：单独的 x 就是 1x，单独的 −x 就是 −1x。把看不见的系数 1 标在项上，合并时可靠得多。
5. Variables on Both Sides | 未知数在两边
Once both sides are simplified, decide which side should keep the x terms. The standard strategy is to move the smaller x term to the other side so that the resulting coefficient stays positive — this avoids negative coefficients and reduces sign errors.
当两边都化简之后，先决定让哪一边保留 x 项。标准策略是把较小的 x 项移到另一边，使得到的系数保持为正——这样能避免负系数，减少变号错误。
x, so subtract 4x from both sides. Then move the constant −2 to the right by adding 2 to both sides. Finally divide both sides by the remaining coefficient 2.
对于 6x − 2 = 4x + 8，较小的 x 项是 4x，所以两边同时减去 4x。然后把常数 −2 通过两边同时加 2 移到右边。最后两边同时除以剩下的系数 2。
6x − 2 = 4x + 8
2x − 2 = 8
2x = 10
x = 5
A useful mental shortcut is the move-and-flip rule: a term crossing the equals sign changes its sign. But do not rely on it until you are fluent, because examiners award method marks for showing the same operation on both sides.
一个有用的心算口诀是移项变号：项跨过等号时符号改变。但在熟练之前不要依赖它，因为阅卷老师是按两边同做同一运算的步骤给方法分的。
6. Worked Example: 2(3x − 1) = 4(x + 2) | 例题完整解答
This is the model question in the style of page 124 question 2. Follow the five steps and write every line down, because each line carries method marks.
这是与第 124 页第 2 题同类型的示范题。请按五步流程解题，并把每一行都写出来，因为每一行都对应方法分。
Step 1 — expand both brackets. Step 2 — there are no further like terms on either side. Step 3 — subtract 4x from both sides. Step 4 — add 2 to both sides. Step 5 — divide both sides by 2.
第一步——展开两边的括号。第二步——两边各自已无同类项可合并。第三步——两边同时减去 4x。第四步——两边同时加 2。第五步——两边同时除以 2。
2(3x − 1) = 4(x + 2)
6x − 2 = 4x + 8
6x − 2 − 4x = 4x + 8 − 4x
2x − 2 = 8
2x = 10
x = 5
= 5 with extra brackets or a stray equals sign, which can make the final answer ambiguous.
答案是 x = 5。最后一行务必让未知数单独出现，不要多写括号或多余的等号，否则最终答案会变得含糊。
7. Checking Your Answer | 检验答案
Put x = 5 back into both sides of the original equation, working out each side separately.
代入检验是代数中最划算的保险，只需大约十五秒。把 x = 5 代回原方程的左右两边，分别计算。
Left: 2(3 × 5 − 1) = 2(14) = 28
Right: 4(5 + 2) = 4(7) = 28
Both sides give 28, so the solution is confirmed. Because the check uses the original equation rather than your rearranged lines, it also catches expansion errors that a second reading might miss.
两边都得 28，解得到验证。因为检验使用的是原方程而不是你变形后的式子，所以它还能抓出重读时容易忽略的展开错误。
If the two sides disagree, do not simply rewrite the answer. Go back to the expansion step first — roughly seven out of ten KS3 errors happen in those first two lines.
如果两边结果不一致，不要直接改答案。先回到展开那一步——KS3 阶段大约七成的错误都发生在前两行。
8. Fractional Coefficients | 分数系数
he addition first, then undo the multiplication. Dividing by a fraction is the same as multiplying by its reciprocal.
第 124 页的一些题目中，x 前面带有分数系数，例如 (2/3)x + 1 = 5。方法不变：先解除加法，再解除乘法。除以一个分数等于乘以它的倒数。
(2/3)x + 1 = 5
(2/3)x = 4
x = 4 × (3/2) = 6
ying (2/3)x + 1 = 5 by 3 gives 2x + 3 = 15, hence 2x = 12 and x = 6. The same answer appears with far fewer fractions to handle.
另一种往往更稳妥的做法是先乘以分母，在移项之前就把分数清掉。把 (2/3)x + 1 = 5 整体乘以 3，得 2x + 3 = 15，于是 2x = 12，x = 6。答案相同，但需要处理的分数少得多。
nown sits in the denominator, however, the equation is no longer linear and is outside the scope of this exercise.
当未知数位于分子、分母是数字时，取倒数这一步依然有效：对于 x/3 = 7，两边同时乘以 3，得 x = 21。但如果未知数出现在分母上，方程就不再是一次的，超出了本练习的范围。
9. Equations with Fractions on Both Sides | 两边都是分数的方程
. The cleanest method is cross-multiplication: multiply the numerator of each side by the denominator of the other side.
同一组练习中还会出现较难的变式：(x + 1)/2 = (2x − 3)/5。最干净的方法是用交叉相乘：把每一边的分子乘以另一边的分母。
5(x + 1) = 2(2x − 3)
5x + 5 = 4x − 6
x = −11
After cross-multiplying, the question is exactly the same type as question 2, so the five-step routine applies unchanged. The negative answer is perfectly acceptable and should be checked in the usual way.
交叉相乘之后，题目就变成了与第 2 题完全相同的类型，五步流程照常适用。答案是负数完全没有问题，按常规方法检验即可。
Left: (−11 + 1)/2 = −5
Right: (2 × −11 − 3)/5 = −25/5 = −5
st be expanded in full; writing 2 × 2x − 3 instead of 2(2x − 3) = 4x − 6 is a common trap.
两边都等于 −5，所以 x = −11 正确。请注意 2x − 3 外面的括号必须完整展开；把 2(2x − 3) 写成 2 × 2x − 3 而不等于 4x − 6，是常见的陷阱。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Mark schemes at KS3 are generous with method marks but unforgiving about sign slips. Knowing the usual suspects in advance is the fastest way to raise your accuracy.
KS3 的评分标准对方法分很宽容，但对符号错误毫不留情。提前熟悉这些常见嫌疑犯，是提高正确率最快的办法。
- Only multiplying the first term inside the bracket, e.g. writing 2(3x − 1) = 6x − 1. | 只乘括号内的第一项，例如把 2(3x − 1) 写成 6x − 1。
- Forgetting that a minus sign in front of a bracket flips every sign inside it. | 忘记括号前的负号会让括号内所有项的符号都变号。
- Dividing only one side of the equation by the coefficient. | 只把等式的一边除以系数。
- Moving a term across the equals sign without changing its sign. | 移项时没有变号。
- Treating x and x², or x and a constant, as like terms. | 把 x 与 x²，或 x 与常数，当成同类项。
- Skipping the substitution check and losing a mark to a careless slip. | 省略代入检验，因为粗心而丢分。
A simple habit defeats most of these: after every line, ask whether the equation is still balanced and whether every term on the left is still accounted for on the right.
一个简单的习惯就能击败其中大部分错误：每写完一行，就问自己等号是否仍然平衡，左边的每一项是否都还能在右边找到对应。
11. Exam-Style Practice | 考试风格练习
Work through the following items without looking at the answers. They cover the full range of question 2 on page 124: brackets, variables on both sides, fractional coefficients and a negative solution.
请在不看答案的情况下完成下面各题。它们覆盖了第 124 页第 2 题的全部考查范围：括号、两边含未知数、分数系数以及负数解。
- (a) 3(x − 4) − 2(x + 1) = 5 | 提示：展开时第二个括号全部变号
- (b) 5x + 3 = 2x + 18 | 提示：两边同时减去 2x
- (c) 4(2x + 1) = 3(x + 8) | 提示：先展开再合并
- (d) (3/4)x − 2 = 7 | 提示：两边同时乘以 4，或先加 2
- (e) (x + 3)/4 = (2x − 1)/3 | 提示：交叉相乘
ves x = 12. (e) 3(x + 3) = 4(2x − 1) gives 3x + 9 = 8x − 4, so 5x = 13 and x = 13/5 = 2.6.
答案：(a) 3x − 12 − 2x − 2 = 5，得 x = 19。(b) 3x = 15，得 x = 5。(c) 8x + 4 = 3x + 24，得 5x = 20，x = 4。(d) (3/4)x = 9，得 x = 12。(e) 3(x + 3) = 4(2x − 1)，得 3x + 9 = 8x − 4，即 5x = 13，x = 13/5 = 2.6。
Time yourself at roughly one minute per part. If a part takes much longer, the difficulty is almost always in the expansion line, not in the arithmetic at the end.
给自己计时，大约每小问一分钟。如果某一小问耗时明显更长，难点几乎总是出在展开那一行，而不是最后的算术。
12. Summary and Revision Tips | 小结与复习建议
Everything in Cambridge KS3 Maths page 124 exercise 2 rests on one routine: expand the brackets, simplify each side, gather the x terms on one side and the numbers on the other, divide by the coefficient, then substitute back to check. Brackets, unknowns on both sides and fractional coefficients are only variations on that routine.
剑桥 KS3 数学第 124 页练习 2 的全部内容都建立在同一套流程上：展开括号、分别化简两边、把 x 项集中到一边、常数集中到另一边、除以系数，最后代入检验。括号、两边含未知数、分数系数，都只是这套流程的不同变体。
For revision, keep three things in mind. First, always write the operation you perform on both sides, because method marks reward the process. Second, practise the bracket expansion with a negative sign in front until it becomes automatic, since that is where most errors begin. Third, never hand in an unchecked answer — a single substitution line is worth the fifteen seconds every time.
复习时请记住三点。第一，把两边所做的运算明确写出来，因为方法分奖励的是过程。第二，反复练习括号前带负号的展开，直到变成条件反射，因为大多数错误都从这里开始。第三，绝不要交上未经检验的答案——一行代入检验只花十五秒，永远划得来。
Once these habits are in place, page 124 stops being a challenge and becomes a reliable source of easy marks in the algebra section of the KS3 assessment.
一旦养成这些习惯，第 124 页就不再是难题，而会成为 KS3 测评代数部分中稳定拿分的来源。
Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图