📚 Solving Linear Equations: Cambridge KS3 p54_2 | 解一元一次方程：剑桥 KS3 第54页练习2
In this revision guide, written in the style of Cambridge KS3 page 54 exercise 2, you will learn how to solve one-step, two-step, bracket and variable-on-both-sides equations using the balance principle and inverse operations. The unknown, usually written as x, appears only to the power of 1, so every equation can be solved step by step and checked at the end.
在这篇剑桥 KS3 第54页练习2 风格的复习指南中，你将学习如何利用等式平衡原理和逆运算来解一步方程、两步方程、含括号的方程以及两边都含未知数的方程。未知数通常写作 x，只出现一次方，因此每一道方程都可以按步骤逐步求解，并在最后进行检验。
1. What Is a Linear Equation? | 什么是一元一次方程？
A linear equation is an equation in which the unknown, usually written as x, has degree 1. This means you will only see terms such as x, 2x, −3x or x/4, and never x², x³ or 1/x.
一元一次方程是指未知数（通常写作 x）的次数为 1 的方程。这意味着你只会看到 x、2x、−3x 或 x/4 这样的项，而不会出现 x²、x³ 或 1/x。
. The solution is the value of x that makes the equation true when it is substituted back into the original equation.
它的一般形式是 ax + b = c，其中 a ≠ 0。方程的解就是代入原方程后能使方程成立的 x 值。
ax + b = c, a ≠ 0
Use this checklist to decide whether an equation is linear.
你可以用下面的清单判断一个方程是否为一元一次方程。
- 3x + 1 = 10 is linear, because x has degree 1. | 3x + 1 = 10 是一元一次方程，因为 x 的次数为 1。
- x/5 − 2 = 0 is linear, because x/5 means (1/5)x. | x/5 − 2 = 0 是一元一次方程，因为 x/5 就是 (1/5)x。
- x² + 1 = 5 is not linear, because of the x² term. | x² + 1 = 5 不是一元一次方程，因为有 x² 项。
- 1/x = 4 is not linear, because the unknown is in the denominator. | 1/x = 4 不是一元一次方程，因为未知数出现在分母上。
2. Balance Principle and Inverse Operations | 等式平衡原理与逆运算
An equation is balanced like a set of scales. If you add, subtract, multiply or divide one side, you must do exactly the same to the other side, otherwise the equation becomes false.
方程像一架平衡的天平。如果你对一边进行加、减、乘、除，就必须对另一边做完全相同的运算，否则方程就不再成立。
The key is to use inverse operations. Addition and subtraction are inverse operations; multiplication and division are inverse operations. Your aim is to isolate x step by step on one side of the equals sign.
关键是使用逆运算。加法与减法互为逆运算；乘法与除法互为逆运算。你的目标是逐步把 x 单独留在等号的一边。
The table below shows the common inverse pairs you will use when solving equations.
下表展示了你在解方程时常用的逆运算配对。
| Operation 运算 | Inverse Operation 逆运算 |
|---|---|
| + 5 | − 5 |
| − 7 | + 7 |
| × 3 | ÷ 3 |
| ÷ 4 | × 4 |
Keep the equals signs lined up in a column as you work downwards. This makes your method clear to the examiner and reduces careless slips.
解题时把等号上下对齐排成一列。这样既能让阅卷老师看清你的方法，也能减少粗心错误。
3. One-Step Equations | 一步方程
A one-step equation needs only a single inverse operation to isolate x. Decide what has been done to x, then undo it.
一步方程只需要一次逆运算就能把 x 单独留下来。先判断 x 被做了什么运算，再把它反做回去。
For addition and subtraction, undo with the opposite operation.
对于加法和减法，用相反的运算抵消。
x + 9 = 15 → x = 15 − 9 = 6
x − 8 = 4 → x = 4 + 8 = 12
For multiplication and division, undo with the opposite operation as well.
对于乘法和除法，同样用相反的运算抵消。
4x = 28 → x = 28 ÷ 4 = 7
x/5 = 3 → x = 3 × 5 = 15
ltiply both sides by −1 to get x = −7. In −3x = 12, divide both sides by −3 to get x = −4.
注意 x 前面的负号。在 −x = 7 中，两边同乘 −1 得到 x = −7。在 −3x = 12 中，两边同除以 −3 得到 x = −4。
4. Two-Step Equations | 两步方程
A two-step equation has both a multiplication (or division) and an addition (or subtraction) acting on x. Undo the addition or subtraction first, then undo the multiplication or division.
两步方程中，x 同时受到乘法（或除法）和加法（或减法）的作用。先抵消加法或减法，再抵消乘法或除法。
Example 1: solve 3x + 5 = 20.
例 1：解 3x + 5 = 20。
3x + 5 = 20 → 3x = 15 → x = 5
Example 2: solve 2x − 7 = 11.
例 2：解 2x − 7 = 11。
2x − 7 = 11 → 2x = 18 → x = 9
Example 3: solve x/4 + 1 = 6.
例 3：解 x/4 + 1 = 6。
x/4 + 1 = 6 → x/4 = 5 → x = 20
Remember the order: work backwards through the operations, from the outside inwards, until x stands alone.
记住顺序：像剥洋葱一样从外向内逆向操作，直到 x 单独留在一边。
5. Equations with Brackets | 含括号的方程
When x appears inside a bracket, you have two valid methods. Method A divides both sides by the number in front of the bracket first, which is usually faster. Method B expands the bracket first, which always works.
当 x 出现在括号里时，有两种都正确的方法。方法 A 先把两边同除以括号前的数，通常更快；方法 B 先展开括号，虽然步骤多但总是可行。
Method A example: solve 3(x + 4) = 27.
方法 A 例子：解 3(x + 4) = 27。
3(x + 4) = 27 → x + 4 = 9 → x = 5
Method B example on the same equation: expand first, then solve.
方法 B 用同一个方程演示：先展开，再求解。
3x + 12 = 27 → 3x = 15 → x = 5
Brackets with a negative result: solve 2(x − 3) = −8.
括号内结果为负的情况：解 2(x − 3) = −8。
2(x − 3) = −8 → x − 3 = −4 → x = −1
number, take extra care with signs, for example −2(x + 1) = 6 gives x + 1 = −3, so x = −4.
如果括号前乘的是负数，要格外注意符号。例如 −2(x + 1) = 6 得到 x + 1 = −3，所以 x = −4。
6. Variables on Both Sides | 两边都含未知数的方程
If x appears on both sides, first collect all the x terms on one side and all the numbers on the other side. A useful trick is to subtract the smaller x term so that the remaining x term is positive.
如果等号两边都出现 x，先把所有含 x 的项集中到一边，把所有数字集中到另一边。一个实用技巧是减去较小的 x 项，使剩下的 x 项为正。
Example: solve 5x − 3 = 2x + 9.
例：解 5x − 3 = 2x + 9。
5x − 3 = 2x + 9 → 3x − 3 = 9 → 3x = 12 → x = 4
Example with brackets on both sides: solve 2(x + 5) = 4x − 2.
两边都含括号的例子：解 2(x + 5) = 4x − 2。
2x + 10 = 4x − 2 → 10 = 2x − 2 → 12 = 2x → x = 6
Check each step: subtracting 2x from both sides is allowed, and so is adding 2 to both sides. The balance principle never changes.
每一步都要检查：两边同减 2x 是允许的，两边同加 2 也是允许的。平衡原理始终不变。
7. Equations with Fractions | 含分数的方程
Fractions look harder than they are. If the equation contains a single fraction, multiply both sides by the denominator. If it contains several fractions, multiply every term by the lowest common multiple of the denominators.
分数看起来比实际难。如果方程只含一个分数，两边同乘分母；如果含多个分数，就把每一项都乘以各分母的最小公倍数。
Single fraction containing x: solve (x + 1)/2 = 3.
分子含 x 的单个分数：解 (x + 1)/2 = 3。
(x + 1)/2 = 3 → x + 1 = 6 → x = 5
Fraction with a coefficient: solve (2/3)x = 8.
未知数带分数系数：解 (2/3)x = 8。
(2/3)x = 8 → x = 8 × 3 ÷ 2 = 12
Two fractions: solve x/2 + x/3 = 5.
两个分数的情况：解 x/2 + x/3 = 5。
x/2 + x/3 = 5 → 3x + 2x = 30 → 5x = 30 → x = 6
The most common error here is forgetting to multiply every single term, including terms that do not contain a fraction, by the common denominator.
这里最常见的错误是忘记把每一个项都乘以公分母，包括那些本来不含分数的项。
8. Checking Your Solution | 检验你的解
Once you have a value for x, substitute it back into the original equation and evaluate both sides separately. If the left-hand side equals the right-hand side, your solution is correct.
求出 x 的值后，把它代回原方程，分别计算等号两边。如果左边等于右边，你的解就是正确的。
Check 3x + 5 = 20 with x = 5.
用 x = 5 检验 3x + 5 = 20。
LHS = 3 × 5 + 5 = 20, RHS = 20, LHS = RHS
Check 5x − 3 = 2x + 9 with x = 4.
用 x = 4 检验 5x − 3 = 2x + 9。
LHS = 5 × 4 − 3 = 17, RHS = 2 × 4 + 9 = 17, LHS = RHS
Checking takes about ten seconds and it turns a careless mistake into a mark you keep. In KS3 tests and later GCSE papers, showing the check can also earn credit for accuracy.
检验只需大约十秒钟，却能把一个粗心错误变成保住的一分。在 KS3 测验以及后来的 GCSE 考试中，写出检验过程还可能获得准确性方面的分数。
9. Turning Word Problems into Equations | 把应用题转化为方程
Most word problems follow the same pattern: choose a letter for the unknown, translate each sentence into algebra, solve the equation, then answer the question in words.
大多数应用题都遵循同样的套路：用字母表示未知数，把每句话翻译成代数式，解方程，最后用文字回答问题。
Example 1: a number is multiplied by 4 and then increased by 7; the result is 31. Find the number.
例 1：某个数乘以 4 再加上 7，结果是 31。求这个数。
4x + 7 = 31 → 4x = 24 → x = 6
Example 2: the length of a rectangle is 3 cm more than its width, and the perimeter is 26 cm. Find the width.
例 2：一个长方形的长比宽多 3 cm，周长是 26 cm。求宽。
2(w + w + 3) = 26 → 4w + 6 = 26 → w = 5 cm
Example 3: the sum of three consecutive integers is 42. Find the smallest one.
例 3：三个连续整数之和是 42。求最小的那个数。
n + (n + 1) + (n + 2) = 42 → 3n + 3 = 42 → n = 13
Always define your letter in writing at the start, for example let x be the number, and always give units in your final answer.
一定要在开头用文字说明字母的含义，例如"设 x 为这个数"，并在最终答案中写上单位。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Careless slips lose more marks than genuine misunderstanding. The list below collects the errors that appear most often in KS3 exercise books.
粗心失误丢的分往往比真正不会丢的还多。下面的清单汇总了 KS3 作业本中最常出现的错误。
- Doing an operation to one side only. | 只对等式的一边进行运算。
- Forgetting that × 2 undoes ÷ 2, and that + 3 undoes − 3. | 忘记 × 2 抵消 ÷ 2，+ 3 抵消 − 3。
- Moving a term across the equals sign without changing its sign. | 把项移到等号另一边时忘记变号。
- Multiplying only the fractional term by the denominator. | 只把分数项乘以分母，漏掉其他项。
- Solving −x = 7 as x = 7 instead of x = −7. | 把 −x = 7 解成 x = 7，而正确结果是 x = −7。
- Forgetting to expand a bracket multiplied by a negative number correctly. | 括号前是负数时展开出错。
In an exam, write one line per step, keep the equals signs aligned, and finish with the check. If time allows, re-read the question to make sure you answered what was asked rather than only finding x.
考试时每步写一行，等号上下对齐，最后做检验。如果时间允许，再读一遍题目，确保你回答的是题目所问，而不只是求出 x。
11. Quick Summary and Practice | 快速总结与练习
The whole topic rests on three ideas: keep the equation balanced, undo operations in reverse order, and check the answer by substitution.
整个主题建立在三个要点上：保持方程平衡，按相反顺序抵消运算，以及用代入法检验答案。
ax + b = c → ax = c − b → x = (c − b) ÷ a, a ≠ 0
Try these five questions, then check your answers against the list below.
试着做下面五道题，然后与答案清单核对。
- x + 12 = 20 | 答案：x = 8
- 5x − 4 = 21 | 答案：x = 5
- 4(x + 2) = 36 | 答案：x = 7
- 7x + 1 = 3x + 17 | 答案：x = 4
- x/3 − 1 = 4 | 答案：x = 15
If every answer matched, you have mastered the balance method. If not, revisit section 2 and section 4, since almost every error in linear equations traces back to those two ideas.
如果全部答对，说明你已经掌握了平衡法。如果有错，请回到第 2 节和第 4 节复习，因为一元一次方程中的绝大多数错误都源于这两个要点。
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