📚 Cambridge KS3 Maths: Solving Linear Equations (p141 Q2) | 剑桥 KS3 数学：解一元一次方程（第141页第2题）
Welcome to this Cambridge KS3 Maths revision guide. We will work through a key algebra topic: solving linear equations in one variable. This includes one-step and two-step equations, equations with brackets, equations with fractions, and equations with unknowns on both sides. The methods shown here are directly relevant to Cambridge Checkpoint questions, including the exercise found on page 141 question 2. Every technique is explained in a step-by-step way, with checks and common errors highlighted, so that you can apply the same structure to any linear equation you meet in a test.
欢迎阅读本篇剑桥 KS3 数学复习指南。我们将系统学习一个核心代数主题：解一元一次方程。内容包括一步和两步方程、含括号的方程、含分数的方程以及未知数在等式两边的方程。这里展示的方法与 Cambridge Checkpoint 试题直接相关，包括第141页第2题的练习。每一种方法都会按步骤讲解，并附上检验方法和常见错误提醒，让你在考试中遇到任何一元一次方程都能套用同样的思路。
1. What Is a Linear Equation? | 什么是一元一次方程？
re are no squares, cubes, roots or reciprocal powers of the unknown. For example, 3x + 5 = 20 and 4(x - 2) = 16 are linear equations. The word 'linear' comes from the fact that if you plotted y = 3x + 5, the graph would be a straight line.
一元一次方程是一个代数等式，表示两个表达式相等，而且未知数只以一次方的形式出现，不含未知数的平方、立方、根号或负指数。例如 3x + 5 = 20 和 4(x - 2) = 16 都是一元一次方程。单词 'linear'（线性）源于这样一个事实：如果画出 y = 3x + 5 的图像，你会得到一条直线。
g 5 gives 3 × 5 + 5 = 20, which is true. A linear equation can have exactly one solution, no solution, or infinitely many solutions, but at KS3 we mostly meet equations with exactly one solution.
解方程的意思是找出能让等式两边相等的未知数的值。在这个例子中，x = 5 是解，因为代入 5 得到 3 × 5 + 5 = 20，等式成立。线性方程可能有一个解、无解或无穷多个解，但在 KS3 阶段我们最常遇到的是恰好有一个解的方程。
3x + 5 = 20 → 3x = 15 → x = 5
2. The Golden Rule: Keep Both Sides Balanced | 黄金法则：保持两边平衡
An equation is like a balanced scale. Whatever you do to the left-hand side, you must also do to the right-hand side. If you add 4 to one side, you add 4 to the other. If you divide one side by 3, you divide the other side by 3 as well. This single rule is the reason every solving method works.
方程就像一台平衡的天平。你对左边做了什么，就必须对右边做同样的事。左边加 4，右边也要加 4；左边除以 3，右边也要除以 3。正是这条简单的规则，让所有解方程的方法都成立。
The operations we use are called inverse operations, because they undo each other. Addition is undone by subtraction, multiplication is undone by division. To solve an equation, we undo the operations around the unknown one at a time, working in reverse order, until the unknown stands alone.
我们使用的运算叫做逆运算，因为它们能互相抵消。加法用减法抵消，乘法用除法抵消。解方程时，我们围绕未知数逐个抵消运算，按照相反的顺序进行，直到未知数单独留在等号一边。
If a = b, then a ± c = b ± c and a ÷ c = b ÷ c (c ≠ 0)
3. One-Step Equations | 一步方程
A one-step equation needs only one inverse operation to solve. If the unknown is added to a number, subtract that number from both sides. If the unknown is subtracted, add. If the unknown is multiplied, divide. If the unknown is divided, multiply.
一步方程只需要一个逆运算就能解出。如果未知数被加上某个数，就在两边同时减去这个数；如果未知数被减去，就加上；如果未知数被乘以某个数，就除以这个数；如果未知数被除以某个数，就乘以这个数。
both sides by 4 to get x = 8. To solve x ÷ 5 = 6, multiply both sides by 5 to get x = 30.
例如，解 x + 9 = 15，两边同时减 9，得到 x = 6。解 x - 7 = 4，两边同时加 7，得到 x = 11。解 4x = 32，两边同时除以 4，得到 x = 8。解 x ÷ 5 = 6，两边同时乘以 5，得到 x = 30。
Notice that we always write the operation on both sides on the same line, so the examiner can see your method. A correct answer with no working often loses method marks in Checkpoint-style questions.
注意，我们总是把同一个运算同时写在等号两边，让阅卷老师看到你的方法。在 Checkpoint 风格的题目中，只写答案而不写过程往往会丢掉方法分。
x + 9 = 15 → x = 15 - 9 → x = 6
4. Two-Step Equations | 两步方程
A two-step equation has two operations acting on the unknown. The standard order is to deal with addition or subtraction first, then multiplication or division. Think of the order of operations in reverse: if the unknown was multiplied and then added to, we undo the addition first.
两步方程中有两个运算作用于未知数。标准顺序是先处理加或减，再处理乘或除。可以把它想成运算顺序的逆过程：如果未知数先被乘、再被加，我们就先抵消那个加法。
en divide both sides by 2: x = 6. Check by substituting: 2 × 6 + 7 = 19, which is correct.
以 2x + 7 = 19 为例。两边同时减 7，得到 2x = 12；再两边同时除以 2，得到 x = 6。代回检验：2 × 6 + 7 = 19，成立。
3 = 6. Multiply both sides by 3: x = 18. Check: 18 ÷ 3 - 4 = 6 - 4 = 2, which is correct.
再看 x ÷ 3 - 4 = 2。两边同时加 4，得到 x ÷ 3 = 6；再两边同时乘以 3，得到 x = 18。检验：18 ÷ 3 - 4 = 6 - 4 = 2，成立。
2x + 7 = 19 → 2x = 12 → x = 6
5. Equations with Brackets | 含括号的方程
When an equation contains brackets, you have two valid routes. The first route is to expand the brackets first, then solve the resulting two-step or three-step equation. The second route is to divide both sides by the number in front of the bracket immediately, which is usually faster when the bracket is a simple factor.
当方程中含有括号时，你有两条可行的路线。第一条是先展开括号，再解得到的两步或三步方程；第二条是立刻把两边同时除以括号前面的数，当括号外是一个简单的因数时，这条路线通常更快。
, then add 2 to get x = 6. Both routes give the same answer, and both are acceptable in a Cambridge KS3 examination.
看 4(x - 2) = 16。第一条路线：展开得到 4x - 8 = 16，再加 8 得到 4x = 24，最后除以 4 得到 x = 6。第二条路线：两边同时除以 4 得到 x - 2 = 4，再加 2 得到 x = 6。两条路线答案相同，在剑桥 KS3 考试中都是可接受的写法。
you expand, remember to multiply every term inside the bracket, including the second one.
第二个例子是 3(2x + 1) = 27。两边除以 3 得到 2x + 1 = 9，于是 2x = 8，x = 4。若选择展开，则得到 6x + 3 = 27，于是 6x = 24，x = 4。展开时一定要把括号内的每一项都乘到，包括第二项。
4(x - 2) = 16 → 4x - 8 = 16 → 4x = 24 → x = 6
6. Equations with Fractions | 含分数的方程
e, (x + 1) ÷ 3 = 4 becomes x + 1 = 12, so x = 11. Check: (11 + 1) ÷ 3 = 4, which is correct.
含分数的方程看起来比实际难。如果未知数只出现在一个分数里，就把两边同时乘以分母。例如 (x + 1) ÷ 3 = 4 变为 x + 1 = 12，所以 x = 11。检验：(11 + 1) ÷ 3 = 4，成立。
Consider x ÷ 2 + x ÷ 3 = 5. The lowest common multiple of 2 and 3 is 6, so multiply every term by 6: 3x + 2x = 30, giving 5x = 30 and x = 6.
如果两边都有分数，或者有两个不同的分母，就把每一项都乘以分母的最小公倍数。看 x ÷ 2 + x ÷ 3 = 5。2 和 3 的最小公倍数是 6，所以每项乘以 6：3x + 2x = 30，得到 5x = 30，x = 6。
ion: multiply the numerator on the left by the denominator on the right, and set it equal to the numerator on the right multiplied by the denominator on the left. This is simply multiplying both sides by the product of the two denominators.
对于形如 a ÷ b = c ÷ d 的方程，常用技巧是交叉相乘：把左边的分子乘以右边的分母，令其等于右边的分子乘以左边的分母。这本质上就是把两边同时乘以两个分母的乘积。
x ÷ 2 + x ÷ 3 = 5 → 3x + 2x = 30 → 5x = 30 → x = 6
7. Unknowns on Both Sides | 未知数在等式两边
When the unknown appears on both sides, first collect all the unknown terms on one side and all the number terms on the other. It is conventional to move the smaller unknown term, so that the coefficient of the unknown stays positive and the arithmetic is easier.
当未知数出现在等式两边时，先把所有含未知数的项集中到一边，把所有常数项集中到另一边。习惯上我们移动较小的未知数项，这样未知数的系数保持为正，运算更简单。
equation: 5 × 4 - 3 = 17 and 2 × 4 + 9 = 17. Both sides match, so the solution is correct.
解 5x - 3 = 2x + 9。两边同时减 2x，得到 3x - 3 = 9；两边同时加 3，得到 3x = 12；除以 3，得到 x = 4。代回原方程检验：5 × 4 - 3 = 17，2 × 4 + 9 = 17，两边相等，所以解是正确的。
x. Divide by 3: x = 5. Check: 2(5 + 3) = 16 and 5 × 5 - 9 = 16, which confirms the answer.
再看一个既有括号又有两边未知数的例子：2(x + 3) = 5x - 9。展开得 2x + 6 = 5x - 9；减 2x 得 6 = 3x - 9；加 9 得 15 = 3x；除以 3 得 x = 5。检验：2(5 + 3) = 16，5 × 5 - 9 = 16，答案正确。
5x - 3 = 2x + 9 → 3x = 12 → x = 4
8. Checking Answers and Counting Solutions | 检验答案与判断解的个数
Checking is not optional; it is the fastest way to catch a sign error. Substitute your answer back into the original equation and evaluate both sides separately. If the two values are equal, your solution is right. If not, the error is usually in the first line of your working.
检验不是可做可不做的步骤，它是发现符号错误最快的方法。把答案代回原方程，分别计算两边的值。如果两个值相等，解就是正确的；如果不相等，错误通常出在解题的第一行。
. If the unknown terms cancel and you are left with a true statement such as 6 = 6, the equation is an identity and has infinitely many solutions.
不是每个线性方程都有解。如果未知数项被消掉，剩下像 3 = 7 这样的假命题，那么这个方程无解。如果未知数项被消掉，剩下像 6 = 6 这样的真命题，那么这个方程是恒等式，有无数多个解。
g gives 3x + 6 = 3x + 6, which is always true, so every value of x satisfies the equation.
对于 2x + 3 = 2x + 7，两边同时减 2x 后剩下 3 = 7，这是假命题，所以方程无解。对于 3(x + 2) = 3x + 6，展开得 3x + 6 = 3x + 6，永远成立，所以任何 x 值都满足该方程。
| Case | 情形 | After simplifying | 化简后 | Number of solutions | 解的个数 |
|---|---|---|
| x = a number | x = 5 | Exactly one | 恰好一个 |
| False statement | 3 = 7 | None | 无解 |
| True statement | 6 = 6 | Infinitely many | 无穷多个 |
9. Common Mistakes at KS3 | KS3 阶段的常见错误
The most frequent error is a sign slip when expanding a negative bracket. For instance, -2(x - 5) becomes -2x + 10, not -2x - 10, because a negative multiplied by a negative gives a positive. Write the multiplication out slowly in the margin if you are unsure.
最常见的错误是展开带负号的括号时出现符号失误。例如 -2(x - 5) 应展开为 -2x + 10，而不是 -2x - 10，因为负数乘以负数得正数。如果不确定，就在旁边把乘法慢慢写出来。
est habit is to subtract or add first, and only divide when the unknown term stands alone.
第二个常见错误是只除了某一边的一部分。在 2x + 7 = 19 中，若两边除以 2，必须写成 x + 3.5 = 9.5，而不是 x + 7 = 9.5。最稳妥的习惯是先做完加减，等到未知数项单独存在时再除。
estion incorrectly. Always re-read the equation from the paper before you start, and write your final answer as x = ... on its own line.
其他常见失误包括移项时忘记变号、乘以公分母时漏乘某一项，以及抄错题目。动笔前先把题目重新读一遍，最后把答案单独写一行，形式为 x = ...。
10. Worked Example in the Style of p141 Q2 | 第141页第2题同类题的解题示范
The exercise at page 141 question 2 asks you to solve a set of linear equations of mixed type. The worked example below follows exactly the same structure: brackets, unknowns on both sides, a check, and a clear final answer. Use the numbered steps as a template for your own working.
第141页第2题要求你解一组混合类型的一元一次方程。下面的解题示范采用了完全相同的结构：含括号、未知数在两边、检验，以及清晰的最终答案。你可以把这些编号步骤当作自己写过程的模板。
giving 2x = 14. Step 4: divide both sides by 2, giving x = 7. Step 5: check by substitution: 5(7 - 2) = 25 and 3 × 7 + 4 = 25, so the solution is correct.
解 5(x - 2) = 3x + 4。第 1 步：展开括号，得到 5x - 10 = 3x + 4。第 2 步：两边同时减 3x，得到 2x - 10 = 4。第 3 步：两边同时加 10，得到 2x = 14。第 4 步：两边同时除以 2，得到 x = 7。第 5 步：代入检验，5(7 - 2) = 25，3 × 7 + 4 = 25，两边相等，解正确。
) = 3(x + 4). Expand: 4x + 2 = 3x + 12. Subtract 3x: x + 2 = 12. Subtract 2: x = 10. Check
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