📚 Cambridge KS3 Maths: Linear Equations Practice (p259_2) | 剑桥 KS3 数学：一元一次方程练习（p259_2）
quations are one of the most important algebraic skills at Key Stage 3. We will cover the balancing method, two-step equations, brackets, fractions, variables on both sides, checking your answer, word problems and the most common errors that cost marks in tests. Work through each section slowly and attempt the practice questions yourself before reading the solutions.
欢迎阅读这篇基于练习卷 p259_2 的剑桥 KS3 数学复习文章。重点是一元一次方程，这是 Key Stage 3 代数中最重要的技能之一。我们将学习天平法、两步方程、去括号、分数系数、未知数在两边、检验解、文字题以及考试中最常见的失分错误。请慢慢完成每个部分，并自己尝试练习题后再看解答。
1. Understanding Linear Equations | 理解一元一次方程
A linear equation is an equation in which the unknown, often written as x, appears only to the power of 1. There are no squares, cubes, square roots or unknowns in denominators. When you plot the solutions of a linear equation on a graph, they form a straight line - which is exactly why we call it linear. At KS3, most equations contain one unknown and have exactly one solution.
一元一次方程是未知数（通常写作 x）只出现一次幂的方程。式中没有平方、立方、根号，未知数也不出现在分母上。把线性方程的对应点画在图像上，会形成一条直线——这正是它被称为 linear（线性）的原因。在 KS3 阶段，大多数方程只含一个未知数，并且恰好有一个解。
Typical examples include the equations below. The goal is always to find the value of the unknown that makes the equation true.
常见的例子包括下面这些方程。目标始终是求出使方程成立的未知数的值。
= 12 and 3y − 2 = 16 and 4(n + 1) = 28
The left-hand side (LHS) and right-hand side (RHS) are separated by the equals sign.
等号左边称为左边（LHS），右边称为右边（RHS）。
A solution makes the LHS equal to the RHS. Any other value does not.
代入后能使左边等于右边的值才是解，其他任何值都不行。
Letters other than x are allowed: y, n, a, t and so on all behave in the same way.
未知数不一定是 x：y、n、a、t 等字母的运算规则完全相同。
| Term 术语 | Meaning 含义 |
| Unknown 未知数 | The letter whose value we must find 需要求值的字母 |
| Coefficient 系数 | The number multiplying the unknown, e.g. 3 in 3y 乘在未知数前面的数，如 3y 中的 3 |
| Constant 常数项 | A number on its own, e.g. −2 in 3y − 2 单独的数值，如 3y − 2 中的 −2 |
| Solution 解 | The value that satisfies the equation 使方程成立的值 |
2. Balancing Method | 天平法
Think of an equation as a balance scale. Whatever you do to one side, you must do to the other side. This keeps the equation true. This single idea is the foundation of all equation solving at KS3 and beyond.
把方程想像成一座天平。你对一边做的任何操作，也必须对另一边做，这样方程才能保持成立。这一个思想就是 KS3 乃至更高级阶段解方程的全部基础。
To solve x + 7 = 15, subtract 7 from both sides: x + 7 − 7 = 15 − 7, so x = 8.
要解 x + 7 = 15，两边同时减去 7：x + 7 − 7 = 15 − 7，因此 x = 8。
x + 7 = 15 → x = 15 − 7 → x = 8
n by division, and vice versa. To solve x ÷ 4 = 6, multiply both sides by 4 to get x = 24.
总是使用逆运算：加法由减法抵消，乘法由除法抵消，反之亦然。要解 x ÷ 4 = 6，两边同时乘以 4，得到 x = 24。
Write one operation per line so your working can be marked.
每行只写一步运算，这样过程才能被评分。
Keep the equals signs lined up vertically - it makes errors obvious.
把等号竖直对齐——一旦写错会很容易看出来。
Never move a term without changing its sign.
移项时一定要改变符号。
3. Solving Two-Step Equations | 解两步方程
. For example, in 2x + 3 = 11, the unknown x is first multiplied by 2 and then 3 is added.
两步方程中有两个运算作用在未知数上。例如，在 2x + 3 = 11 中，未知数 x 先乘以 2，然后加 3。
Undo the addition or subtraction first, then undo the multiplication or division. This order is the reverse of the order of operations, sometimes called BIDMAS or BODMAS.
先消去加法或减法，再消去乘法或除法。这个顺序与运算顺序（有时称为 BIDMAS 或 BODMAS）正好相反。
2x + 3 = 11 → 2x = 8 → x = 4
soning written out in full: 2x + 3 − 3 = 11 − 3, so 2x = 8; then 2x ÷ 2 = 8 ÷ 2, so x = 4.
完整的推理过程如下：2x + 3 − 3 = 11 − 3，得 2x = 8；再 2x ÷ 2 = 8 ÷ 2，得 x = 4。
3y − 2 = 16 → 3y = 18 → y = 6
3y − 2 = 16 → 3y = 18 → y = 6
5t + 4 = 39 → 5t = 35 → t = 7
5t + 4 = 39 → 5t = 35 → t = 7
x ÷ 3 + 2 = 9 → x ÷ 3 = 7 → x = 21
x ÷ 3 + 2 = 9 → x ÷ 3 = 7 → x = 21
4. Equations with Brackets | 带括号的方程
When an equation contains brackets, expand them first using the distributive law. Multiply the term outside the bracket by each term inside the bracket, paying careful attention to negative signs.
当方程中含有括号时，先用分配律展开。用括号外的项分别乘以括号内的每一项，并特别注意负号。
a(b + c) = ab + ac and a(b − c) = ab − ac
ce paper: expand to 4n + 4 = 28, subtract 4 to get 4n = 24, then divide by 4 to get n = 6.
解练习卷中的 4(n + 1) = 28：展开得 4n + 4 = 28，两边减 4 得 4n = 24，再除以 4 得 n = 6。
4(n + 1) = 28 → 4n + 4 = 28 → 4n = 24 → n = 6
= 7. A very common mistake is to write 3x − 2 = 15, forgetting to multiply the −2 as well.
要小心括号前的负号：3(x − 2) = 15 展开为 3x − 6 = 15，再得 3x = 21，所以 x = 7。最常见的错误是写成 3x − 2 = 15，忘了 −2 也要乘以 3。
Alternative for 4(n + 1) = 28: divide both sides by 4 first to get n + 1 = 7, then n = 6.
4(n + 1) = 28 的另一种解法：两边先除以 4 得 n + 1 = 7，再得 n = 6。
Both methods are valid; choose the one that produces fewer steps.
两种方法都对，选择步骤更少的那一种。
If brackets appear on both sides, expand both sides before collecting terms.
如果两边都有括号，先把两边都展开，再合并同类项。
5. Equations with Fractions | 分数系数方程
Fractions make equations look harder than they are. The trick is to multiply every term by the denominator, which clears the fraction and leaves a linear equation you already know how to solve.
分数会让方程看起来比实际难。技巧是把每一项都乘以分母，这样就能消去分数，留下一个你已经会解的线性方程。
e x ÷ 3 + 2 = 7: subtract 2 to get x ÷ 3 = 5, then multiply both sides by 3 to get x = 15.
解 x ÷ 3 + 2 = 7：先减 2 得 x ÷ 3 = 5，再两边乘以 3 得 x = 15。
x ÷ 3 + 2 = 7 → x ÷ 3 = 5 → x = 15
erent denominators, multiply every term by the lowest common multiple of the denominators.
当分子含有多项时，先乘再展开：(x + 1) ÷ 4 = 3 变为 x + 1 = 12，所以 x = 11。当出现两个不同分母时，把每一项都乘以分母的最小公倍数。
(2x) ÷ 5 = 4 → 2x = 20 → x = 10
(2x) ÷ 5 = 4 → 2x = 20 → x = 10
x ÷ 2 + x ÷ 3 = 5 → multiply by 6 → 3x + 2x = 30 → 5x = 30 → x = 6
x ÷ 2 + x ÷ 3 = 5 → 乘以 6 → 3x + 2x = 30 → 5x = 30 → x = 6
Always write the multiplier on both sides of the equation, not just next to the fraction.
乘数要写在等号两边，而不是只写在分数旁边。
6. Unknowns on Both Sides | 未知数在等号两边
strategy is to collect all the unknown terms on one side and all the constant terms on the other side.
有些方程两边都含未知数，例如 5x − 3 = 2x + 9。策略是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。
rom both sides: 3x − 3 = 9. Then add 3 to both sides: 3x = 12. Finally divide by 3: x = 4.
两边同时减去 2x：3x − 3 = 9。再两边同时加 3：3x = 12。最后除以 3：x = 4。
5x − 3 = 2x + 9 → 3x − 3 = 9 → 3x = 12 → x = 4
= −3x + 9, which still works but produces negative coefficients and more chances to slip.
你也可以减去较小的那个未知数项，而不是较大的。两边同时减 5x 会得到 −3 = −3x + 9，结果虽然正确，但会产生负系数，出错的机会更多。
7a + 2 = 4a + 17 → 3a = 15 → a = 5
7a + 2 = 4a + 17 → 3a = 15 → a = 5
8 − x = 3x − 4 → 12 = 4x → x = 3
8 − x = 3x − 4 → 12 = 4x → x = 3
If the coefficient ends up negative, multiply the whole equation by −1 at the end.
如果最后的系数是负数，可以在最后把整个方程乘以 −1。
7. Checking Your Solution | 检验你的解
Checking is not optional - it is a mark-winning habit. Substitute your answer back into the original equation and confirm that the left-hand side equals the right-hand side. If it does not, you have made an arithmetic or sign error somewhere.
检验不是可做可不做的步骤，而是能拿分的习惯。把答案代回原方程，确认左边等于右边。如果不相等，说明某处出现了计算错误或符号错误。
= 11: LHS = 2 × 4 + 3 = 8 + 3 = 11, and RHS = 11. Since 11 = 11, the solution is correct.
把 x = 4 代入 2x + 3 = 11 检验：左边 = 2 × 4 + 3 = 8 + 3 = 11，右边 = 11。因为 11 = 11，所以解正确。
| Equation 方程 | Solution 解 | Check 检验 |
| x + 7 = 15 | x = 8 | 8 + 7 = 15 ✓ |
| 3y − 2 = 16 | y = 6 | 3 × 6 − 2 = 16 ✓ |
| 4(n + 1) = 28 | n = 6 | 4 × 7 = 28 ✓ |
| 5x − 3 = 2x + 9 | x = 4 | 17 = 17 ✓ |
8. Turning Word Problems into Equations | 把文字题转化为方程
Word problems test whether you can translate English into algebra. Follow four fixed steps: define the unknown with a letter, write an equation from the words, solve the equation, then write the answer in the context of the question with the correct units.
文字题考查的是你能否把语言翻译成代数式。遵循四个固定步骤：用字母定义未知数，根据文字写出方程，解方程，最后结合题目情境和单位写出答案。
Example: "A number is multiplied by 3 and then 4 is subtracted. The result is 20." Let the number be n. Then 3n − 4 = 20, so 3n = 24 and n = 8. The number is 8.
例：一个数乘以 3 再减去 4，结果是 20。设这个数为 n，则 3n − 4 = 20，所以 3n = 24，n = 8。这个数是 8。
3n − 4 = 20 → 3n = 24 → n = 8
2x + 3) = 26, so 2x + 3 = 13, 2x = 10 and x = 5. The width is 5 cm and the length is 8 cm.
几何类题目：一个长方形的长为 (x + 3) 厘米，宽为 x 厘米，周长是 26 厘米。因为周长 = 2(长 + 宽)，所以 2(2x + 3) = 26，得 2x + 3 = 13，2x = 10，x = 5。宽是 5 厘米，长是 8 厘米。
Consecutive numbers: n, n + 1, n + 2. Their sum is 3n + 3.
连续整数：n、n + 1、n + 2，它们的和是 3n + 3。
Age problems: "in 5 years" means add 5; "3 years ago" means subtract 3.
年龄问题："5 年后"就是加 5，"3 年前"就是减 3。
Always answer the question that was actually asked, not just the value of x.
最后要回答题目真正问的内容，而不是只写 x 的值。
9. Common Errors to Avoid | 常见错误与避免方法
Most lost marks in linear equations come from a small number of repeated mistakes. Learning to recognise them is faster than learning more theory.
一元一次方程的失分大多来自少数几个反复出现的错误。学会识别这些错误，比多学理论更有效。
Doing the same operation to only one side. The balance must stay balanced.
只对一边做运算。天平必须始终保持平衡。
Forgetting to multiply every term inside a bracket, especially a negative one.
忘记乘以括号内的每一项，尤其是负项。
ors when moving a term across the equals sign, e.g. writing 3x = 11 + 2 instead of 11 − 2.
移项时符号出错，例如把 3x = 11 − 2 写成 3x = 11 + 2。
Dividing by the coefficient before undoing the addition or subtraction.
在消去加减之前就先除以系数。
Giving the answer as a fraction when the question asks for a whole number, or forgetting units in a word problem.
题目要求整数却给出分数答案，或文字题中漏写单位。
A useful habit: after each line, ask yourself "is this line still equivalent to the one above?" If you cannot justify the step, it is probably wrong.
一个有用的习惯：每写一行，问自己"这一行和上一行还等价吗？"如果无法说明这一步的理由，它多半是错的。
10. Mixed Practice Set | 综合练习
Try all eight questions without a calculator, then check your answers. They follow the same order as the sections above, so you can see exactly which skill each question tests.
请在不使用计算器的情况下完成全部八题，然后核对答案。题目顺序与上面的小节一致，方便看出每题考查的技能。
(a) x + 9 = 21
(a) x + 9 = 21
(b) 4y − 5 = 23
(b) 4y − 5 = 23
(c) 2(x + 4) = 30
(c) 2(x + 4) = 30
(d) 5(x − 3) = 2(x + 3)
(d) 5(x − 3) = 2(x + 3)
(e) x ÷ 5 + 3 = 8
(e) x ÷ 5 + 3 = 8
(f) (x + 2) ÷ 3 = 6
(f) (x + 2) ÷ 3 = 6
(g) 7m + 1 = 4m + 19
(g) 7m + 1 = 4m + 19
(h) Three consecutive integers add up to 81. Find the largest.
(h) 三个连续整数之和为 81，求最大的那个数。
| Question 题号 | Answer 答案 | Key step 关键步骤 |
| (a) | x = 12 | Subtract 9 两边减 9 |
| (b) | y = 7 | Add 5, divide by 4 |
| (c) | x = 11 | Divide by 2 or expand |
| (d) | x = 7 | Expand both sides 两边展开 |
| (e) | x = 25 | Multiply by 5 两边乘 5 |
| (f) | x = 16 | Multiply by 3 两边乘 3 |
| (g) | m = 6 | Collect m on one side |
| (h) | 28 | n + (n + 1) + (n + 2) = 81 |
11. Exam Technique and Summary | 应试技巧与总结
In a KS3 test or an end-of-topic assessment, method marks are usually available even when the final answer is wrong. Show every step, keep the equals signs aligned down the page, and never erase your working to save space.
在 KS3 测验或单元测评中，即使最终答案错了，过程分通常仍然可以拿到。把每一步都写出来，让等号沿页面对齐，也不要为了省地方擦掉过程。
Here is the whole topic in four lines: undo addition and subtraction first, then undo multiplication and division; expand brackets before collecting terms; clear fractions by multiplying by the denominator; check by substituting back into the original equation.
整个专题可以浓缩成四句话：先消去加减，再消去乘除；先去括号再合并同类项；通过乘以分母来消去分数；把解代回原方程进行检验。
Balance → Inverse operations → Check → Answer in context
Practice a handful of equations every day rather than twenty in one sitting. Fluency with linear equations is not only worth marks on its own - it is the foundation for simultaneous equations, inequalities, rearranging formulae and straight-line graphs later in KS3 and at IGCSE.
每天练习几道方程，比一次做二十
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