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Solving Linear Equations (Cambridge KS3, p187, Exercise 2) | 解一元一次方程（剑桥 KS3，第 187 页练习 2）

2026-08-21 11:59 · 本文部分内容由AI辅助创作

📚 Solving Linear Equations (Cambridge KS3, p187, Exercise 2) | 解一元一次方程（剑桥 KS3，第 187 页练习 2）

Linear equations are the workhorses of KS3 algebra. In this revision guide, based on Cambridge KS3 page 187, Exercise 2, you will learn how to isolate the unknown, handle brackets and fractions, deal with variables on both sides, set up equations from word problems, and check your work so that careless errors never cost you marks again.

线性方程是 KS3 代数中最重要的工具之一。本文以剑桥 KS3 第 187 页练习 2 为起点，带你复习解一元一次方程所需的全部关键技能：如何分离未知数、如何处理括号和分数、如何应对等号两边都有未知数的情况、如何根据应用题列方程，以及如何检验答案，让粗心错误不再丢分。

1. What is a Linear Equation? | 什么是线性方程？

y + 5 are both linear equations. There are no x² terms, no square roots, and no variables in the denominator.

线性方程是一种代数等式，其中未知变量只以一次幂出现。例如 4x + 1 = 9 和 2y − 3 = y + 5 都是线性方程。方程中没有 x² 项，没有平方根，也没有变量出现在分母中。

= 2 because 4(2) + 1 = 9. Solving an equation means finding every value that satisfies it.

线性方程的解就是使等式成立的变量取值。在 4x + 1 = 9 中，解是 x = 2，因为 4(2) + 1 = 9。解方程意味着找出所有使等式成立的值。

e left side is 2(8) − 3 = 13 and the right side is 8 + 5 = 13, so the solution is correct.

对于 2y − 3 = y + 5，两边同时减去 y 得到 y − 3 = 5，再加 3 得 y = 8。检验：左边 2(8) − 3 = 13，右边 8 + 5 = 13，等式成立，解答正确。

On a graph, a linear equation in two variables always produces a straight line. That is exactly why it is called "linear". In KS3, however, most linear equations you solve contain just one unknown, such as x or y.

在图像上，含有两个变量的线性方程总是画出一条直线，这正是它被称为"线性"的原因。不过在 KS3 中，你遇到的大多数线性方程只含一个未知数，例如 x 或 y。

2. Using Inverse Operations | 使用逆运算

To solve a linear equation, rearrange it so that the variable stands alone on one side of the equals sign. This is done by applying inverse operations to both sides. Addition and subtraction undo each other; multiplication and division undo each other.

解线性方程时，要重新整理方程，使变量单独位于等号一边。做法是对方程两边同时施加逆运算。加法与减法互为逆运算，乘法与除法互为逆运算。

The four inverse relationships you must know by heart are shown below. Whatever you do to one side, you must do exactly the same to the other side.

下面四种逆运算关系必须熟记于心。你对等号一边做了什么运算，就必须对另一边做完全相同的运算。

A useful habit is to work in reverse order of BIDMAS. Undo the addition or subtraction first, then undo the multiplication or division. This keeps the arithmetic simple and the signs manageable.

一个很实用的习惯是：按 BIDMAS 的相反顺序操作。先消去加法和减法，再消去乘法和除法。这样计算更简单，符号也不容易出错。

3. The Balance Method | 天平法

Think of an equation as a balanced pair of scales. The two sides are equal, so anything you add, subtract, multiply or divide must be applied to both pans at once, otherwise the scales tip.

把方程想象成一架平衡的天平。等号两边重量相等，因此你加、减、乘、除的任何量都必须同时施加到两个托盘上，否则天平就会倾斜。

3x + 7 = 22 → 3x = 15 → x = 5

a simpler equation that is equivalent to the one before it, so the solution never changes.

在上面这个例子中，两边同时减 7 得 3x = 15，再两边同时除以 3 得 x = 5。每一行都是与前一行等价的更简单的方程，因此解始终不变。

x ÷ 4 − 3 = 2 → x ÷ 4 = 5 → x = 20

Here the subtraction is undone first, then the division. Writing one clear line of working per step, with the equals signs aligned, makes your reasoning easy for an examiner to follow and easy for you to check.

这里先消去减法，再消去除法。每一步写一行清晰的运算过程，并且把等号对齐，既方便阅卷老师看懂你的思路，也方便你自己检查。

4. Equations with Brackets | 含括号的方程

When an equation contains brackets, you have two safe strategies. Strategy A is to expand the brackets first and then solve. Strategy B, which is often faster, is to divide both sides by the number in front of the bracket first.

当方程中含有括号时，有两种稳妥的策略。策略 A 是先展开括号再解方程；策略 B 往往更快，即先把两边同时除以括号前的系数。

3(x + 2) = 15 → 3x + 6 = 15 → 3x = 9 → x = 3

gives x + 2 = 5, so x = 3 directly. Both routes are valid, and both give the same answer.

展开得 3x + 6 = 15，这就是策略 A。若先把两边同时除以 3，得 x + 2 = 5，立刻得到 x = 3。两种方法都正确，答案完全一致。

The most common mistake with brackets is failing to multiply every term inside. For example, 5(2x − 1) means 10x − 5, not 10x − 1. Always multiply the whole bracket by the number outside it.

与括号有关的常见错误是漏乘括号内的某一项。例如 5(2x − 1) 等于 10x − 5，而不是 10x − 1。一定要用括号外的数乘遍括号内每一项。

4(x − 3) = 2x + 2 → 4x − 12 = 2x + 2 → 2x = 14 → x = 7

Notice the negative sign inside the bracket: 4(x − 3) expands to 4x − 12. Forgetting that minus sign turns a correct method into a wrong answer, so write the expansion line out in full every time.

注意括号内的负号：4(x − 3) 展开为 4x − 12。漏掉这个负号会让正确的方法得出错误答案，所以每次都要把展开那一步完整写出来。

5. Equations with Fractions | 含分数的方程

Fractions are not a different type of equation; they just need one extra opening move. Multiply every term on both sides by the lowest common multiple of the denominators. This clears all the fractions and leaves a familiar linear equation.

含分数的方程并不是另一种方程，只是需要多一步开场动作：把两边每一项都乘以各分母的最小公倍数。这样就能一次性清掉所有分母，留下熟悉的线性方程。

x ÷ 3 + x ÷ 4 = 7 → 4x + 3x = 84 → 7x = 84 → x = 12

ich matches the original equation. Always test the answer in the original fractional form, not in the cleared form.

两边同时乘 12 就消去了两个分母。检验：12 ÷ 3 + 12 ÷ 4 = 4 + 3 = 7，与原方程一致。检验时一定要代回原来的分数形式，而不是消去分母后的形式。

(2x + 1) ÷ 3 = 5 → 2x + 1 = 15 → 2x = 14 → x = 7

When a single fraction equals a whole number, simply multiply both sides by the denominator. When a fraction equals another fraction, cross-multiply instead, which is the same as multiplying by both denominators at once.

当单个分数等于一个整数时，直接两边同时乘分母即可。当分数等于分数时，改用交叉相乘，这等价于同时乘以两个分母。

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(x + 1) ÷ 2 = (x − 2) ÷ 3 → 3(x + 1) = 2(x − 2) → 3x + 3 = 2x − 4 → x = −7

is correct. Negative solutions are perfectly normal and should never be rejected on sight.

代入检验：(−7 + 1) ÷ 2 = −3，(−7 − 2) ÷ 3 = −3，两边相等，故 x = −7 正确。负数解是完全正常的，不要一看到负号就否定答案。

6. Variables on Both Sides | 未知数在等号两边

Many KS3 equations, including several on page 187, place the unknown on both sides. The rule is simple: collect all the variable terms on the side with the larger coefficient, and all the number terms on the other side.

许多 KS3 方程（包括第 187 页的若干题）在等号两边都含未知数。规则很简单：把所有的未知数项移到系数较大的那一边，把所有的常数项移到另一边。

5x − 4 = 2x + 11 → 3x = 15 → x = 5

. Keeping the coefficient positive avoids unnecessary negative signs later in the working.

两边同时减 2x 得 3x − 4 = 11，再加 4 得 3x = 15。让未知数系数保持为正，可以避免后续出现不必要的负号。

7 − 2x = 3x − 8 → 15 = 5x → x = 3

ng the variable on the right-hand side is perfectly acceptable; the answer is still x = 3.

这里更简单的做法是两边同时加 2x，再加 8，得到 15 = 5x。把未知数写在等号右边完全可以接受，答案仍然是 x = 3。

7. Forming Equations from Word Problems | 从应用题中列方程

Examination questions rarely say "solve 4x + 1 = 9" to choose a letter for the unknown, translate each sentence into symbols, and then solve.

考试题目很少直接给出"解 4x + 1 = 9"，而是描述一个情境，让你自己列出方程。这项技能的关键是：用一个字母表示未知量，把每句话翻译成符号，然后再解方程。

w + w + 3) = 26, so 4w + 6 = 26, 4w = 20, w = 5. The width is 5 cm and the length is 8 cm.

一个长方形的长比宽多 3 厘米，周长是 26 厘米。设宽为 w，则长为 w + 3，由周长得 2(w + w + 3) = 26，即 4w + 6 = 26，4w = 20，w = 5。所以宽 5 厘米，长 8 厘米。

(n + 2) = 72, giving 3n + 3 = 72, 3n = 69 and n = 23. The three numbers are 23, 24 and 25.

三个连续整数之和为 72。设最小的为 n，则另外两个是 n + 1 和 n + 2。于是 n + (n + 1) + (n + 2) = 72，得 3n + 3 = 72，3n = 69，n = 23。三个数是 23、24 和 25。

Word-problem questions almost always carry a "communicate" mark. Write a clear definition of your letter, show the equation you formed, solve it, and finish with a sentence that answers the question in context, complete with units.

应用题通常带有"表达与交流"的得分点。要清楚写出所设字母的含义，列出方程，解出结果，最后用一句话结合情境和单位回答问题。

8. Checking Your Answer | 检验答案

Checking is not optional. It takes fifteen seconds and it catches the majority of errors made under exam pressure. Substitute your value back into the original equation, evaluate both sides separately, and confirm they are equal.

检验不是可选项。它只花十五秒，却能抓住考场上因紧张而犯的大多数错误。把你的答案代回原方程，分别算出两边的值，确认它们相等。

x = 7: LHS = 4(7 − 3) = 16, RHS = 2(7) + 2 = 16 ✓

Write "LHS" and "RHS" above your working and show the two numbers side by side. If they differ, do not simply change your answer: go back through the steps, because the mistake may lie in the checking itself.

在做题过程中标注"左边 LHS"和"右边 RHS"，把两个数值并排写出。如果两者不等，不要急着改答案，而应回头逐行检查，因为错误也可能出在检验这一步。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Most lost marks in linear equations come from a small number of predictable slips. Knowing them in advance is the cheapest way to raise your accuracy.

线性方程中丢失的分数，大多来自少数几种可预见的失误。提前了解它们，是提高准确率最省力的方法。

Mistake 错误 Example 例子 Fix 改正方法
Operating on only one side 只对一边运算 3x + 7 = 22 → 3x = 22 Subtract 7 from both sides 两边同时减 7
Sign error when moving terms 移项时符号出错 2x + 5 = 11 → 2x = 11 + 5 Moving a term changes its sign 移项要变号
Incomplete expansion 展开不完整 5(2x − 1) = 10x − 1 Multiply every term: 10x − 5 每项都要乘
Losing a negative sign 丢负号 −3(x − 4) = −3x − 12 Correct: −3x + 12 正确应为 −3x + 12
Multiplying only some terms 未乘遍每一项 x ÷ 3 + x ÷ 4 = 7 → 4x + x = 84 Multiply by 12 throughout 每一项都乘 12
Stopping before dividing 忘记最后一步除法 4x = 20, answer 20 Divide by 4: x = 5 除以 4 得 x = 5

Another frequent issue is working too fast in the head. Written equations are not mental arithmetic tests; writing every step protects you from exactly the slips listed above.

另一个常见问题是过度依赖心算。解方程不是口算测验，把每一步都写下来，正好能防止上表列出的那些失误。

10. Worked Examples in the Style of p187, Exercise 2 | 第 187 页练习 2 典型例题精讲

The questions below follow the pattern of Cambridge KS3 page 187, Exercise 2. Try each one on paper before reading the solution, then compare your working line by line.

下面几道题完全按照剑桥 KS3 第 187 页练习 2 的题型设计。请先自己在纸上完成，再看解析，并逐行对比解题过程。

ne reads 3x = 15 rather than x = 5, the answer is incomplete and will not earn full marks.

注意上面每道题的解答最后都让未知数单独留下，系数为 1。如果你的最后一行写的是 3x = 15 而不是 x = 5，就说明解答尚未完成，拿不到满分。

11. Why "Linear"? Graphs and Meaning | 为什么叫"线性"？图像与意义

lution pairs. Plot them and they line up perfectly along a straight line, with gradient 2 and y-intercept 1.

含有两个变量的线性方程，如 y = 2x + 1，有无穷多组解。把解画在坐标系中，它们恰好排成一条直线，斜率为 2，y 轴截距为 1。

y = mx + c (m = gradient, c = y-intercept)

+ 1 reaches a height of 7. The answer x = 3 is the x-coordinate of that point on the line.

解一元方程 2x + 1 = 7，等价于问直线 y = 2x + 1 在何处高度达到 7。答案 x = 3 就是那个点的横坐标。

This connection matters at IGCSE and A Level, where you solve simultaneous equations by finding the intersection of two straight lines. A solid grounding in KS3 linear equations makes that later work far easier.

这个联系在 IGCSE 和 A Level 阶段非常重要，因为那里要通过求两条直线的交点来解联立方程。在 KS3 打牢一元一次方程的基础，之后的学习会轻松得多。

12. Practice Set and Next Steps | 练习题与下一步

Work through the ten equations below without a calculator. Aim for full written solutions, one line per step, with a check at the end of each question.

不用计算器完成下面十道方程。要求写出完整过程，每一步一行，并在每题末尾做检验。