📚 Mastering Linear Equations | 掌握一次方程
A linear equation is an equation in which the highest power of the variable is 1. In KS3 Cambridge Mathematics, you are expected to solve one-step, two-step and multi-step linear equations confidently, including those with brackets, fractions and variables on both sides.
一次方程是指未知数的最高次数为 1 的方程。在剑桥 KS3 数学中，你需要熟练地求解一步、两步和多步一次方程，包括含有括号、分数以及两边都有未知数的方程。
1. What Is a Linear Equation? | 什么是一次方程
, 3x + 4 = 19 is a linear equation because the variable x appears only to the first power.
方程是说明两个表达式相等的一个陈述。一次方程的一般形式为 ax + b = c，其中 a、b、c 是常数，x 是未知数。例如，3x + 4 = 19 是一次方程，因为变量 x 只出现一次方。
You can recognise a linear equation by checking that there are no terms such as x², x³, 1/x or √x. These terms make the equation non-linear and require different methods.
你可以通过检查方程中是否没有 x²、x³、1/x 或 √x 这样的项来识别一次方程。如果出现这些项，方程就是非线性的，需要用不同的方法求解。
2. The Balance Method | 天平法
The balance method is a visual way to solve linear equations. Imagine an old-fashioned balance scale: the left pan and the right pan must have the same mass. If you add, subtract, multiply or divide both sides by the same amount, the balance is preserved.
天平法是一种求解一次方程的直观方法。想象一台老式天平秤：左盘和右盘必须有相同的质量。如果你在两边同时加、减、乘或除以同一个数，天平仍然保持平衡。
7 − 7 = 15 − 7, so x = 8. Doing the same operation on both sides keeps the equation true.
例如，要解方程 x + 7 = 15，需要在两边同时减去 7。这样得到 x + 7 − 7 = 15 − 7，所以 x = 8。在两边同时进行相同的运算可以保证方程仍然成立。
3. Basic Steps for Solving a Linear Equation | 解一次方程的基本步骤
When solving a linear equation, follow a clear order: simplify each side if possible, then collect like terms, then move variable terms to one side and constant terms to the other. Finally, divide by the coefficient of the variable.
解一次方程时，要遵循清晰的步骤：先尽可能化简每一边，然后合并同类项，接着把含有未知数的项移到一边，把常数项移到另一边，最后除以未知数的系数。
s to get 3x − 3 = 9. Then add 3 to both sides to get 3x = 12. Divide by 3 to obtain x = 4.
例如：解方程 5x − 3 = 2x + 9。首先两边同时减去 2x，得到 3x − 3 = 9；然后两边同时加 3，得到 3x = 12；再除以 3，得到 x = 4。
Always write each step on a new line and keep the equals signs lined up. This helps you avoid mistakes and makes your working easy to follow.
每一步都要换行书写，并让等号上下对齐。这有助于避免错误，也让解题过程更容易被理解。
4. Removing Brackets | 去括号
Before solving an equation that contains brackets, expand them using the distributive law. Multiply each term inside the bracket by the number or variable outside it. For example, 3(x + 4) becomes 3x + 12.
在解含有括号的方程之前，要先用乘法分配律去括号。把括号外的数或变量分别乘以括号内的每一项。例如，3(x + 4) 应展开为 3x + 12。
h sides: 2x = 28. Divide by 2: x = 14. Checking: 2(14 − 5) = 2 × 9 = 18, which is correct.
例如：解方程 2(x − 5) = 18。展开得到 2x − 10 = 18；两边同时加 10，得到 2x = 28；再除以 2，得到 x = 14。检验：2(14 − 5) = 2 × 9 = 18，结果正确。
3), which gives −2x + 3. A common error is to write −2x − 3, so watch the signs carefully.
当括号前有负号时要特别小心。例如，−(2x − 3) 表示 −1 × (2x − 3)，结果为 −2x + 3。常见错误是写成 −2x − 3，因此一定要仔细处理符号。
5. Unknowns on Both Sides | 两边都有未知数
The key is to collect the variable terms on one side first. You can do this by adding or subtracting the same term on both sides.
许多 KS3 方程的两边都有含有未知数的项，例如 7x − 2 = 4x + 10。关键是先把未知项集中到一边。你可以在两边同时加上或减去同一个含未知数的项。
om both sides: 3x − 2 = 10. Then add 2 to both sides: 3x = 12. Finally divide by 3: x = 4.
解方程 7x − 2 = 4x + 10 时，先在两边同时减去 4x，得到 3x − 2 = 10；再在两边同时加 2，得到 3x = 12；最后除以 3，得到 x = 4。
Some equations require you to move the variable to the right-hand side if the left side becomes too complicated. Both directions are valid as long as you perform the same operation on both sides.
有些方程如果把未知数移到右边会更简单。只要对两边进行相同的运算，把未知数移到左边或右边都可以。
6. Equations Involving Fractions | 含分数的方程
When an equation contains fractions, multiply every term by the lowest common denominator to clear the fractions. This turns the equation into an integer equation that is easier to solve.
当方程中含有分数时，可以把每一项都乘以最小公分母，以消去分数。这样方程就转化为更容易求解的整数方程。
x/3 = 3. Multiply both sides by 3: x = 9. Checking: 9/3 + 2 = 3 + 2 = 5, which is correct.
例如：解方程 x/3 + 2 = 5。两边同时减去 2，得到 x/3 = 3；两边同时乘以 3，得到 x = 9。检验：9/3 + 2 = 3 + 2 = 5，结果正确。
iple of 2 and 3, to get 3(x + 1) = 2(x − 1). Expand and solve: 3x + 3 = 2x − 2, so x = −5.
对于像 (x + 1)/2 = (x − 1)/3 这样的方程，可以交叉相乘。将两边同时乘以 2 和 3 的最小公倍数 6，得到 3(x + 1) = 2(x − 1)。展开并求解：3x + 3 = 2x − 2，所以 x = −5。
7. Forming Equations from Word Problems | 根据文字题列方程
Word problems require you to translate a real-life situation into an algebraic equation. Read the problem carefully, identify the unknown quantity, and use a letter such as x to represent it. Then build the equation step by step.
文字题要求你把实际问题转化为代数方程。仔细读题，找出未知量，并用 x 等字母表示它，然后一步步建立方程。
uation is 2(x + 2x) = 36, or 6x = 36, so x = 6. The width is 6 cm and the length is 12 cm.
例如：一个矩形的长是宽的两倍，周长为 36 厘米。求宽。设宽为 x 厘米，则长为 2x 厘米。周长方程为 2(x + 2x) = 36，即 6x = 36，所以 x = 6。宽为 6 厘米，长为 12 厘米。
When forming equations from word problems, define the variable clearly and check that the equation matches the given relationship. This will help you avoid misinterpretation.
根据文字题列方程时，要清楚地定义变量，并检查方程是否符合题目给出的关系。这有助于避免理解错误。
8. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Many students lose marks because of small mistakes. The most common errors include forgetting to change signs when moving terms, expanding brackets incorrectly, and dividing by the wrong number.
许多学生因为一些小的错误而失分。最常见的错误包括：移项时忘记变号、去括号时展开错误、除以了错误的数。
| Mistake | 错误 | Correction | 正确做法 |
| From 3x + 5 = 20, writing 3x = 25 | Subtract 5: 3x = 15 |
| Expanding −2(x − 4) as −2x − 8 | Write −2x + 8 |
| From 4x = 20, writing x = 80 | Divide by 4: x = 5 |
To avoid sign errors, write the operation you are performing in the margin or above the equals sign. This makes each step visible and easier to check.
为了避免符号错误，可以在等号上方或旁边写出你正在进行的运算。这样每一步都清晰可见，也更容易检查。
9. Checking Your Answer | 检验答案
Always substitute your answer back into the original equation to check that it works. This habit can catch small mistakes and give you confidence in your final answer.
一定要把你的答案代回到原方程中进行检验。这个习惯可以发现小的错误，并让你对最终答案更有信心。
, substitute: 4(3) + 1 = 12 + 1 = 13. Since both sides are equal, the solution is correct.
例如，如果你解方程 4x + 1 = 13 得到 x = 3，就代入检验：4(3) + 1 = 12 + 1 = 13。两边相等，说明解是正确的。
If the left-hand side does not equal the right-hand side after substitution, retrace your steps carefully. The error may be in expansion, collecting like terms or dividing at the end.
如果代入后左边与右边不相等，就要仔细回看解题步骤。错误可能出现在去括号、合并同类项或最后除以系数的过程中。
10. Worked Practice Examples | 精讲练习示例
Below are three worked examples that show the full method. Read each line and make sure you understand why each operation is used.
下面是三道展示完整方法的精讲例题。请逐行阅读，并确保理解每一步为什么这样做。
Example 1: Solve 6x + 9 = 45.
Subtract 9: 6x = 36.
Divide by 6: x = 6.
例题 1：解方程 6x + 9 = 45。
两边减 9：6x = 36。
两边除以 6：x = 6。
Example 2: Solve 3(2x − 1) = 27.
Expand: 6x − 3 = 27.
Add 3: 6x = 30.
Divide by 6: x = 5.
例题 2：解方程 3(2x − 1) = 27。
展开：6x − 3 = 27。
两边加 3：6x = 30。
两边除以 6：x = 5。
Example 3: Solve 2x + 5 = x + 12.
Subtract x: x + 5 = 12.
Subtract 5: x = 7.
例题 3：解方程 2x + 5 = x + 12。
两边减 x：x + 5 = 12。
两边减 5：x = 7。
These examples show that if you act systematically, even multi-step equations become manageable.
这些例题说明，只要按部就班地进行，多步方程也能轻松解决。
11. Real-Life Uses of Linear Equations | 一次方程的实际应用
Linear equations are used in everyday situations such as calculating mobile phone bills, comparing taxi fares, converting units, or working out how many items you can buy with a fixed amount of money.
一次方程常用于日常生活中，例如计算手机话费、比较出租车费用、换算单位，或者计算用固定金额可以购买多少物品。
For example, a taxi company charges a fixed starting fee of £2 and then £ ilometre. The total cost C for d kilometres can be written as C = 2 + 1.5d. If the cost is £11, solve 2 + 1.5d = 11.
例如，一家出租车公司收取 2 英镑起步价，之后每公里 1.50 英镑。行驶 d 公里的总费用 C 可以写成 C = 2 + 1.5d。如果费用为 11 英镑，则解方程 2 + 1.5d = 11。
So the journey was 6 kilometres. This shows how algebra helps answer practical questions.
两边同时减去 2：1.5d = 9；两边除以 1.5：d = 6。因此行驶了 6 公里。这说明代数可以帮助解决实际生活中的问题。
12. Summary and Key Points | 总结与要点
When solving linear equations, simplify each side, collect like terms, isolate the variable, and check your answer. Practise regularly with different types of equations to build speed and accuracy.
解一次方程时，要化简每一边、合并同类项、把未知数单独放在一边，并检验答案。要经常练习不同类型的方程，以提高速度和准确性。
Remember these key points: variables must be raised only to the power 1; operations must be applied to both sides; brackets are expanded using the distributive law; fractions can be cleared using the lowest common denominator; substitution checks your answer.
请牢记以下要点：未知数的次数必须为 1；运算必须同时作用于方程的两边；括号用乘法分配律展开；分数可以通过最小公分母消去；代入法可以检验答案。
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