📚 Solving Linear Equations (Cambridge KS3 p115) | 一元一次方程求解（剑桥初中数学第115页）
Linear equations are one of the most important building blocks in Key Stage 3 algebra. On page 115 of the Cambridge Mathematics course, you will meet a range of techniques designed to help you solve equations confidently and accurately. This article walks you through each method, from simple one-step equations to those involving brackets, fractions and variables on both sides.
一元一次方程是 KS3 代数中最重要的基石之一。在剑桥数学课程的第115页，你将学习一系列旨在帮助你自信且准确地解方程的方法。本文将从简单的一步方程到含有括号、分数和未知数在两边的情况，逐步讲解每一种解法。
1. What is a Linear Equation? | 什么是一元一次方程？
A linear equation in one variable is an equality that contains a variable, usually x, raised only to the power 1. There are no terms like x², x³ or 1/x. The solution is the value that makes the equation true when substituted for the variable.
一元一次方程是含有一个变量（通常是 x）且变量的指数仅为 1 的等式。式中没有 x²、x³ 或 1/x 这样的项。方程的解是代入变量后使等式成立的数值。
Example: 2x + 3 = 11
示例：2x + 3 = 11
2. Solving One-Step Equations | 解一步方程
One-step equations can be solved by performing a single inverse operation. If the equation is x + a = b, subtract a from both sides. If it is x – a = b, add a to both sides.
一步方程可以通过执行单一的逆运算来求解。如果方程是 x + a = b，则两边同时减去 a；如果是 x – a = b，则两边同时加上 a。
For multiplication and division, if the equation is ax = b, divide both sides by a. If it is x/a = b, multiply both sides by a.
对于乘法和除法，若方程为 ax = b，则两边除以 a；若方程为 x/a = b，则两边乘以 a。
x + 7 = 15 → x = 8
4x = 20 → x = 5
3. Solving Two-Step Equations | 解两步方程
Two-step equations require two inverse operations, usually undoing addition/subtraction first, then multiplication/division. For instance, in 2x + 3 = 11, subtract 3 from both sides to get 2x = 8, then divide by 2 to obtain x = 4.
两步方程需要两次逆运算，通常先处理加法/减法，再处理乘法/除法。例如，对于 2x + 3 = 11，两边先减 3 得到 2x = 8，再除以 2 得到 x = 4。
Always remember the order of operations is reversed when solving: we undo any addition or subtraction before dealing with the coefficient of x.
始终记住，求解时运算顺序是相反的：先解除加减，再处理 x 的系数。
4. Equations with Brackets | 带括号的方程
When brackets appear, expand them first using the distributive law. For example, 3(x + 2) = 15 becomes 3x + 6 = 15. Then solve as a two-step equation: subtract 6 to get 3x = 9, and divide by 3 to find x = 3.
当出现括号时，首先用分配律展开。例如，3(x + 2) = 15 变为 3x + 6 = 15。然后当作两步方程求解：减 6 得到 3x = 9，除以 3 得到 x = 3。
If there is a negative sign in front of a bracket, be careful to multiply every term inside by -1: -(2x – 5) = -2x + 5.
如果括号前有负号，要小心将里面的每一项都乘以 -1：-(2x – 5) = -2x + 5。
5. Equations with Variables on Both Sides | 含未知数在两边
When the variable appears on both sides of the equation, collect all variable terms on one side and constant terms on the other. For 5x + 2 = 3x + 10, subtract 3x from both sides to obtain 2x + 2 = 10. Then subtract 2: 2x = 8, so x = 4.
当未知数出现在方程两边时，将所有含变量的项移到一边，常数项移到另一边。对于 5x + 2 = 3x + 10，两边减去 3x 得 2x + 2 = 10。再减 2：2x = 8，因此 x = 4。
Always aim to have a positive coefficient for x. If you end up with -x, multiply both sides by -1.
始终争取让 x 的系数为正。如果最终得到 -x，就将两边乘以 -1。
6. Equations with Fractions | 分数方程
To solve equations containing fractions, eliminate the denominators by multiplying every term by the lowest common denominator. For x/3 + 1/2 = 5/6, multiply through by 6: 2x + 3 = 5, then 2x = 2, so x = 1.
要解含有分数的方程，可以通过将每一项都乘以最小公分母来消去分母。对于 x/3 + 1/2 = 5/6，两边乘以 6：2x + 3 = 5，然后 2x = 2，得 x = 1。
Another common type is when a fraction equals a constant: (2x)/5 = 4. Multiply both sides by 5 to give 2x = 20, then x = 10.
另一种常见类型是分数等于一个常数：(2x)/5 = 4。两边乘以 5 得 2x = 20，x = 10。
7. Checking Your Solution | 验算你的解
After finding a value for x, always substitute it back into the original equation to check both sides are equal. This habit catches arithmetic mistakes and reinforces your understanding of what a solution means.
求出 x 的值后，一定要将其代回原方程，检查两边是否相等。这个习惯可以捕捉算术错误，并加深你对解的含义的理解。
For 2x + 3 = 11, if x = 4, then left side = 2(4) + 3 = 8 + 3 = 11 = right side. Correct!
对于 2x + 3 = 11，如果 x = 4，左边 = 2(4) + 3 = 8 + 3 = 11 = 右边。正确！
8. Real-life Applications | 实际应用
Linear equations model many everyday situations. If a taxi charges a fixed fee of pound 3 plus pound 2 per mile, and the total fare is pound 15, the equation is 2m + 3 = 15. Solving gives m = 6 miles.
一元一次方程可以模拟许多日常情景。如果出租车收取 3 英镑固定费用加上每英里 2 英镑，总车费为 15 英镑，方程就是 2m + 3 = 15。解得 m = 6 英里。
Interpreting the solution in context ensures you answer the original problem, not just manipulate symbols.
在情境中解释解的含义，能确保你回答的是最初的问题，而不仅仅是进行符号运算。
9. Common Mistakes and How to Avoid Them | 常见错误及如何避免
A frequent error is forgetting to perform an operation on both sides. If you subtract 5 from the left, you must subtract 5 from the right as well. Another is mishandling negative signs when expanding brackets.
一个常见错误是忘记对两边执行相同的运算。如果你在左边减 5，右边也必须减 5。另一个错误是展开括号时处理负号不当。
When solving two-step equations, some students divide before subtracting. Always reverse the order of operations: do addition/subtraction first, then multiplication/division.
在解两步方程时，有些学生先除后减。一定要逆转运算顺序：先进行加/减，再进行乘/除。
10. Summary | 总结
To solve any linear equation reliably, follow a clear sequence: eliminate brackets, clear fractions, collect variable terms on one side and constants on the other, then isolate the variable by inverse operations. Check your answer by substitution. Page 115 of the Cambridge course gives you plenty of practice to turn these steps into a confident routine.
要可靠地求解任何一元一次方程，请遵循明确的顺序：去括号，消分数，将变量项和常数项分别集中到两边，然后通过逆运算求出变量。通过代入检查答案。剑桥课程的第115页为你提供了大量练习，帮助你将这些步骤变成自信的习惯。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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