📚 Solving Linear Equations: A Cambridge KS3 Guide | 解一元一次方程：剑桥KS3指南
Linear equations are the foundation of algebra at Cambridge KS3. A linear equation is a statement that two expressions are equal, and it usually contains one unknown value, often written as x or n. Solving the equation means finding the value of the unknown that makes the statement true.
一元一次方程是剑桥 KS3 代数的基础。一个线性方程表示两个表达式相等，通常含有一个未知数，常用 x 或 n 表示。解方程就是找到使等式成立的未知数值。
1. What Is a Linear Equation? | 什么是一元一次方程？
A linear equation has at least one algebraic expression and an equal sign. The unknown appears only to the first power, so you will not see x², x³ or 1/x in a linear equation at this stage. Common forms include x + 7 = 15, 3n − 4 = 8 and 2(a + 3) = 12.
一元一次方程至少含有一个代数表达式和一个等号。未知数只出现一次方，因此在这个阶段你不会看到 x²、x³ 或 1/x 这样的项。常见形式包括 x + 7 = 15、3n − 4 = 8 和 2(a + 3) = 12。
x + 7 = 15 is a linear equation because x has power 1.
x + 7 = 15 是一元一次方程，因为 x 的指数为 1。
2. Equation or Expression? | 方程还是表达式？
Many KS3 mistakes come from confusing an expression with an equation. An expression such as 4x + 9 does not contain an equal sign, so it cannot be solved. An equation such as 4x + 9 = 21 makes a complete statement that can be true for a particular value of x.
许多 KS3 阶段的错误来自混淆表达式和方程。像 4x + 9 这样的表达式没有等号，因此无法求解。像 4x + 9 = 21 这样的方程是一个完整陈述，对于某个特定的 x 值可以成立。
- Expression: 4x + 9, 2n − 5, a/3 + 7. These are not solved. 表达式：4x + 9、2n − 5、a/3 + 7。这些不需要求解。
- Equation: 4x + 9 = 21, 2n − 5 = 3, a/3 + 7 = 10. These can be solved. 方程：4x + 9 = 21、2n − 5 = 3、a/3 + 7 = 10。这些可以求解。
3. The Balance Principle | 天平原理
An equation behaves like a balance scale. If you add, subtract, multiply or divide one side, you must do exactly the same to the other side. The goal is to isolate the unknown on one side of the equal sign.
方程就像一个天平。如果你对一边进行加、减、乘、除，必须对另一边进行完全相同的操作。目标是把未知数单独留在等式一边。
If a = b, then a + c = b + c, a − c = b − c, a × c = b × c, a ÷ c = b ÷ c
若 a = b，则 a + c = b + c，a − c = b − c，a × c = b × c，a ÷ c = b ÷ c（c ≠ 0）
Inverse operations are used to undo operations around the unknown. Addition is undone by subtraction, subtraction by addition, multiplication by division, and division by multiplication.
逆运算用于消除未知数周围的运算。加法用减法消除，减法用加法消除，乘法用除法消除，除法用乘法消除。
| Operation in equation | 方程中的运算 | Inverse operation | 逆运算 |
| + 5 | − 5 |
| − 3 | + 3 |
| × 4 | ÷ 4 |
| ÷ 2 | × 2 |
4. One-Step Equations | 一步方程
In a one-step equation, only one operation separates the unknown from being alone. You apply one inverse operation to both sides and immediately find the solution.
在一步方程中，只有一个运算使未知数不孤立。你对两边应用一次逆运算，就能立即求出解。
- Solve x + 9 = 14: subtract 9 from both sides, so x = 14 − 9 = 5. 解 x + 9 = 14：两边减去 9，得到 x = 14 − 9 = 5。
- Solve x − 5 = 11: add 5 to both sides, so x = 11 + 5 = 16. 解 x − 5 = 11：两边加上 5，得到 x = 11 + 5 = 16。
- Solve 3x = 18: divide both sides by 3, so x = 18 ÷ 3 = 6. 解 3x = 18：两边除以 3，得到 x = 18 ÷ 3 = 6。
- Solve x/4 = 6: multiply both sides by 4, so x = 6 × 4 = 24. 解 x/4 = 6：两边乘以 4，得到 x = 6 × 4 = 24。
5. Two-Step Equations | 两步方程
For a two-step equation such as 2x + 3 = 11, first remove the added or subtracted term, then remove the coefficient of x by division. This order makes the working clear and reduces mistakes.
对于像 2x + 3 = 11 这样的两步方程，先处理加减项，再通过除法处理 x 的系数。这个顺序使步骤清晰并减少错误。
2x + 3 = 11 → 2x = 11 − 3 = 8 → x = 8 ÷ 2 = 4
Another example is 5x − 4 = 16. Add 4 to both sides to get 5x = 20, then divide by 5 to get x = 4.
另一个例子是 5x − 4 = 16。两边加上 4 得到 5x = 20，再除以 5 得到 x = 4。
For x/3 + 2 = 7, subtract 2 first, then multiply by 3:
对于 x/3 + 2 = 7，先减去 2，再乘以 3：
x/3 + 2 = 7 → x/3 = 5 → x = 5 × 3 = 15
6. Equations with Brackets | 带括号的方程
When an equation contains brackets, you can either expand the brackets first or divide both sides by the coefficient outside. Both methods are valid, but division first often gives smaller numbers.
当方程中含有括号时，你可以先展开括号，也可以两边先除以括号外的系数。两种方法都正确，但先除以系数通常会使数字更小。
Example: solve 2(x + 4) = 14.
例题：解 2(x + 4) = 14。
Method 1: expand first.
方法一：先展开。
2(x + 4) = 14 → 2x + 8 = 14 → 2x = 6 → x = 3
Method 2: divide first.
方法二：先除以 2。
2(x + 4) = 14 → x + 4 = 7 → x = 3
Always check that the bracket has been applied to every term inside when expanding.
展开时始终要检查括号是否作用于里面的每一项。
7. Unknowns on Both Sides | 未知数在两边
When the unknown appears on both sides of the equation, collect like terms first. Add or subtract the same unknown term from both sides so that x appears on only one side, then solve as usual.
当未知数出现在方程两边时，先合并同类项。在两边同时加上或减去相同的未知项，使 x 只出现在一边，然后按常规步骤求解。
Example: solve 5x − 2 = 2x + 7.
例题：解 5x − 2 = 2x + 7。
5x − 2 = 2x + 7 → 5x = 2x + 9 → 3x = 9 → x = 3
Step by step: add 2 to both sides, then subtract 2x from both sides, then divide by 3.
分步说明：两边加上 2，然后两边减去 2x，再除以 3。
A common mistake is to move a term without changing its sign. Always use inverse operations, not ‘moving to the other side’ without reason.
常见错误是移项时没有改变符号。始终使用逆运算，而不是毫无理由地 ‘ 把项移到另一边 ‘。
8. Equations with Fractions and Decimals | 含分数和小数的方程
Equations with fractions or decimals can be solved by using the same balance principle. To clear a fraction, multiply both sides by the denominator. To clear a decimal, multiply both sides by a power of 10 if it makes the working easier.
含有分数或小数的方程可以用相同的天平原理求解。要消去分数，两边乘以分母。要消去小数，两边乘以 10 的幂，如果这样能使计算更简单的话。
Example: solve x/4 + 2 = 5.
例题：解 x/4 + 2 = 5。
x/4 + 2 = 5 → x/4 = 3 → x = 3 × 4 = 12
Example: solve 0.5x + 1 = 3.2.
例题：解 0.5x + 1 = 3.2。
0.5x + 1 = 3.2 → 0.5x = 2.2 → x = 2.2 ÷ 0.5 = 4.4
You can also multiply every term by 10 to get 5x + 10 = 32, then solve to get 5x = 22 and x = 4.4.
你也可以每一项乘以 10，得到 5x +
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