📚 Mastering Linear Equations | 掌握一元一次方程
Linear equations are one of the most important building blocks in KS3 Cambridge Mathematics. They appear in almost every topic, from geometry problems to real-life word questions, and they prepare you for harder algebra in IGCSE. A strong understanding of how to solve linear equations will help you work faster and more accurately in exams.
一元一次方程是剑桥中学数学 KS3 阶段最重要的基础内容之一。它们几乎出现在每个主题中，从几何问题到现实生活应用题，为你进入 IGCSE 更难的代数学习做好准备。扎实掌握一元一次方程的解法，能让你在考试中做得更快、更准。
1. What is a Linear Equation? | 什么是一元一次方程？
n. The equation can be written in forms such as x + 3 = 7, 2x - 5 = 9, or 4x + 1 = 2x + 9.
一元一次方程是指未知数（通常用 x 表示）的指数仅为 1 的方程。基本的一元一次方程中没有 x²、x³ 或平方根等项。方程可以写成 x + 3 = 7、2x - 5 = 9 或 4x + 1 = 2x + 9 这样的形式。
In KS3, you are expected to solve these equations by finding the value of the unknown that makes the equation true. This value is called the solution or root of the equation. Solving an equation means isolating the variable on one side of the equals sign.
在 KS3 阶段，你需要通过求出使方程成立的未知数的值来解方程。这个值叫做方程的解或根。解方程意味着把变量单独放在等号的一边。
2. The Balance Method | 天平法
Think of an equation as a balance scale: the left side and the right side must always weigh the same. If you add, subtract, multiply, or divide one side, you must do exactly the same to the other side to keep the equation balanced.
把方程想象成一个天平：左边和右边必须始终保持平衡。如果你在一边加、减、乘或除，就必须在另一边做完全相同的运算，才能保持方程平衡。
left side is currently 4 more than x. To restore balance and find x, subtract 4 from both sides.
这个思路叫做天平法。它是所有解方程方法的基础，因为它能在保持等式成立的同时，帮助你分离出未知数。例如，如果 x + 4 = 10，左边目前比 x 多 4。为了恢复平衡并求出 x，需要两边同时减去 4。
x + 4 - 4 = 10 - 4 → x = 6
3. Solving One-Step Equations | 解一步方程
A one-step equation needs only one inverse operation to solve. For addition, subtract; for subtraction, add; for multiplication, divide; for division, multiply. The table below shows the four basic types of one-step equations and how to undo each operation.
一步方程只需要一个逆运算就能解出来。加法要用减法来解；减法要用加法；乘法要用除法；除法要用乘法。下表展示了一步方程的四种基本类型以及如何逆运算。
| Type | 类型 | Example | 示例 | Inverse Operation | 逆运算 |
|---|---|---|
| Addition | 加法 | x + 5 = 12 | Subtract 5 | 减去 5 |
| Subtraction | 减法 | x - 3 = 8 | Add 3 | 加上 3 |
| Multiplication | 乘法 | 5x = 20 | Divide by 5 | 除以 5 |
| Division | 除法 | x ÷ 4 = 3 | Multiply by 4 | 乘以 4 |
f the balance. For example, in 5x = 20, divide both sides by 5: 5x ÷ 5 = 20 ÷ 5, so x = 4.
一定要在两边清楚地写出运算，这样就不会失去平衡的线索。例如，在 5x = 20 中，两边同时除以 5：5x ÷ 5 = 20 ÷ 5，所以 x = 4。
4. Solving Two-Step Equations | 解两步方程
A two-step equation involves two operations, such as multiplication and addition. To solve it, you must undo the operations in the correct order: reverse the addition or subtraction first, then reverse the multiplication or division. This is the reverse of BIDMAS.
两步方程包含两个运算，例如乘法和加法。解这类方程时，必须按正确的顺序进行逆运算：先处理加法或减法，再处理乘法或除法。这就是 BIDMAS 的逆向顺序。
he x is multiplied by 2, and then 3 is added. To undo, subtract 3 first, then divide by 2.
考虑 2x + 3 = 11。x 先乘以 2，然后加上 3。要逆运算，先减去 3，再除以 2。
2x + 3 - 3 = 11 - 3 → 2x = 8 → 2x ÷ 2 = 8 ÷ 2 → x = 4
7 = 14, add 7 first because the 7 is subtracted. Then divide both sides by 3 to get x = 7.
对于 3x - 7 = 14，先加上 7，因为 7 是被减去的。然后两边除以 3，得到 x = 7。
3x - 7 = 14 → 3x = 21 → x = 7
It can help to write each new line directly below the previous one and keep the equals signs lined up. This makes your working easy to follow and reduces mistakes.
把每一步新算式直接写在上一步下面，并对齐等号，会很有帮助。这样书写清晰易懂，也能减少错误。
5. Equations with Brackets | 含括号的方程
ckets first using the distributive law. For example, 3(x + 2) = 18 expands to 3x + 6 = 18.
当方程中含有括号时，最可靠的方法是先用乘法分配律展开括号。例如，3(x + 2) = 18 展开后得到 3x + 6 = 18。
the two-step equation: subtract 6 from both sides and divide by 3. The solution is x = 4.
然后解两步方程：两边减去 6，再除以 3。解为 x = 4。
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