📚 Solving Linear Equations | 解一元一次方程
Linear equations are one of the most important topics in KS3 Cambridge Mathematics. They appear in algebra, word problems, and even geometry. This article explains how to solve one-variable linear equations step by step, with clear examples and common mistakes to avoid.
一元一次方程是剑桥 KS3 数学中最重要的主题之一。它们在代数、应用题甚至几何中都会出现。本文将逐步讲解如何解一元一次方程，并给出清晰的例子和需要避免的常见错误。
1. What Is an Equation? | 什么是方程？
ssion on the left, x + 3, has the same value as the expression on the right, 7. The letter x is called the unknown or variable. Solving the equation means finding the value of x that makes the statement true.
方程是表示两个表达式相等的数学语句，必须含有等号。例如 x + 3 = 7 就是一个方程。左边的表达式 x + 3 与右边的表达式 7 具有相同的值。字母 x 称为未知数或变量。解方程就是求出使等式成立的 x 的值。
In KS3, equations are usually linear. This means the unknown is not raised to a power like x² or x³. Linear equations form a straight line when graphed. The highest power of x is 1.
在 KS3 阶段，方程通常是一次的。这意味着未知数不会出现 x² 或 x³ 这样的幂。一次方程的图像是一条直线。x 的最高次数是 1。
x + 3 = 7
2. The Balance Method | 天平法
The balance method is a visual way to keep equations true. Imagine an old-fashioned scale. The left pan and right pan must stay level. If you add weight to one side, you must add the same weight to the other side.
天平法是一种直观保持方程成立的方法。想象一架旧式天平，左右两盘必须保持水平。如果你在一侧增加重量，就必须在另一侧也增加相同的重量。
In algebra, whatever operation you apply to one side, you must apply to the other. This could be adding, subtracting, multiplying, dividing, or expanding brackets. The goal is to isolate the unknown on one side of the equals sign.
在代数中，无论你对一边进行什么运算，都必须对另一边进行同样的运算。这些运算可以是加、减、乘、除或展开括号。目标是使未知数单独出现在等号的一边。
If x + 4 = 10, then x + 4 − 4 = 10 − 4
3. Solving One-Step Equations | 解一步方程
A one-step equation needs only one inverse operation. To undo addition, subtract. To undo subtraction, add. To undo multiplication, divide. To undo division, multiply. The table below shows common operations and their inverses.
一步方程只需要一次逆运算。要抵消加法，就用减法；要抵消减法，就用加法；要抵消乘法，就用除法；要抵消除法，就用乘法。下表列出了常见运算及其逆运算。
| Operation | Inverse Operation |
|---|---|
| + 5 | − 5 |
| − 7 | + 7 |
| × 3 | ÷ 3 |
| ÷ 4 | × 4 |
For addition: x + 5 = 12. Subtract 5 from both sides.
对于加法：x + 5 = 12。两边同时减去 5。
x + 5 = 12 → x = 12 − 5 → x = 7
For multiplication: 4x = 20. Divide both sides by 4.
对于乘法：4x = 20。两边同时除以 4。
4x = 20 → x = 20 ÷ 4 → x = 5
For division: x/3 = 6. Multiply both sides by 3.
对于除法：x/3 = 6。两边同时乘以 3。
x/3 = 6 → x = 6 × 3 → x = 18
4. Solving Two-Step Equations | 解两步方程
11. To solve, undo the addition first, then undo the multiplication. This follows the order of operations in reverse. The rules of BIDMAS are applied backwards.
两步方程中未知数包含两次运算。例如 2x + 3 = 11。求解时，先抵消加法，再抵消乘法。这相当于逆用运算顺序。BIDMAS 规则要反过来使用。
2x + 3 = 11
Step 1: Subtract 3 from both sides.
第一步：两边同时减去 3。
2x = 8
Step 2: Divide both sides by 2.
第二步：两边同时除以 2。
x = 4
Here is another example: 5x − 2 = 13.
再看一个例子：5x − 2 = 13。
5x − 2 = 13 → 5x = 15 → x = 3
5. Equations with Brackets | 含括号的方程
ombine like terms if needed. After expanding, solve the equation using the balance method.
当方程含有括号时，先展开括号。使用分配律：a(b + c) = ab + ac。然后根据需要合并同类项。展开后，用天平法解方程。
2(x + 3) = 10
and the bracket: 2x + 6 = 10. Then subtract 6 from both sides: 2x = 4. Divide by 2: x = 2.
展开括号：2x + 6 = 10。然后两边同时减去 6：2x = 4。除以 2：x = 2。
f the bracket, remember to change the sign of every term inside. Example: 3 − (x + 1) = 1.
如果括号前有负号，记得改变括号内每一项的符号。例：3 − (x + 1) = 1。
3 − (x + 1) = 1 → 3 − x − 1 = 1 → 2 − x = 1 → x = 1
6. Unknown on Both Sides | 未知数在两边
If the unknown appears on both sides, collect like terms first. Move the smaller unknown term to the other side by using inverse operations. Always do the same to both sides of the equation. This keeps the equation balanced.
如果未知数出现在两边，首先合并同类项。将较小的未知数项移到另一边，使用逆运算。两边必须进行相同操作。这样才能保持方程平衡。
5x = 2x + 9
Subtract 2x from both sides: 3x = 9. Then divide by 3: x = 3.
两边同时减去 2x：3x = 9。然后除以 3：x = 3。
from both sides: 2x + 4 = 10. Then subtract 4 from both sides: 2x = 6. Divide by 2: x = 3.
对于 3x + 4 = x + 10，两边同时减去 x：2x + 4 = 10。再两边同时减去 4：2x = 6。除以 2：x = 3。
7. Fractions in Equations | 方程中的分数
ere are two fractions, find a common denominator or multiply by the least common multiple.
要去掉分数，两边同时乘以分母。对于 x/2 = 6，两边乘以 2：x = 12。如果有两个分数，先找公分母或乘以最小公倍数。
x/2 = 6 → x = 6 × 2 → x = 12
For x/3 + 1 = 5, first subtract 1 from both sides, then multiply by 3.
对于 x/3 + 1 = 5，先两边同时减去 1，再两边乘以 3。
x/3 + 1 = 5 → x/3 = 4 → x = 12
For (x + 2)/3 = 4, multiply both sides by 3: x + 2 = 12. Then subtract 2: x = 10.
对于 (x + 2)/3 = 4，两边同时乘以 3：x + 2 = 12。然后减去 2：x = 10。
8. Checking Your Answer | 检验答案
Always substitute your solution back into the original equation. Replace the unknown with your value. If the left side equals the right side, your answer is correct. If not, check your working carefully.
永远将解代回原方程进行检验。用求得的值替换未知数。如果左边等于右边，答案就是正确的。如果不是，请仔细检查计算过程。
Example: Equation is 2x + 3 = 11. Solution is x = 4.
示例：方程为 2x + 3 = 11，解为 x =
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