📚 Solving Linear Equations: One-Step and Two-Step Methods | 解一元一次方程：一步与两步方法
tes, graphs, simultaneous equations and functions - depends on them. This article explains what a linear equation is, teaches the balance method, works carefully through one-step and two-step examples such as x + 5 = 12, 4x − 3 = 17 and 2(x + 1) = 10, and finishes with common mistakes plus exam-style practice. By the end, you will be able to solve one-step and two-step equations confidently.
一元一次方程是剑桥 KS3 数学中最重要的代数主题之一，之后的公式变形、坐标、图像、联立方程与函数几乎都依赖它。本文讲解什么是一元一次方程，介绍天平法，详细演算诸如 x + 5 = 12、4x − 3 = 17 和 2(x + 1) = 10 这样的方程，最后指出常见错误并附考试风格练习。学完后，你将能自信地解一步和两步方程。
1. What is a Linear Equation? | 什么是一元一次方程？
A linear equation is an equation in which the variable, usually x, appears only to the power of 1. There are no terms such as x², x³ or 1/x. The word "linear" comes from the fact that the graph of such an equation is a straight line.
一元一次方程是指变量（通常为 x）的指数只为 1 的方程。方程中没有 x²、x³ 或 1/x 这样的项。"一次"这个名字来自这类方程对应的图像是一条直线。
variable is squared, and 1/x = 5 is not linear because the variable is in the denominator.
例如，x + 7 = 15 和 3x − 2 = 13 都是一元一次方程。方程 2x² = 8 不是一次方程，因为变量被平方；1/x = 5 也不是一次方程，因为变量出现在分母中。
type normally have exactly one solution, and we write the answer in the form x = a number.
解方程就是求出使左边等于右边的 x 值。在剑桥 KS3 中，这类方程通常恰好有一个解，我们把答案写成 x = 某个数 的形式。
x + 7 = 15 → x = 8 | 3x − 2 = 13 → x = 5
2. The Balance Method | 天平法
Think of an equation as a perfectly balanced pair of scales. The left-hand side sits on one pan and the right-hand side sits on the other. If you add, subtract, multiply or divide on one side only, the scales tip and the equation is no longer true.
把方程想象成一台完全平衡的天平。左边放在一个托盘上，右边放在另一个托盘上。如果你只对一边做加、减、乘、除，天平就会倾斜，等式不再成立。
The golden rule of the balance method is: whatever you do to one side, you must do exactly the same to the other side. This keeps the two sides equal, so every new equation you write has the same solution as the original one.
天平法的黄金法则是：你对一边做什么运算，就必须对另一边做完全相同的运算。这样两边始终保持相等，因此你写下的每一个新方程都与原方程有相同的解。
To free the variable, we use inverse operations. Addition is undone by subtraction, subtraction by addition, multiplication by division, and division by multiplication. Solving an equation is simply a process of peeling off these operations one at a time until x stands alone.
为了把变量"解放"出来，我们使用逆运算。加法用减法抵消，减法用加法抵消，乘法用除法抵消，除法用乘法抵消。解方程的过程，就是把这些运算一层层剥离，直到 x 独自留下。
x + 5 = 12 → x + 5 − 5 = 12 − 5 → x = 7
3. Key Vocabulary and Notation | 关键术语与符号
Before solving, make sure you understand the words examiners use. Being able to name the parts of an equation helps you explain your reasoning and read questions accurately.
在动手解题之前，先确认你理解考官使用的术语。能够说出方程各部分的名字，有助于你解释思路并准确读懂题目。
| Term | 术语 | Meaning | 含义 | Example | 例子 |
| Variable 变量 | The unknown letter 未知的字母 | x |
| Coefficient 系数 | The number multiplying the variable 与变量相乘的数 | 4 in 4x |
| Constant 常数项 | A number on its own 单独的数字 | −3 in 4x − 3 |
| Term 项 | A part separated by + or − 被加减号分开的部分 | 4x and −3 |
| LHS / RHS 左式 / 右式 | Left-hand side and right-hand side 等号左右两边 | 4x − 3 = 17 |
| Solution / root 解 / 根 | The value of x that works 使等式成立的 x 值 | x = 5 |
| Inverse operation 逆运算 | The operation that undoes another 抵消另一种运算的运算 | + and −, × and ÷ |
Notice that "solve" and "evaluate" are different instructions. Solve means find x; evaluate means substitute a value and work out the number. Always check which one the question is asking for.
注意 "solve（解）" 和 "evaluate（求值）" 是两种不同的指令。Solve 意为求 x；evaluate 意为代入一个数值并算出结果。做题时一定要看清题目问的是哪一个。
4. One-Step Equations: Addition and Subtraction | 解一步方程：加与减
A one-step equation needs only a single operation to isolate the variable. When a number is added to x, we subtract that number from both sides. When a number is subtracted from x, we add it to both sides.
一步方程只需一次运算就能把变量孤立出来。当 x 加上某个数时，我们在两边同时减去这个数；当 x 减去某个数时，我们在两边同时加上这个数。
Example 1: solve x + 5 = 12. Subtract 5 from both sides: x + 5 − 5 = 12 − 5, so x = 7.
例 1：解 x + 5 = 12。两边同时减 5：x + 5 − 5 = 12 − 5，所以 x = 7。
x + 5 = 12 → x = 12 − 5 → x = 7
Example 2: solve x − 4 = 9. Add 4 to both sides: x − 4 + 4 = 9 + 4, so x = 13.
例 2：解 x − 4 = 9。两边同时加 4：x − 4 + 4 = 9 + 4，所以 x = 13。
x − 4 = 9 → x = 9 + 4 → x = 13
may sit on the right. Subtract 3 from both sides to get 4 = x, which is the same as x = 4.
例 3：解 7 = x + 3。变量也可能出现在右边。两边同时减 3，得到 4 = x，也就是 x = 4。
me rule. For x + 2.5 = 6, subtract 2.5 to get x = 3.5. For x − ½ = 2, add ½ to get x = 2½.
小数与分数遵循完全相同的规则。对于 x + 2.5 = 6，两边减 2.5 得 x = 3.5；对于 x − ½ = 2，两边加 ½ 得 x = 2½。
5. One-Step Equations: Multiplication and Division | 解一步方程：乘与除
When the variable is multiplied by a coefficient, divide both sides by that coefficient. When the variable is divided by a number, multiply both sides by that number.
当变量被某个系数相乘时，两边同时除以这个系数；当变量被某个数除时，两边同时乘以这个数。
Example 1: solve 3x = 15. Divide both sides by 3: 3x ÷ 3 = 15 ÷ 3, so x = 5.
例 1：解 3x = 15。两边同时除以 3：3x ÷ 3 = 15 ÷ 3，所以 x = 5。
3x = 15 → x = 15 ÷ 3 → x = 5
Example 2: solve x ÷ 4 = 7. Multiply both sides by 4: x = 7 × 4 = 28.
例 2：解 x ÷ 4 = 7。两边同时乘 4：x = 7 × 4 = 28。
x ÷ 4 = 7 → x = 7 × 4 → x = 28
the halving, so x = 12. Equivalently, divide both sides by ½, which gives the same result.
例 3：解 ½x = 6。乘以 2 可抵消取半，所以 x = 12。也可以两边同时除以 ½，结果相同。
3 ÷ 5 = 0.6. Answers are allowed to be fractions or decimals; do not force a whole number.
例 4：解 5x = 3。此时答案不是整数：x = 3 ÷ 5 = 0.6。答案可以是分数或小数，不要强行凑整数。
ront of x tells you what to divide by. In 6x = 42 the coefficient is 6, so x = 42 ÷ 6 = 7.
这里的要点很简单：x 前面的系数就是你要除的数。在 6x = 42 中系数是 6，所以 x = 42 ÷ 6 = 7。
6. Worked Example: x + 5 = 12 | 例题详解：x + 5 = 12
Let us write out a full solution in the style expected in a Cambridge KS3 answer booklet. Layout matters as much as the arithmetic, because marks are often given for showing each step.
下面按剑桥 KS3 答题册要求的形式写出完整解答。书写格式与计算同样重要，因为每一步通常都有分数。
Step 1 - Write the equation: x + 5 = 12.
第 1 步 - 写出方程：x + 5 = 12。
verse operation. Since 5 is added to x, we subtract 5 from both sides: x + 5 − 5 = 12 − 5.
第 2 步 - 确定逆运算。因为 x 上加的是 5，所以两边同时减 5：x + 5 − 5 = 12 − 5。
Step 3 - Simplify both sides: x = 7.
第 3 步 - 化简两边：x = 7。
the original equation: LHS = 7 + 5 = 12, RHS = 12. Since LHS = RHS, the answer is correct.
第 4 步 - 代入检验。把原方程中的 x 换成 7：左边 = 7 + 5 = 12，右边 = 12。因为左边 = 右边，所以答案正确。
x + 5 = 12 → x = 7 check: 7 + 5 = 12 ✓
x + 5 = −12, giving x = −17. The method never changes; only the arithmetic becomes harder.
常见的变式是常数项为负或数字更大，例如 x + 5 = −12，则 x = −17。方法始终不变，只是计算变得更难。
7. Two-Step Equations: Undoing Operations in Reverse Order | 解两步方程：按相反顺序抵消运算
ied by 4 and then 3 is subtracted. To solve, we undo those operations in the reverse order.
两步方程中，变量受到两种运算的作用。在 4x − 3 = 17 中，变量先乘以 4，再减去 3。解题时，我们要按相反的顺序把这些运算逐一抵消。
The order of operations (BIDMAS) says multiplication before addition or subtraction. Undoing therefore works backwards: undo the addition or subtraction first, then undo the multiplication or division. This is why the constant term is always dealt with before the coefficient.
运算顺序（BIDMAS）规定先乘除后加减。因此逆运算要倒着来：先抵消加减，再抵消乘除。这就是为什么总是先处理常数项，再处理系数。
4x − 3 = 17: add 3 to both sides to get 4x = 20, then divide both sides by 4 to get x = 5.
对于 4x − 3 = 17：两边同时加 3，得 4x = 20；再两边同时除以 4，得 x = 5。
4x − 3 = 17 → 4x = 20 → x = 5
ample x ÷ 3 + 4 = 9, we still subtract first: x ÷ 3 = 5, then multiply by 3 to get x = 15.
如果方程中是除法，例如 x ÷ 3 + 4 = 9，我们仍然先减：x ÷ 3 = 5，再乘以 3 得 x = 15。
A quick habit that saves marks: always write the new equation underneath the old one, keeping the equals signs aligned in a single column.
一个能省下分数的好习惯：把新方程写在原方程下面，让等号对齐在同一列上。
8. Worked Example: 4x − 3 = 17 | 例题详解：4x − 3 = 17
Step 1 - Write the equation: 4x − 3 = 17.
第 1 步 - 写出方程：4x − 3 = 17。
2 - Add 3 to both sides to remove the constant term: 4x − 3 + 3 = 17 + 3, giving 4x = 20.
第 2 步 - 两边同时加 3，消去常数项：4x − 3 + 3 = 17 + 3，得 4x = 20。
Step 3 - Divide both sides by the coefficient 4: 4x ÷ 4 = 20 ÷ 4, giving x = 5.
第 3 步 - 两边同时除以系数 4：4x ÷ 4 = 20 ÷ 4，得 x = 5。
Step 4 - Check: LHS = 4 × 5 − 3 = 20 − 3 = 17, RHS = 17. The solution is verified.
第 4 步 - 检验：左边 = 4 × 5 − 3 = 20 − 3 = 17，右边 = 17。解得到验证。
4x − 3 = 17 → 4x = 20 → x = 5 check: 4 × 5 − 3 = 17 ✓
hen divide by 4 to get 3 = x, so x = 3. Always finish by writing the variable on the left.
再看一个常数项在前面的变式：20 = 4x + 8。两边减 8 得 12 = 4x，再除以 4 得 3 = x，所以 x = 3。最后请习惯把变量写在左边。
6, so x = −3. Negative answers are perfectly normal and should be checked in the same way.
另一个变式是 2x + 7 = 1，答案是负数：2x = −6，所以 x = −3。负答案是正常的，同样要代入检验。
9. Equations with Brackets: 2(x + 1) = 10 | 含括号的方程：2(x + 1) = 10
When a bracket is multiplied by a number, there are two correct routes to the solution. Choosing the quicker one saves time in an exam, but both earn full marks if the working is clear.
当括号被一个数相乘时，有两条正确的解题路线。选择更快的那条可以节省考试时间，但只要步骤清晰，两种方法都能拿满分。
the bracket is multiplied by 2, divide both sides by 2: x + 1 = 5. Then subtract 1: x = 4.
方法 A - 先除。因为括号被 2 乘，所以两边同时除以 2：x + 1 = 5。再减 1：x = 4。
2(x + 1) = 10 → x + 1 = 5 → x = 4
x + 2 = 10. Subtract 2: 2x = 8. Divide by 2: x = 4. The answer is the same, as it must be.
方法 B - 先展开。把括号乘开：2x + 2 = 10。减 2：2x = 8。除以 2：x = 4。答案完全相同，也必须相同。
or example 3(x + 2) = 10, expanding is safer: 3x + 6 = 10, so 3x = 4 and x = 4 ÷ 3 ≈ 1.33.
当括号外的数能整除右边时，方法 A 通常更快。若不能整除，例如 3(x + 2) = 10，展开更稳妥：3x + 6
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