📚 Solving Linear Equations | 解一元一次方程
Page 116 of the Cambridge KS3 Mathematics textbook introduces solving linear equations — a core algebraic skill that forms the foundation for later work in graphs, formulae and problem solving. The page covers how to balance equations, use inverse operations, and handle equations with brackets or fractions. Mastery of this topic is essential for progressing to more complex algebra in IGCSE Mathematics.
剑桥 KS3 数学教材第 116 页介绍了一元一次方程的解法，这是代数中的核心技能，为后续的图像、公式和应用题学习奠定基础。该页内容涵盖如何保持等式的平衡、使用逆运算，以及如何处理含有括号或分数的方程。掌握该主题对于衔接 IGCSE 数学中更复杂的代数内容至关重要。
1. What Is a Linear Equation? | 什么是一元一次方程？
A linear equation is an equation that contains an unknown variable (usually x) raised only to the power of 1. There are no terms like x², x³, √x or 1/x. Its graph is always a straight line.
一元一次方程是只含有一个未知数（通常用 x 表示）且该未知数的最高次数为 1 的方程。方程中不会出现 x²、x³、√x 或 1/x 这样的项。它的图像总是一条直线。
Standard form: ax + b = c, where a, b, c are constants and a ≠ 0.
标准形式：ax + b = c，其中 a、b、c 为常数，且 a ≠ 0。
Examples: 3x + 5 = 11, 2(4 – x) = 10, ½x – 3 = 7. All of these can be solved by keeping the equation balanced and isolating x.
例如：3x + 5 = 11、2(4 – x) = 10、½x – 3 = 7。这些方程都可以通过保持等式平衡并分离 x 来求解。
2. The Balancing Method | 等式平衡法
An equation works like a pair of balanced scales. Whatever you do to one side, you must do exactly the same to the other side to keep them equal. This principle — using inverse operations — is the foundation of solving any linear equation.
方程就像一架平衡的天平。无论你对某一边做什么运算，另一边也必须做完全相同的运算，才能保持相等。这个原则——使用逆运算——是解所有一元一次方程的基础。
Inverse operations: addition ↔ subtraction, multiplication ↔ division, squaring ↔ square root. Always apply the inverse operation to undo the current step.
逆运算：加法与减法互为逆运算，乘法与除法互为逆运算，平方与开平方互为逆运算。始终用逆运算来消除当前的运算步骤。
| Operation | Inverse Operation |
| + | – |
| – | + |
| × | ÷ |
| ÷ | × |
若左侧为某运算，则在两边做其逆运算，保持平衡。
3. Solving One‑Step Equations | 解一步方程
One-step equations require only a single inverse operation to isolate x. For example, if x + 7 = 12, subtract 7 from both sides to get x = 5.
一步方程只需进行一次逆运算就能得到 x。例如对于 x + 7 = 12，两边同时减去 7，得到 x = 5。
For x – 4 = 9, add 4 to both sides: x = 13. For 3x = 18, divide both sides by 3: x = 6. For x/5 = 2, multiply both sides by 5: x = 10.
对于 x – 4 = 9，两边加 4：x = 13。对于 3x = 18，两边除以 3：x = 6。对于 x/5 = 2，两边乘 5：x = 10。
The key is to identify the operation acting on x and then apply its inverse. Always check your answer by substituting it back into the original equation.
关键是要找出作用于 x 的运算，然后施加其逆运算。始终通过将答案代回原方程进行检验。
4. Solving Two‑Step Equations | 解两步方程
When x has both a multiplier (or divisor) and an added or subtracted constant, you need two steps. Most students work from the outside in — reverse the order of operations: undo addition/subtraction first, then undo multiplication/division.
当 x 既有乘数（或除数）又有加减的常数时，就需要两步求解。多数学生遵循从外向内的原则——逆向运算顺序：先消去加法或减法，再消去乘法或除法。
Example: 2x + 3 = 15. Step 1: subtract 3 from both sides → 2x = 12. Step 2: divide both sides by 2 → x = 6.
例子：2x + 3 = 15。第一步：两边减去 3 → 2x = 12。第二步：两边除以 2 → x = 6。
For 7 – x/3 = 4, first subtract 7 from both sides: –x/3 = –3. Then multiply both sides by –3: x = 9. When the coefficient or constant is negative, keep careful track of signs.
对于 7 – x/3 = 4，首先两边减 7：–x/3 = –3。然后两边乘 –3：x = 9。当系数或常数为负数时，务必仔细留意符号。
5. Equations with the Unknown on Both Sides | 未知数在等式两边的方程
Some linear equations have the variable term on both sides of the equals sign. The strategy is to remove the smaller variable term first, then solve the resulting one- or two-step equation.
有些一元一次方程在等号两边都含有未知数项。此时的方法是先消去较小的未知数项，然后求解得到的一步或两步方程。
Example: 5x – 4 = 2x + 8. Subtract 2x from both sides → 3x – 4 = 8. Add 4 to both sides → 3x = 12. Divide by 3 → x = 4.
例子：5x – 4 = 2x + 8。两边减去 2x → 3x – 4 = 8。两边加 4 → 3x = 12。除以 3 → x = 4。
If the equation is 3 – 4x = 2x + 9, adding 4x to both sides collects the x terms on the right: 3 = 6x + 9. Then subtract 9: –6 = 6x. Divide by 6: x = –1.
如果是 3 – 4x = 2x + 9，两边加上 4x 可将 x 项集中到右边：3 = 6x + 9。然后减去 9：–6 = 6x。除以 6：x = –1。
6. Expanding Brackets First | 先展开括号
When the equation contains brackets, the first job is to expand them using the distributive law. Then simplify by collecting like terms before applying the balancing method.
当方程中含有括号时，第一步是使用分配律展开括号。然后通过合并同类项进行化简，再运用平衡法求解。
Example: 3(2x – 1) = 15. Expand to 6x – 3 = 15. Add 3 → 6x = 18. Divide by 6 → x = 3.
例子：3(2x – 1) = 15。展开得 6x – 3 = 15。加 3 → 6x = 18。除以 6 → x = 3。
For 4(5 – x) = 2(x + 1), expand both: 20 – 4x = 2x + 2. Add 4x to both sides: 20 = 6x + 2. Subtract 2: 18 = 6x. x = 3.
对于 4(5 – x) = 2(x + 1)，两边展开：20 – 4x = 2x + 2。两边加 4x：20 = 6x + 2。减 2：18 = 6x。x = 3。
7. Equations Involving Fractions | 含分数的方程
Fraction equations may look tricky, but one powerful technique is to multiply every term by the lowest common denominator (LCD) to eliminate fractions immediately. This transforms the equation into an integer form, which is much easier to solve.
含分数的方程看起来棘手，但一个有效的方法是用最小公分母 (LCD) 乘以每一项，立即消去分数。这便将方程转化为整数形式，求解变得容易得多。
Example: (x+1)/3 = 5. Multiply both sides by 3: x + 1 = 15. x = 14.
例子：(x+1)/3 = 5。两边乘 3：x + 1 = 15。x = 14。
Example: x/2 + x/3 = 5. The LCD of 2 and 3 is 6. Multiply every term by 6: 3x + 2x = 30 → 5x = 30 → x = 6.
例子：x/2 + x/3 = 5。2 和 3 的最小公分母是 6。每一项乘 6：3x + 2x = 30 → 5x = 30 → x = 6。
If a fraction has a binomial numerator like (2x – 1)/4, treat the whole numerator as a quantity. Multiply both sides by 4 first: 2x – 1 = 4 × 3 = 12. Then solve.
如果分数的分子是二项式，例如 (2x – 1)/4，则将整个分子视为一个整体。先两边乘 4：2x – 1 = 4 × 3 = 12。然后求解。
8. Checking Your Solution | 检验答案
In mathematics, a solution is not complete until it has been verified. Substitute your found value back into the original equation (the one before any steps were taken) and confirm that both sides are equal.
在数学中，不经过检验的解是不完整的。把你求得的值代回原方程（未做任何操作之前的那个方程），确认两边相等。
For x = 6 in 2x + 3 = 15: left side = 2(6) + 3 = 15, right side = 15. ✓
对于 2x + 3 = 15 中 x = 6：左边 = 2(6) + 3 = 15，右边 = 15。✓
Checking catches algebraic mistakes and builds confidence. On exams, it is a powerful way to gain marks — even a quick mental check can show whether your answer is reasonable.
检验可以发现代数运算的错误，并增强解题信心。在考试中，这是获取分数的一种有效方式——即使是快速的心算检验，也能看出答案是否合理。
9. Common Errors to Avoid | 常见错误分析
Even strong students make predictable mistakes when solving linear equations. Being aware of these can help you avoid losing marks.
即使程度不错的学生在解一元一次方程时也会犯一些常见的错误。了解这些错误有助于避免失分。
- Forgetting to apply an operation to both sides — e.g. only adding 3 to the left side.
忘记对两边同时运算——例如只在左边加了 3。 - Mishandling negative coefficients — e.g. –x = 5 leads to x = –5, not x = 5.
错误处理负系数——例如 –x = 5 应该得出 x = –5，而不是 x = 5。 - Incorrect expansion of brackets — e.g. 3(2x – 5) becomes 6x – 5 instead of 6x – 15.
括号展开错误——例如 3(2x – 5) 变成 6x – 5，而正确的应是 6x – 15。 - Adding unlike terms — e.g. 3x + 2 treated as 5x.
合并了不同类项——例如把 3x + 2 当作 5x。
Always write each step clearly and line up your work vertically. This makes it easier to spot mistakes.
务必清晰地写出每一步，并把演算内容垂直对齐排版。这样更容易发现错误。
10. Word Problems Leading to Linear Equations | 由实际问题建立一元一次方程
Many real-world problems can be expressed as linear equations. The challenge is to translate English statements into algebraic language. Identify the unknown, assign a variable, and build an equation from the given relationships.
许多实际问题都可以用一元一次方程来表示。难点在于将文字描述转化为代数语言。找出未知量，设一个变量，然后根据所给关系建立方程。
Example: “Three times a number plus 7 equals 22.” Let the number be n. Then 3n + 7 = 22 → 3n = 15 → n = 5.
例子：「一个数的 3 倍加 7 等于 22。」设这个数为 n。则有 3n + 7 = 22 → 3n = 15 → n = 5。
Example: “The perimeter of a rectangle is 36 cm. Its length is 4 cm more than its width. Find the dimensions.” Let width = w, length = w + 4. Perimeter: 2w + 2(w + 4) = 36 → 4w + 8 = 36 → w = 7, length = 11.
例子：「一个长方形的周长是 36 cm，长比宽多 4 cm。求长和宽。」设宽为 w，则长为 w + 4。周长：2w + 2(w + 4) = 36 → 4w + 8 = 36 → w = 7，长 = 11。
Always define your variable clearly and ensure the equation reflects exactly what the problem states.
务必清晰地定义变量，并确保方程准确反映题目所给的描述。
11. Linking to Graphs | 与图像的关联
Every linear equation y = mx + c graphs as a straight line. The solution to an equation like 2x + 1 = 5 can also be seen as the x‑coordinate where the line y = 2x + 1 intersects the horizontal line y = 5. This visual link helps reinforce the meaning of solving an equation.
每一个线性方程 y = mx + c 的图像都是一条直线。方程 2x + 1 = 5 的解也可以理解为直线 y = 2x + 1 与水平线 y = 5 交点的 x 坐标。这种可视化联系有助于加深对解方程意义的理解。
By plotting both sides of an equation on the same coordinate grid, you can estimate solutions even without exact algebra. This is especially useful when checking work or when equations become more complex.
通过在同一坐标系中画出方程两边的图像，你即使不进行精确的代数计算，也可以估算出解。这在检验答案或当方程变得更复杂时特别有用。
12. Practice Set from Page 116 | 来自第 116 页的练习
The exercises on page 116 are designed to build fluency step by step. They move from simple one-step equations to those demanding expansion and handling of fractions. Consistent practice with these problems solidifies the balancing technique.
第 116 页的练习旨在逐步提高熟练度，从简单的一步方程过渡到需要展开和处理分数的方程。持续练习这些题目能巩固平衡法技巧。
Try these examples:
请尝试以下例题：
a) 4x – 7 = 13
b) 5(2 – x) = 3x + 2
c) (x+2)/5 – 1 = 3
d) 2x/3 + x/4 = 11
Solutions: a) x = 5, b) x = 1, c) x = 18, d) x = 12. Use the methods from the sections above to verify each step.
答案：a) x = 5，b) x = 1，c) x = 18，d) x = 12。请用上面各节的方法逐步验证。
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