📚 Solving Linear Equations | 解一元一次方程
Linear equations are the foundation of algebra at KS3. They appear in almost every Cambridge Lower Secondary Mathematics assessment, and the skills you build here will be used again in graphs, formulae and problem-solving. This article explains how to solve linear equations step by step, with clear methods and common pitfalls.
一元一次方程是 KS3 代数的基础。它们几乎出现在每一次剑桥初中数学测评中，你在这里建立的技能将再次用于图像、公式和问题解决。本文逐步讲解如何解一元一次方程，并给出清晰的方法和常见误区。
1. What Are Linear Equations? | 什么是一元一次方程？
ms like x², x³ or 1/x. The general form is ax + b = c, where a, b and c are known numbers.
一元一次方程是指未知数（通常写作 x）只以 1 次幂出现的方程。这意味着方程中没有 x²、x³ 或 1/x 这样的项。一般形式为 ax + b = c，其中 a、b 和 c 为已知数。
11 and 5(x + 2) = 35. Solving means finding every value of x that makes the equation true.
在 KS3 阶段你会遇到诸如 3x + 5 = 14、2x − 7 = 11 和 5(x + 2) = 35 这样的方程。解方程意味着求出使方程成立的每一个 x 值。
- 3x + 5 = 14 → linear, because x has power 1；3x + 5 = 14 → 一元一次，因为 x 的指数为 1
- x² + 5 = 14 → not linear, because x has power 2；x² + 5 = 14 → 不是一元一次，因为 x 的指数为 2
- 1/x + 5 = 14 → not linear, because x is in the denominator；1/x + 5 = 14 → 不是一元一次，因为 x 在分母中
2. Balancing Method: The Golden Rule | 天平法：黄金法则
An equation behaves like a balance scale. Whatever you do to one side, you must also do to the other side. This keeps the equation true. You can add, subtract, multiply or divide both sides by the same non-zero number.
方程就像一个天平。你对一边做的任何操作，也必须对另一边做同样的操作。这样才能保持方程成立。你可以在等号两边同时加、减、乘或除以同一个非零数。
If a = b, then a + c = b + c and a × c = b × c
This rule is the foundation of every method in this article. If you forget it, the equation will lose its balance and the solution will be wrong.
这条规则是本文所有方法的基础。如果你忘记它，方程就会失去平衡，解也会出错。
3. Solving Equations by Adding or Subtracting | 通过加减解方程
example, in the equation x + 7 = 15, the + 7 is added to x, so subtract 7 from both sides.
为了单独得到 x，需要消去与 x 相连的加法或减法。例如，在方程 x + 7 = 15 中，+ 7 加在 x 上，所以两边同时减去 7。
x + 7 = 15
x + 7 − 7 = 15 − 7
x = 8
The answer is x = 8. You can check it by replacing x with 8: 8 + 7 = 15, which is correct.
答案是 x = 8。你可以用 8 代替 x 来检验：8 + 7 = 15，结果正确。
equation is x − 9 = 6, add 9 to both sides because −9 is attached to x. This gives x = 15.
如果方程是 x − 9 = 6，因为 −9 与 x 相连，所以两边同时加上 9，得到 x = 15。
4. Solving Equations by Multiplying or Dividing | 通过乘除解方程
ite operation. For 5x = 30, divide both sides by 5. For x/4 = 3, multiply both sides by 4.
当 x 被乘以或除以一个数时，就使用相反的运算。对于 5x = 30，两边同时除以 5；对于 x/4 = 3，两边同时乘以 4。
5x = 30
5x ÷ 5 = 30 ÷ 5
x = 6
x/4 = 3
x/4 × 4 = 3 × 4
x = 12
negative coefficients, such as −3x = 12, where you divide both sides by −3 to get x = −4.
始终在两边使用相同的运算，并且永远不要除以零。这个方法也适用于负系数，例如 −3x = 12，两边除以 −3 得到 x = −4。
5. Equations with Unknowns on Both Sides | 未知数在等号两边的方程
o collect all x terms on one side and the numbers on the other. A good strategy is to remove the smaller x term first.
有些方程的等号两边都有 x，例如 5x + 2 = 3x + 10。第一步是把所有含 x 的项移到一边，把数字移到另一边。一个好的策略是先去消去较小的 x 项。
5x + 2 = 3x + 10
5x − 3x + 2 = 3x − 3x + 10
2x + 2 = 10
2x + 2 − 2 = 10 − 2
2x = 8
x = 4
In this example, subtracting 3x from both sides removed the x term on the right. Then the equation became simple enough to solve in two more steps.
在这个例子中，两边同时减去 3x 消去了右边的 x 项。然后方程变得足够简单，再经过两步就可以解出。
6. Equations Involving Brackets | 含括号的方程
ckets first. This means multiplying each term inside the bracket by the number outside it.
当方程含有括号时，例如 3(x + 2) = 18，首先要展开括号。这意味着用括号外的数乘以括号内的每一项。
3(x + 2) = 18
3x + 6 = 18
3x + 6 − 6 = 18 − 6
3x = 12
x = 4
+ 1) = 3, expand −2(x + 1) to get −2x − 2, not −2x + 2. Then collect like terms and solve.
注意负号。在方程 5 − 2(x + 1) = 3 中，展开 −2(x + 1) 得到 −2x − 2，而不是 −2x + 2。然后合并同类项并求解。
7. Equations with Fraction Coefficients | 含分数系数的方程
move the constant term on the same side as the fraction, then multiply by the denominator.
如果 x 出现在分数中，例如 x/3 + 2 = 7，先消去与分数同一边的常数项，然后乘以分母。
x/3 + 2 = 7
x/3 + 2 − 2 = 7 − 2
x/3 = 5
x/3 × 3 = 5 × 3
x = 15
u can also multiply by the reciprocal of 2/5, but writing every step helps avoid mistakes.
对于像 2x/5 = 8 这样的方程，先两边乘以 5，然后除以 2。这样得到 2x = 40，然后 x = 20。你也可以乘以 2/5 的倒数，但写出每一步有助于避免错误。
8. Checking Your Solution | 检验你的解
ample, if you solve 3x + 5 = 14 and get x = 3, check by calculating 3(3) + 5 = 9 + 5 = 14.
求出解后，总要把它代回原方程。两边必须得到相同的值。例如，如果你解 3x + 5 = 14 得到 x = 3，就通过计算 3(3) + 5 = 9 + 5 = 14 来检验。
This habit is especially useful in exams because it catches sign errors and arithmetic slips before you move on. If the check fails, go back and look at the steps one by one.
这个习惯在考试中特别有用，因为它能在你继续之前发现符号错误和算术失误。如果检验失败，就回去逐步检查。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
One common mistake is applying an operation to only one side of the equation. Remember the balance rule: both sides must change in the same way.
一个常见错误是只对等式的一边进行某种运算。记住天平规则：两边必须进行同样的变化。
| Mistake | Correction |
| x + 5 = 12 → x = 12 + 5 | x + 5 = 12 → x = 12 − 5 |
| 3x = 18 → x = 18 × 3 | 3x = 18 → x = 18 ÷ 3 |
| 2(x + 4) = 10 → 2x + 4 = 10 | 2(x + 4) = 10 → 2x + 8 = 10 |
Another common error is losing a negative sign when moving terms across the equals sign. Write each line carefully and check that every sign stays with the correct term.
另一个常见错误是在把项移过等号时丢失负号。仔细书写每一行，并检查每个符号是否与正确的项保持在一起。
10. Exam-Style Practice and Worked Examples | 考试风格练习与例题
Worked example 1: Solve 7 − 2x = 1.
例题 1：解方程 7 − 2x = 1。
7 − 2x = 1
7 − 7 − 2x = 1 − 7
−2x = −6
−2x ÷ (−2) = −6 ÷ (−2)
x = 3
Worked example 2: Solve 4(x − 3) = 2x + 6.
例题 2：解方程 4(x − 3) = 2x + 6。
4(x − 3) = 2x + 6
4x − 12 = 2x + 6
4x − 2x − 12 = 2x − 2x + 6
2x − 12 = 6
2x − 12 + 12 = 6 + 12
2x = 18
x = 9
These examples show that a clear sequence of inverse operations leads to the solution. Practise similar questions until the steps feel automatic, and always check your answer in the original equation.
这些例子表明，清晰的逆运算顺序可以得到解。练习类似的题目直到步骤变得自然，并且总是把答案代回原方程进行检验。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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