📚 Solving Linear Equations | 解一元一次方程
Welcome to this Cambridge KS3 Mathematics revision guide on solving linear equations. A linear equation is one of the most important building blocks in algebra, and it appears throughout the Cambridge Lower Secondary curriculum. In this article, you will learn how to balance equations, apply inverse operations, remove brackets, handle fractions, and solve word problems with confidence.
欢迎阅读这篇剑桥 KS3 数学复习指南，主题是解一元一次方程。一元一次方程是代数中最重要的基础模块之一，贯穿剑桥初中数学课程。在本文中，你将学习如何平衡方程、使用逆运算、去括号、处理分数，并自信地解决文字应用题。
1. What is a Linear Equation? | 什么是一元一次方程？
ally be written in the form ax + b = c, where a, b and c are numbers and x is the unknown.
一元一次方程是指未知数的最高次数为 1 的方程。它通常可以写成 ax + b = c 的形式，其中 a、b、c 是数字，x 是未知数。
d 3x − 4 = 11 are linear equations because the variable x appears only to the first power.
例如，x + 5 = 12 和 3x − 4 = 11 都是一元一次方程，因为变量 x 只出现一次方。
x + 5 = 12
3x − 4 = 11
2. Balancing Method: The Golden Rule | 天平法：黄金法则
An equation is like a balance scale. Whatever you do to one side of the equation, you must also do to the other side. This is called the balancing method.
方程就像一架天平。无论你对等式的一边做什么操作，都必须对另一边做同样的操作。这称为天平法。
If you add, subtract, multiply or divide one side by a number, you must do exactly the same to the other side to keep the equation true.
如果你对一边加、减、乘或除以一个数，你必须对另一边做完全相同的操作，才能保持等式成立。
x + 3 = 10 ⇒ x + 3 − 3 = 10 − 3 ⇒ x = 7
3. One-Step Equations | 一步方程
One-step equations require only one inverse operation to isolate the unknown. For addition, subtract the same number; for subtraction, add the same number.
一步方程只需要一次逆运算就能求出未知数。加法方程要减去同一个数；减法方程要加上同一个数。
For multiplication, divide both sides by the coefficient. For division, multiply both sides by the denominator.
对于乘法方程，两边除以系数；对于除法方程，两边乘以分母。
Example 1: x + 7 = 15. Subtract 7 from both sides to get x = 8.
例1：x + 7 = 15。两边减去 7，得到 x = 8。
x + 7 = 15 ⇒ x = 15 − 7 ⇒ x = 8
Example 2: 5x = 40. Divide both sides by 5 to get x = 8.
例2：5x = 40。两边除以 5，得到 x = 8。
5x = 40 ⇒ x = 40 ÷ 5 ⇒ x = 8
Example 3: x ÷ 4 = 6. Multiply both sides by 4 to get x = 24.
例3：x ÷ 4 = 6。两边乘以 4，得到 x = 24。
x ÷ 4 = 6 ⇒ x = 6 × 4 ⇒ x = 24
4. Two-Step Equations | 两步方程
Two-step equations involve two operations, usually a multiplication or division combined with an addition or subtraction. Always reverse the order of operations: deal with addition or subtraction first, then multiplication or division.
两步方程包含两种运算，通常是乘除与加减的组合。一定要按照运算顺序的逆序来解：先处理加减，再处理乘除。
First subtract 3 from both sides to get 2x = 8. Then divide both sides by 2 to get x = 4.
例：解方程 2x + 3 = 11。先两边减去 3，得到 2x = 8。然后两边除以 2，得到 x = 4。
2x + 3 = 11 ⇒ 2x = 11 − 3 ⇒ 2x = 8 ⇒ x = 4
lve x ÷ 5 − 2 = 3. Add 2 to both sides to get x ÷ 5 = 5, then multiply by 5 to get x = 25.
另一个例子：解方程 x ÷ 5 − 2 = 3。两边加 2，得到 x ÷ 5 = 5，然后两边乘以 5，得到 x = 25。
x ÷ 5 − 2 = 3 ⇒ x ÷ 5 = 5 ⇒ x = 25
5. Equations with Brackets | 含括号的方程
When an equation contains brackets, you can either expand the brackets first or divide both sides by the factor in front of the brackets if it divides evenly.
当方程含有括号时，你可以先展开括号，或者如果括号前的因数能整除另一边，也可以先用这个因数去除两边。
ve 3(x + 2) = 21. Expand to get 3x + 6 = 21, then subtract 6 and divide by 3 to get x = 5.
例：解方程 3(x + 2) = 21。展开得到 3x + 6 = 21，然后减去 6 再除以 3，得到 x = 5。
3(x + 2) = 21 ⇒ 3x + 6 = 21 ⇒ 3x = 15 ⇒ x = 5
es by 3 first: x + 2 = 7, then subtract 2 to get x = 5. Both methods give the same answer.
或者先两边除以 3：x + 2 = 7，再减去 2，得到 x = 5。两种方法结果相同。
3(x + 2) = 21 ⇒ x + 2 = 7 ⇒ x = 5
6. Equations with Unknowns on Both Sides | 未知数在等号两边的方程
If the unknown appears on both sides of the equation, collect the unknown terms on one side and the number terms on the other side. Move the smaller unknown term first to keep the coefficient positive.
如果未知数出现在等号两边，把含未知数的项移到一边，把数字项移到另一边。先移较小的未知数项，使系数保持为正。
s to get 3x + 2 = 14. Subtract 2 from both sides to get 3x = 12. Divide by 3 to get x = 4.
例：解方程 5x + 2 = 2x + 14。两边减去 2x，得到 3x + 2 = 14。两边减去 2，得到 3x = 12。除以 3，得到 x = 4。
5x + 2 = 2x + 14 ⇒ 3x + 2 = 14 ⇒ 3x = 12 ⇒ x = 4
Always check that you have moved terms correctly by substituting your answer back into the original equation.
始终将答案代回原方程，检查移项是否正确。
7. Equations with Fractions | 含分数的方程
To solve an equation with fractions, first eliminate the denominators by multiplying every term on both sides by the lowest common denominator. Then solve the resulting equation as usual.
解含分数的方程时，先用最小公分母乘以两边每一项，消去分母。然后按通常方法解所得方程。
3 + 1 = 5. Subtract 1 from both sides to get x ÷ 3 = 4, then multiply by 3 to get x = 12.
例1：解方程 x ÷ 3 + 1 = 5。两边减去 1，得到 x ÷ 3 = 4，然后两边乘以 3，得到 x = 12。
x ÷ 3 + 1 = 5 ⇒ x ÷ 3 = 4 ⇒ x = 12
h sides by 3 to get 2x + 1 = 15. Subtract 1 to get 2x = 14, then divide by 2 to get x = 7.
例2：解方程 (2x + 1) ÷ 3 = 5。两边乘以 3，得到 2x + 1 = 15。减去 1，得到 2x = 14，再除以 2，得到 x = 7。
(2x + 1) ÷ 3 = 5 ⇒ 2x + 1 = 15 ⇒ 2x = 14 ⇒ x = 7
8. Word Problems | 文字应用题
Word problems require you to translate a real-life situation into an algebraic equation. Read the problem carefully, identify the unknown, and write an equation that represents the given information.
文字应用题要求你把现实情境转化为代数方程。仔细阅读题目，确定未知数，并写出表示已知信息的方程。
Example: A rectangle has length (x + 4) cm and width 3 cm. Its perimeter is 26 cm. Write and solve an equation for x.
例：一个矩形的长为 (x + 4) cm，宽为 3 cm，周长为 26 cm。写出并解关于 x 的方程。
2(x + 4 + 3) = 26 ⇒ 2(x + 7) = 26 ⇒ x + 7 = 13 ⇒ x = 6
Always define your variable clearly and check that your answer makes sense in the context of the problem.
始终清楚地定义变量，并检查答案在题目情境中是否合理。
9. Checking Your Solution | 检验你的解
After solving an equation, substitute your answer back into the original equation to verify that both sides are equal. This is an essential habit for exam success.
解完方程后，把答案代回原方程，验证两边是否相等。这是考试成功的重要习惯。
substituting: 2 × 4 + 3 = 8 + 3 = 11. Since both sides equal 11, the solution is correct.
例：对于方程 2x + 3 = 11，如果你得到 x = 4，代入检验：2 × 4 + 3 = 8 + 3 = 11。因为两边都等于 11，所以解是正确的。
2(4) + 3 = 11 ⇒ 8 + 3 = 11 ⇒ 11 = 11 ✓
10. Common Mistakes to Avoid | 常见错误
Many students lose marks by making small algebraic errors. Here are the most common mistakes and how to avoid them.
许多学生因小的代数错误而丢分。以下是常见错误及避免方法。
Forgetting to do the same operation on both sides of the equation.
忘记对方程两边做同样的操作。
ncorrectly moving terms: for example writing 2x + 3 = 11 as 2x = 8 instead of 2x = 11 − 3.
移项错误：例如把 2x + 3 = 11 写成 2x = 8，而不是 2x = 11 − 3。
Mishandling negative signs, especially when dividing or multiplying by a negative number.
负号处理错误，特别是在除以或乘以负数时。
Expanding brackets incorrectly, such as 3(x + 2) = 3x + 2 instead of 3x + 6.
去括号错误，例如 3(x + 2) = 3x + 2，而不是 3x + 6。
Take your time with each step and check your working carefully.
每一步都要仔细，认真检查你的计算过程。
11. Graphical View of Linear Equations | 线性方程的图像视角
or example, the equation 2x + 1 = 7 is equivalent to finding where y = 2x + 1 meets y = 7.
一元一次方程可以看作求两条直线交点的 x 坐标。例如，方程 2x + 1 = 7 等价于求 y = 2x + 1 与 y = 7 的交点。
3. On a graph, the line y = 2x + 1 crosses the horizontal line y = 7 at the point (3, 7).
解 2x + 1 = 7 得到 x = 3。在图像上，直线 y = 2x + 1 与水平线 y = 7 在点 (3, 7) 处相交。
2x + 1 = 7 ⇒ 2x = 6 ⇒ x = 3
This graphical idea helps you understand why a linear equation has exactly one solution.
这个图像概念帮助你理解为什么一元一次方程恰好有一个解。
12. Practice Challenge | 练习挑战
Try this challenge question to test your understanding. Solve the equation step by step and check your answer.
试试这道挑战题来检验你的理解。逐步解方程并检查答案。
2(x + 3) = 3(x − 2) + 1
= 3x − 5. Subtract 2x from both sides to get 6 = x − 5. Add 5 to both sides to get x = 11.
两边展开得到 2x + 6 = 3x − 6 + 1，化简为 2x + 6 = 3x − 5。两边减去 2x，得到 6 = x − 5。两边加 5，得到 x = 11。
28 and 3(11 − 2) + 1 = 3 × 9 + 1 = 27 + 1 = 28. Both sides equal 28, so x = 11 is correct.
检验：2(11 + 3) = 2 × 14 = 28，而 3(11 − 2) + 1 = 3 × 9 + 1 = 27 + 1 = 28。两边都等于 28，所以 x = 11 是正确的。
Keep practising these methods regularly, and you will become faster and more accurate in solving linear equations.
经常练习这些方法，你在解一元一次方程时就会更快、更准确。
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