📚 Solving Linear Equations: Complete KS3 Guide | 解一元一次方程：KS3 完整指南
Linear equations are one of the most important building blocks in KS3 mathematics, and they sit at the heart of Cambridge Lower Secondary Checkpoint tests. They also prepare students for later topics such as straight-line graphs, sequences and inequalities, where the same balance-and-undo skills are used again and again.
一元一次方程是 KS3 数学中最重要的基础内容之一，也是剑桥初中 Checkpoint 考试的核心考点。它们同时为后续的直线图像、数列和不等式等内容打好基础，因为在这些主题中，反复使用的正是同一套"保持平衡、逐步逆运算"的技能。
This guide walks through the definition, the balancing method, brackets, fractions, variables on both sides, word problems, graphs, inequalities and sequences, and it finishes with exam-style practice and a revision checklist.
本文将从定义出发，依次讲解天平法、括号、分数、未知数在两边、应用题、图像、不等式与数列，最后给出考试练习题和复习清单。
1. What is a Linear Equation? | 什么是一元一次方程？
rranged into the standard form ax + b = c, where a, b and c are numbers and a is not zero.
一元一次方程是指未知数（通常用 x 表示）的指数为 1 的方程。它总可以整理成标准形式 ax + b = c，其中 a、b、c 是常数，且 a 不等于 0。
ax + b = c (a ≠ 0)
1 = 6 are all linear equations, because the variable x is never squared or cubed. The word "linear" comes from the fact that the graph of such a relationship is a straight line.
例如，2x + 3 = 11、5x - 4 = 16 和 x/2 + 1 = 6 都属于一元一次方程，因为未知数 x 没有被平方或立方。"线性"这个词正是来自这类关系在坐标系中的图像是一条直线。
| Expression (no equals sign) | Equation (has an equals sign) |
| 3x + 5 | 3x + 5 = 20 |
| x² - 4 | x² - 4 = 0 (not linear) |
| 7 - 2x | 7 - 2x = 1 |
Notice the difference: an expression can only be simplified, but an equation can be solved, because the equals sign tells us that two quantities have exactly the same value.
注意区别：代数式只能化简，而方程可以求解，因为等号告诉我们两个量的大小完全相同。
2. The Balancing Method | 天平法
An equation behaves like a balance scale. If you add, subtract, multiply or divide one side, you must do exactly the same to the other side to keep the equality true.
方程就像一架平衡的天平。如果对等号一边进行加、减、乘、除，就必须对另一边做完全相同的运算，等式才能继续保持成立。
This single idea is the whole foundation of equation solving. Whatever you do to the left-hand side, you do to the right-hand side, and the value of x never changes.
这一个思想就是解方程的全部基础。无论你对左边做什么，都要对右边做同样的事，而 x 的值始终不变。
If A = B, then A ± k = B ± k and A × k = B × k (k ≠ 0)
4 gives x = 5. Both steps are visualised easily by imagining weights on a two-pan balance.
例如，从 x + 7 = 12 出发，两边同时减 7，得到 x = 5；从 4x = 20 出发，两边同时除以 4，得到 x = 5。把两个步骤想象成天平两边同时取下或分走相同的砝码，就非常直观。
3. Inverse Operations: The Step-by-Step Routine | 逆运算：标准解题步骤
To isolate x, undo the operations in reverse order. Addition and subtraction are undone first; multiplication and division are undone last. This is why we usually "move the number first, then the coefficient".
要把 x 单独留在一边，就要按相反的顺序撤掉各层运算：先撤加法和减法，最后撤乘法和除法。这就是为什么我们通常"先处理常数，再处理系数"。
Follow the standard routine: expand any brackets, collect like terms, move the constants, move the variable terms, then divide by the coefficient. Finally, always check the answer in the original equation.
标准步骤是：先去括号，再合并同类项，然后移常数项，再移未知数项，最后除以系数。做完之后，一定要把答案代回原方程检验。
| Step | Working for 5x - 4 = 16 | Reason |
| 1 | 5x - 4 = 16 | original equation |
| 2 | 5x = 20 | add 4 to both sides |
| 3 | x = 4 | divide both sides by 5 |
btract 3 to get 2x = 8, then divide by 2 to get x = 4. Two clean steps, two clean answers.
同样的步骤可以解 2x + 3 = 11：先减 3 得 2x = 8，再除以 2 得 x = 4。两步运算，答案干净利落。
4. Equations with Brackets | 含括号的方程
When brackets appear, either expand them first or divide both sides by the bracket coefficient. Expanding is the safer habit for beginners because it turns an unfamiliar question into a familiar one.
当方程中出现括号时，要么先展开括号，要么两边同时除以括号前的系数。对初学者来说，先展开更稳妥，因为它把陌生的题目变成熟悉的形式。
ing the same equation by dividing both sides by 2 first gives x + 3 = 7, so x = 4 as well.
解 2(x + 3) = 14：展开得 2x + 6 = 14，两边减 6 得 2x = 8，再除以 2 得 x = 4。若先两边同时除以 2，得到 x + 3 = 7，同样得 x = 4。
A single sign error here is the most common reason for losing marks in Checkpoint papers.
要特别小心符号：3(x - 5) = 12 展开后是 3x - 15 = 12，而不是 3x + 15 = 12。在这里出现一个符号错误，是 Checkpoint 试卷中最常见的失分原因。
a(x + b) = c → ax + ab = c → x = (c - ab) ÷ a
5. Variables on Both Sides | 未知数出现在两边
If x appears on both sides, collect all the variable terms on the side with the larger coefficient, so the coefficient of x stays positive. This avoids negative coefficients and reduces sign mistakes.
如果等号两边都出现 x，就把所有未知数项集中到系数较大的那一边，这样 x 的系数会保持为正数，能避免负系数并减少符号错误。
: 2x = 8. Divide by 2: x = 4. Substituting back gives 17 = 17, so the solution is correct.
解 3x + 5 = x + 13。两边同时减 x，得 2x + 5 = 13；再减 5，得 2x = 8；除以 2，得 x = 4。代回原式得到 17 = 17，说明解正确。
If instead you get 3 = 3, the equation is an identity and every value of x is a solution.
有时未知数项会完全抵消。例如 2x + 3 = 2x + 7 化简成 3 = 7，这是错的，所以方程无解。如果化简成 3 = 3，则等式恒成立，任何 x 值都是解。
6. Fractions and Decimals in Equations | 分数与小数方程
Fractions look frightening but they follow the same rules. Multiply every term by the lowest common multiple of the denominators to clear the fractions in one step.
分数看起来可怕，但规则完全相同。只要把所有项都乘以各分母的最小公倍数，就能一步清掉所有分数。
o get x = 10. Alternatively, multiply every term by 2 at the start: x + 2 = 12, so x = 10.
解 x/2 + 1 = 6：两边减 1 得 x/2 = 5，再两边乘 2 得 x = 10。也可以一开始就把每一项都乘 2：x + 2 = 12，同样得 x = 10。
7x = 84 and x = 12. Always multiply every single term, including the number on the right.
对于 x/3 + x/4 = 7 这样的方程，3 与 4 的最小公倍数是 12。整体相乘得 4x + 3x = 84，即 7x = 84，所以 x = 12。切记每一项都要乘，包括等号右边的常数。
turn them into whole numbers: 0.5x + 1.2 = 3.7 becomes 5x + 12 = 37, so 5x = 25 and x = 5.
遇到小数时，先整体乘以 10 或 100 化为整数：0.5x + 1.2 = 3.7 变成 5x + 12 = 37，于是 5x = 25，x = 5。
7. Checking Your Answer and Common Mistakes | 检验答案与常见错误
Substituting your answer back into the original equation is not optional - it is the fastest way to find careless errors. If the left-hand side and right-hand side are equal, your solution is correct.
把答案代回原方程不是可选步骤，而是发现粗心错误最快的方法。如果左边与右边的值相等，解就是正确的。
ing fractions, mishandling a negative sign, and writing -x = 5 as x = 5 instead of x = -5.
常见错误包括：只对一边做运算；清分母时漏乘某些项；负号处理错误；以及把 -x = 5 写成 x = 5，而正确答案是 x = -5。
- Left side = Right side? If not, check your arithmetic, not your method.
- Does the answer make sense in the context of the word problem?
- Did you divide by the coefficient instead of subtracting it?
- Did you keep the equals signs aligned down the page?
- 左右两边是否相等？不等就检查计算，而不是方法。
- 答案放入应用题情境中是否合理？
- 最后一步是除以系数，而不是减系数，你做对了吗？
- 等号是否整齐对齐、逐行书写？
8. Forming Equations from Word Problems | 由应用题列方程
Most Checkpoint word problems ask you to define a variable, form an equation, solve it, and then answer the actual question asked. The final sentence matters as much as the algebra.
多数 Checkpoint 应用题要求你：设未知数、列出方程、解方程，最后回答题目真正问的问题。最后一句结论和代数过程同样重要。
3) = 30, giving 4x + 6 = 30, 4x = 24 and x = 6. The width is 6 cm and the length is 9 cm.
示例：一个长方形比宽长 3 cm，周长是 30 cm。设宽为 x，则长为 x + 3。于是 2x + 2(x + 3) = 30，化简得 4x + 6 = 30，4x = 24，x = 6。所以宽是 6 cm，长是 9 cm。
(n + 1) + (n + 2) = 51, so 3n + 3 = 51, 3n = 48 and n = 16. The numbers are 16, 17 and 18.
示例：三个连续整数之和为 51。设最小的为 n，则 n + (n + 1) + (n + 2) = 51，得 3n + 3 = 51，3n = 48，n = 16。三个数是 16、17 和 18。
Always define your variable in words first, and remember to convert your answer back into the units used in the question.
一定要先用文字说明你设的未知数代表什么，并且记得把答案还原成题目所使用的单位。
9. Linear Equations and Straight-Line Graphs | 一元一次方程与直线图像
ving ax + b = c is really asking where a related straight line reaches a particular value.
任何含两个未知数的一元一次方程都可以写成 y = mx + c，其中 m 是斜率，c 是 y 轴截距。解 ax + b = c，本质上就是问某条直线在哪里取到某个特定的值。
horizontal line y = 11. Reading x = 4 off the graph matches the algebraic answer exactly.
2x + 3 = 11 的解，就是直线 y = 2x + 3 与水平线 y = 11 交点横坐标。从图像上读出 x = 4，与代数解法完全一致。
Solving simultaneous linear equations means finding the point where two straight lines cross, which is why equation-solving fluency pays off twice in later topics.
解二元一次方程组就是找两条直线的交点，这也是为什么把一元一次方程解熟，会在后面的主题中带来双倍回报。
10. Inequalities, Sequences and Other Connections | 不等式、数列与其他联系
Linear inequalities are solved almost exactly like equations. The one extra rule is that multiplying or dividing both sides by a negative number reverses the inequality sign.
一元一次不等式的解法与方程几乎一样，唯一多出来的规则是：两边同时乘或除以负数时，不等号方向要反转。
-2x > 8 → x < -4
Sequences also depend on this skill. The nth term of an arithmetic sequence is a + (n - 1)d, and finding which term equals a given value means solving a linear equation for n.
数列同样依赖这项技能。等差数列的第 n 项是 a + (n - 1)d，而求"哪一项等于某个给定值"，就是解一个关于 n 的一元一次方程。
15, so 47 is the 15th term. The same balance method from Section 2 is doing all the work.
对于数列 5, 8, 11, 14, ...，第 n 项是 3n + 2。令 3n + 2 = 47，得 3n = 45，n = 15，所以 47 是第 15 项。这里真正起作用的，仍然是第 2 节中的天平法。
t, or A = πr² for r, uses exactly the inverse-operation logic you have already practised.
科学和几何中的公式也一样：把 v = u + at 变形求 t，或把 A = πr² 变形求 r，用到的正是你已经练熟的逆运算逻辑。
11. Exam-Style Practice and Revision Checklist | 考试练习题与复习清单
Try these five questions without looking at the answers, then check your working line by line. Aim for at least four out of five before moving on to harder topics.
先不看答案完成下面五题，然后逐行检查过程。至少做对四题，再进入更难的专题。
- Solve 4x - 7 = 25.
- Solve 5(x - 2) = 3x + 6.
- Solve x/4 + x/3 = 7.
- The sum of three consecutive even numbers is 72. Find them.
- Solve the inequality 3 - 2x ≥ 11 and describe the solution set.
- 解 4x - 7 = 25。
- 解 5(x - 2) = 3x + 6。
- 解 x/4 + x/3 = 7。
- 三个连续偶数之和为 72，求这三个数。
- 解不等式 3 - 2x ≥ 11，并描述解集。
) = 72 gives 3n = 66 and n = 22, so the numbers are 22, 24 and 26; (5) -2x ≥ 8, so x ≤ -4.
答案：(1) x = 8；(2) 5x - 10 = 3x + 6，故 2x = 16，x = 8；(3) 两边乘 12 得 3x + 4x = 84，x = 12；(4) n + (n + 2) + (n + 4) = 72，得 3n = 66，n = 22，三数为 22、24、26；(5) -2x ≥ 8，故 x ≤ -4。
Revision checklist: can you expand brackets accurately, collect like terms, clear fractions, handle negative coefficients, check your answer by substitution, and translate a word problem into an equation? If all six answers are yes, you are ready for the next stage of algebra.
复习清单：你能否准确展开括号、合并同类项、清除分母、处理负系数、用代入法检验答案，并把应用题翻译成方程？如果这六项都能做到，你就可以进入下一阶段的代数学习了。
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