📚 Solving Linear Equations | 解一元一次方程
Linear equations are one of the most important building blocks in KS3 mathematics. They appear in algebra, problem-solving and even real-life situations such as working out costs, distances and unknown quantities. This article explains how to solve linear equations step by step, using clear methods and examples suitable for Cambridge KS3 learners.
线性方程是 KS3 数学中最重要的基础内容之一。它出现在代数、问题解决甚至现实生活中，例如计算费用、距离和未知量。本文将逐步解释如何解一元一次方程，使用适合 Cambridge KS3 学生的清晰方法和示例。
1. What Is a Linear Equation? | 什么是线性方程？
ly to the power of 1. For example, x + 3 = 7 and 2x − 5 = 9 are linear equations. The word 'linear' be written in the general form ax + b = c, where a, b and c are numbers and a is not zero.
线性方程是指未知变量只出现 1 次方的方程。例如 x + 3 = 7 和 2x − 5 = 9 都是线性方程。'线性' 一词来源于这类方程的图像是一条直线。每一个线性方程都可以写成一般形式 ax + b = c，其中 a、b、c 是数，且 a 不等于零。
x + 3 = 7, 2x − 5 = 9, 4 − x = 1, ½x = 6
ear because the variable x has an exponent of 2. Linear equations only contain terms like x, 2x or −5x, never x², x³ or higher powers. The equals sign means that the expression on the left has exactly the same value as the expression on the right - the two sides are balanced.
像 x² + 2 = 11 这样的方程不是线性方程，因为变量 x 的指数是 2。线性方程只包含 x、2x 或 −5x 这样的项，绝不超过 x²、x³ 或更高次幂。等号表示左边的表达式与右边的表达式取值完全相同——两边是平衡的。
The solution of an equation is the value of the unknown that makes the equation true. Solving means finding that value. A linear equation in one unknown normally has exactly one solution, although some equations have no solution and some are true for every value of x.
方程的解就是使等式成立的未知数的值。解方程就是找出这个值。一元一次方程通常恰好有一个解，不过有些方程无解，有些则对 x 的任何取值都成立。
2. The Balancing Method | 等式平衡法
The balancing method treats an equation like a pair of scales in perfect balance. Whatever you do to one side, you must do exactly the same to the other side, so that the balance is never disturbed. If you add 5 to the left, you add 5 to the right; if you divide the left by 3, you divide the right by 3.
等式平衡法把方程看作一架完全平衡的天平。你对一边做什么，就必须对另一边做完全相同的事，这样平衡才不会被破坏。如果左边加 5，右边也要加 5；如果左边除以 3，右边也要除以 3。
If a = b, then a + c = b + c and a − c = b − c
If a = b, then a × c = b × c and a ÷ c = b ÷ c (c ≠ 0)
he equation simpler, not more complicated. Write one step per line so that your working is easy to follow and easy to check.
目标始终是让未知数单独留在一边，写成 x = 某个数 的形式。每一步都应让方程更简单，而不是更复杂。每一步写一行，这样过程清晰、便于检查。
Always remember the golden rule: an equation is not a calculation. You cannot simply move a number across the equals sign and change its sign without justification - what you are really doing is adding or subtracting the same amount on both sides.
永远记住这条黄金法则：方程不是算式。你不能随意把数字搬到等号另一边再改符号而不说明理由——你实际上是在两边同时加上或减去同一个量。
3. Inverse Operations | 逆运算
Inverse operations undo each other. Addition and subtraction are inverses, and multiplication and division are inverses. To free the unknown from the operations surrounding it, we apply the inverse operation to both sides.
逆运算可以互相抵消。加法与减法互为逆运算，乘法与除法互为逆运算。要把未知数从包围它的运算中解放出来，我们就在两边同时施加逆运算。
+ 3 is undone by − 3（+ 3 用 − 3 抵消）
− 7 is undone by + 7（− 7 用 + 7 抵消）
× 4 is undone by ÷ 4（× 4 用 ÷ 4 抵消）
÷ 5 is undone by × 5（÷ 5 用 × 5 抵消）
When more than one operation acts on the unknown, undo them in the reverse order of BIDMAS. If x is first multiplied by 2 and then has 5 subtracted, you undo the subtraction first and the multiplication second.
当有多个运算作用于未知数时，按 BIDMAS 的相反顺序逐一抵消。如果 x 先乘 2 再减 5，你要先抵消减法，再抵消乘法。
2x − 5 = 9 → + 5 → 2x = 14 → ÷ 2 → x = 7
4. Solving One-Step Equations | 解一步方程
A one-step equation needs only a single inverse operation. Identify what is being done to the unknown, then do the opposite to both sides.
一步方程只需要一次逆运算。先看清未知数被施加了什么运算，然后在两边做相反的运算。
| Equation | 方程 | Inverse step | 逆运算步骤 | Solution | 解 |
| x + 3 = 7 | subtract 3 from both sides | 两边减 3 | x = 4 |
| x − 8 = −3 | add 8 to both sides | 两边加 8 | x = 5 |
| 5x = 45 | divide both sides by 5 | 两边除 5 | x = 9 |
| x ÷ 4 = 3 | multiply both sides by 4 | 两边乘 4 | x = 12 |
| −x = 6 | multiply both sides by −1 | 两边乘 −1 | x = −6 |
Notice the last row: when the unknown has a minus sign in front of it, multiply (or divide) both sides by −1 to make the coefficient of x positive. A negative answer is perfectly acceptable and often the correct one.
注意最后一行：当未知数前面有负号时，把两边同时乘（或除）以 −1，使 x 的系数为正。负答案完全合理，而且往往是正确答案。
f x ÷ 3 = ⅔, multiply both sides by 3 to get x = 2. Keep fractions in their simplest form.
分数在一步方程中经常出现。如果 ½x = 6，两边乘 2 得 x = 12。如果 x ÷ 3 = ⅔，两边乘 3 得 x = 2。分数要保持最简形式。
5. Solving Two-Step Equations | 解两步方程
Most KS3 linear equations need two inverse operations. Work through them in the reverse order of BIDMAS: deal with any addition or subtraction first, then multiplication or division.
大多数 KS3 线性方程需要两次逆运算。按 BIDMAS 的相反顺序进行：先处理加法或减法，再处理乘法或除法。
2x − 5 = 9 → 2x = 14 → x = 7
divide both sides by 2 to obtain x = 7. Two operations, two lines of working, one answer.
在这个例子中，2x 被减去了 5，所以两边加 5 得到 2x = 14。然后 x 被乘以 2，所以两边除以 2 得到 x = 7。两次运算，两行过程，一个答案。
Here is a second example with a division as the final operation:
下面是第二个例子，最后一次运算是除法：
3x + 7 = 22 → 3x = 15 → x = 5
x ÷ 5 + 4 = 10 → x ÷ 5 = 6 → x = 30
h is correct but far messier. Always undo the last operation applied to the unknown first.
常见错误是先除后减。在 3x + 7 = 22 中，先除以 3 会得到 x + 7/3 = 22/3，虽然正确但繁琐得多。永远先抵消最后施加在未知数上的那个运算。
6. Equations with Brackets | 含括号的方程
When the unknown is inside brackets, you have two valid strategies. You can expand the brackets first using the distributive law, or you can divide both sides by the number multiplying the bracket. Choose whichever gives simpler numbers.
当未知数在括号内时，有两种都正确的方法。你可以先用分配律展开括号，也可以两边同除以括号前的数。选择能让数字更简单的那一种。
3(x + 4) = 21
3. Both give the same answer, and Method B is faster here because 21 divides exactly by 3.
方法 A——展开：3x + 12 = 21，所以 3x = 9，x = 3。方法 B——两边同除：x + 4 = 7，所以 x = 3。两种方法答案相同，这里方法 B 更快，因为 21 能被 3 整除。
Be very careful with a negative number in front of a bracket. Multiplying a bracket by −2 changes the sign of every term inside it.
括号前是负数时要特别小心。用 −2 乘整个括号会改变括号内每一项的符号。
−2(x − 3) = 10 → −2x + 6 = 10 → −2x = 4 → x = −2
or example, 2(x + 1) = 3(x − 2) becomes 2x + 2 = 3x − 6, and then x = 8 after rearranging.
如果两边都有括号，先把两边都展开，再合并同类项然后求解。例如 2(x + 1) = 3(x − 2) 变成 2x + 2 = 3x − 6，整理后得 x = 8。
7. Equations with the Unknown on Both Sides | 两边都有未知数的方程
When x appears on both sides, your first job is to collect all the x terms on one side and all the numbers on the other. Subtract the smaller x term from both sides to keep the coefficient positive.
当两边都出现 x 时，第一步是把所有含 x 的项集中到一边，所有数字集中到另一边。两边同时减去较小的 x 项，可以让系数保持为正。
5x − 4 = 3x + 8 → 2x − 4 = 8 → 2x = 12 → x = 6
n the right. Finally, dividing by 2 gives x = 6. Each step is a legitimate balancing move.
这里，两边减去 3x 后左边剩下 2x；两边再加 4 后右边剩下 12；最后两边除以 2 得到 x = 6。每一步都是合理的平衡操作。
t side: 5(6) − 4 = 26. Right side: 3(6) + 8 = 26. Both sides agree, so x = 6 is confirmed.
把解代回原方程检验。左边：5(6) − 4 = 26；右边：3(6) + 8 = 26。两边相等，所以 x = 6 得到确认。
If you reach something like 5 = 5, the equation is an identity and every value of x works.
有时方程根本无解。如果你得到 4 = 7 这样的式子，说明该方程是矛盾式，无解。如果你得到 5 = 5 这样的式子，说明该方程是恒等式，x 的任何取值都成立。
8. Equations with Fractions | 含分数的方程
Fractions make equations look harder than they are. The standard trick is to multiply every term by the lowest common multiple of the denominators. This clears all the fractions in one move.
分数会让方程看起来比实际更难。标准技巧是把每一项都乘以各分母的最小公倍数，这样一步就能清除所有分数。
x/3 + x/4 = 7 → × 12 → 4x + 3x = 84 → 7x = 84 → x = 12
Here the LCM of 3 and 4 is 12. Multiplying each term by 12 turns the equation into plain integers. Always multiply every term, including the number on the right-hand side - forgetting one term is a very common slip.
这里 3 和 4 的最小公倍数是 12。把每一项都乘 12，方程就变成整数方程。一定要乘到每一项，包括等号右边的数字——漏乘某一项是非常常见的失误。
When the whole numerator is an expression, treat it as a single block and multiply both sides by the denominator:
当分子本身是一个表达式时，把它整体看作一块，两边同乘分母：
(2x + 1)/3 = 5 → 2x + 1 = 15 → 2x = 14 → x = 7
way. For ½x + 3 = 7, subtract 3 to get ½x = 4, then multiply both sides by 2 to get x = 8.
½ 或 ⅔ 这样的分数系数同理。对 ½x + 3 = 7，先减 3 得 ½x = 4，再两边乘 2 得 x = 8。
9. Forming and Solving Word Problems | 列方程解应用题
Many exam questions do not give you the equation - you have to build it. The routine is always the same: define the unknown with a letter, translate each sentence into algebra, then solve and answer the question that was actually asked.
很多考题不会直接给出方程，而要你自己列。流程总是相同的：用一个字母表示未知数，把每句话翻译成代数式，然后求解，并回答题目真正问的问题。
Step 1: Choose a letter and state clearly what it represents.（第一步：选一个字母，并清楚说明它代表什么。）
Step 2: Write an expression for every other unknown quantity.（第二步：为其他每一个未知量写出表达式。）
Step 3: Form an equation from the information given.（第三步：根据已知信息列出方程。）
Step 4: Solve the equation and check the answer in context.（第四步：解方程，并结合情境检验答案。）
= 26, giving 4w + 6 = 26, so 4w = 20 and w = 5. The width is 5 cm and the length is 8 cm.
例：长方形的长比宽多 3 cm，周长是 26 cm。设宽为 w cm，则长为 w + 3 cm。于是 2(w + w + 3) = 26，即 4w + 6 = 26，所以 4w = 20，w = 5。宽为 5 cm，长为 8 cm。
Example: a pen costs £2 more than a pencil. Three pens and two pencils cost £ en costs p + 2 pounds. So 3(p + 2) + 2p = 11, giving 5p + 6 = 11, so p = 1. A pencil costs £1 and a pen costs £3.
例：一支钢笔比一支铅笔贵 2 英镑。3 支钢笔和 2 支铅笔共 11 英镑。设铅笔每支 p 英镑，则钢笔每支 p + 2 英镑。于是 3(p + 2) + 2p = 11，即 5p + 6 = 11，p = 1。铅笔每支 1 英镑，钢笔每支 3 英镑。
10. The Graphical Meaning of a Solution | 方程解的图像意义
quation 2x − 5 = 9 can be read as: for which value of x does the expression 2x − 5 equal 9? x − 5, the solution is the x-coordinate of the point where the line reaches a height of 9.
每个线性方程都可以联系到一条直线图像。方程 2x − 5 = 9 可以读作：x 取什么值时，表达式 2x − 5 等于 9？如果画出 y = 2x − 5，方程的解就是直线上高度为 9 那一点的横坐标。
its for every 1 unit across, and it crosses the y-axis at −5. Solving the equation algebraically gives exactly the same x-value as reading the graph.
乘在 x 上的数就是斜率，常数项就是 y 轴截距。在 y = 2x − 5 中，直线每向右 1 个单位就上升 2 个单位，并与 y 轴交于 −5。用代数方法解方程得到的 x 值，与从图像上读出的一致。
pressions equal. This is why solving 5x − 4 = 3x + 8 gives the same answer as finding where the lines y = 5x − 4 and y = 3x + 8 intersect.
两条直线的交点处，两个表达式相等——所以交点的横坐标就是令两个表达式相等所得方程的解。这就是为什么解 5x − 4 = 3x + 8，与求直线 y = 5x − 4 和 y = 3x + 8 的交点得到相同答案。
11. Checking Answers and Common Mistakes | 检验答案与常见错误
Substitution is the fastest way to check any solution. Replace x with your answer in the original equation and evaluate both sides separately. If the two values match, the solution is correct.
代入检验是验证任何解最快的方法。把 x 换成你的答案代入原方程，分别计算两边。如果两个值相等，解就是正确的。
Watch out for these frequent errors:
请留意以下常见错误：
Operating on one side only, which destroys the balance.（只对一边做运算，破坏了平衡。）
Sign errors when subtracting a negative term, such as 4 − (−3) = 7, not 1.（减去负数时符号出错，例如 4 − (−3) = 7，而不是 1。）
Forgetting to multiply the constant term when clearing fractions.（去分母时漏乘常数项。）
Expanding −3(x − 2) as −3x − 6 instead of −3x + 6.（把 −3(x − 2) 展开成 −3x − 6，正确的是 −3x + 6。）
Stopping before isolating x completely, for example leaving 2x = 14.（没有把 x 完全隔离就停手，例如停在 2x = 14。）
Marking your own work honestly and writing full, line-by-line working will catch nearly all of these slips - and it also earns method marks even when the final answer is wrong.
诚实地自查并逐行写出完整过程，几乎能发现以上所有失误——而且在最终答案错误时也能拿到方法分。
12. Exam Tips and Summary | 考试要点与小结
Solving linear equations is a skill that underpins almost every later topic in algebra, including simultaneous equations, inequalities, rearranging formulae and straight-line graphs. Mastering the balancing method now pays off many times over.
解线性方程是支撑后续几乎所有代数内容的核心技能，包括联立方程、不等式、公式变形和直线图像。现在掌握好平衡法，收益会成倍体现。
A reliable checklist for any linear equation:
解任何线性方程的可靠清单：
Remove brackets by expanding.（展开去括号。）
Clear fractions by multiplying by the LCM.（用最小公倍数去分母。）
Collect all x terms on one side.（把所有 x 项集中到一边。）
Collect all number terms on the other side.（把所有数字项集中到另一边。）
Divide by the coefficient of x to reach x = ...（除以 x 的系数，得到 x = ...）
Substitute back to check.（代回检验。）
ax + b = c → x = (c − b) ÷ a, a ≠ 0
In the exam, show every step, keep your equals signs aligned, and finish by answering the question in words if it is a word problem. Accuracy in the small steps produces accuracy in the final answer.
考试中要写出每一步，保持等号对齐，如果是应用题最后要用文字回答问题。小步骤上的准确，才能带来最终答案的准确。
Published by TutorHao | KS3 Mathematics Revision Series | aleveler.com
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