Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Linear equations form the backbone of algebra at Key Stage 3. They appear in everything from simple number puzzles to handling unknown quantities in science and finance. Mastering how to solve them gives you a powerful toolkit for reasoning, checking answers, and building confidence for advanced topics such as linear graphs and simultaneous equations. In this article, we will break down the process step by step, working from the balance method up to equations with brackets, unknowns on both sides, and fractions. Every technique is paired with clear examples and common pitfalls to avoid.

一元一次方程是 KS3 阶段代数的基石。从简单的数字谜题到科学与金融中的未知量处理，方程无处不在。掌握解方程的方法会为你提供一套强大的推理工具，帮助你验算答案，并为一次函数图像、联立方程等更高级的主题建立信心。本文将逐步拆解解法，从平衡法开始，一直延伸到含有括号、两边都有未知数以及带分数的方程。每个技巧都配有清晰的示例和需要避免的常见陷阱。

1. What Is an Equation? | 什么是方程？

An equation is a mathematical statement that two expressions are equal. It always includes an equals sign ‘=’. For example, x + 5 = 12 is an equation because it declares that when you add 5 to the unknown number x, you get 12. The left-hand side (x + 5) and the right-hand side (12) are linked by equality.

方程是声明两个表达式相等的数学陈述。它总是包含一个等号 ‘=’。例如，x + 5 = 12 是一个方程，因为它宣称当你把 5 加到未知数 x 上时，会得到 12。左边 (x + 5) 和右边 (12) 由相等关系连接。

The solution of an equation is the value (or values) that make the statement true. Here the solution is x = 7 because 7 + 5 = 12. Solving an equation means finding every possible value of the variable that satisfies the equation. While linear equations usually have one solution, some special cases can have no solution or infinitely many solutions – we will touch on those later.

方程的解是使得该陈述成立的数值。这里解是 x = 7，因为 7 + 5 = 12。解方程意味着求所有满足方程的变量的值。虽然一元一次方程通常只有一个解，但有些特殊情况可能无解或有无穷多解——我们稍后会提及。

2. The Balance Method | 平衡法原理

Think of an equation as a balanced seesaw. Whatever you do to one side, you must do exactly the same to the other side to keep it balanced. This is the golden rule of equation solving. If you add 3 to the left, add 3 to the right. If you divide the left by 2, divide the right by 2 as well. Mathematically, we are applying inverse operations step by step to isolate the variable.

把方程想象成一个平衡的跷跷板。你对一边做的任何操作，都必须对另一边做完全相同的操作，才能保持平衡。这是解方程的黄金法则。如果你在左边加 3，那么右边也要加 3。如果你把左边除以 2，那么右边同样要除以 2。在数学上，我们就是在逐步施加逆运算，以把变量孤立出来。

We use inverse operations in reverse BIDMAS order (or BODMAS order) to unravel the equation. For instance, if the variable is first multiplied by 3 and then 4 is added, we will first subtract 4 (undoing the addition) and then divide by 3 (undoing the multiplication). Keeping the process systematic prevents mistakes and makes the reasoning clear.

我们按照逆 BIDMAS（或 BODMAS）顺序使用逆运算来解开方程。例如，如果变量先乘以 3 然后再加 4，我们就先减去 4（逆加法）再除以 3（逆乘法）。保持过程有条理可以避免错误，并使推理清晰。

Operation	Inverse operation
Addition (+a)	Subtraction (−a)
Subtraction (−a)	Addition (+a)
Multiplication (×a)	Division (÷a)
Division (÷a)	Multiplication (×a)

Operation ⟷ inverse operation

3. One-Step Equations | 一步方程

These are the simplest type: they involve only one operation connected to the variable. To solve them, apply the inverse operation to both sides.

这是最简单的一类：它们只涉及与变量有关的一种运算。要解它们，只需对两边施加逆运算即可。

Example: x + 9 = 15. Subtract 9 from both sides: x + 9 − 9 = 15 − 9 → x = 6. Always check: 6 + 9 = 15 ✓.

例如：x + 9 = 15。两边同时减去 9：x + 9 − 9 = 15 − 9 → x = 6。务必验算：6 + 9 = 15 ✓。

Another example: 5y = 35. Divide both sides by 5: 5y ÷ 5 = 35 ÷ 5 → y = 7. Check: 5 × 7 = 35 ✓.

另一个例子：5y = 35。两边同除以 5：5y ÷ 5 = 35 ÷ 5 → y = 7。验算：5 × 7 = 35 ✓。

A slightly trickier one-step equation involves the variable in the denominator, such as 12 ÷ x = 4. We can think of it as a multiplication by the reciprocal, or multiply both sides by x to get 12 = 4x, then divide by 4 → x = 3.

稍微有技巧的一步方程涉及变量在分母的情况，比如 12 ÷ x = 4。我们可以将其视为乘以倒数，或者两边同乘 x 得到 12 = 4x，再除以 4 → x = 3。

4. Two-Step Equations | 两步方程

When two operations are applied to the variable, we undo them in reverse order. Typically you address addition/subtraction before multiplication/division.

当变量被施加了两种运算时，我们按照相反的顺序逐步消去。通常先处理加减法，再处理乘除法。

Solve 3x + 7 = 22. Step 1: subtract 7 from both sides → 3x = 15. Step 2: divide both sides by 3 → x = 5. Check: 3 × 5 + 7 = 22 ✓.

解 3x + 7 = 22。第一步：两边同时减去 7 → 3x = 15。第二步：两边同时除以 3 → x = 5。验算：3 × 5 + 7 = 22 ✓。

Solve p/4 − 2 = 3. First, add 2 to both sides → p/4 = 5. Then multiply both sides by 4 → p = 20. Here, the subtraction is undone first, then the division. Remember, the bar in p/4 represents division, so its inverse is multiplication.

解 p/4 − 2 = 3。首先，两边加 2 → p/4 = 5。然后两边同乘 4 → p = 20。这里先消去减法，再消去除法。请注意，p/4 中的分数线表示除法，其逆运算是乘法。

5. Equations with Brackets | 含括号的方程

When brackets appear, the two main strategies are expanding the brackets first or, if suitable, dividing both sides by the coefficient outside the bracket. Both approaches obey the balance rule.

当出现括号时，两种主要策略是：先展开括号，或者如果合适的话，两边同除以括号外的系数。两种方法都遵循平衡法则。

Example 1: 4(x + 3) = 28. You can expand to 4x + 12 = 28, subtract 12, divide by 4 → x = 4. Alternatively, divide both sides by 4 first → x + 3 = 7 → x = 4. The second approach is quicker when the right side is divisible by the coefficient.

示例 1：4(x + 3) = 28。可以展开成 4x + 12 = 28，减去 12，再除以 4 → x = 4。也可以先两边同时除以 4 → x + 3 = 7 → x = 4。当右边能被系数整除时，第二种方法更快。

Example 2: 3(2y − 5) = 9y. Expanding gives 6y − 15 = 9y. Now subtract 6y from both sides → −15 = 3y. Divide by 3 → y = −5. Check: 3(2(−5)−5) = 3(−10−5) = 3(−15) = −45, and 9(−5) = −45 ✓.

示例 2：3(2y − 5) = 9y。展开得 6y − 15 = 9y。现在两边同减 6y → −15 = 3y。除以 3 → y = −5。验算：3(2(−5)−5) = 3(−10−5) = 3(−15) = −45，而 9(−5) = −45 ✓。

Where brackets contain more complex expressions, always expand carefully, paying attention to negative signs. For instance, −2(3 − a) = −6 + 2a, because −2 × 3 = −6 and −2 × (−a) = +2a.

当括号内包含更复杂的表达式时，务必仔细展开，注意负号。例如，−2(3 − a) = −6 + 2a，因为 −2 × 3 = −6，而 −2 × (−a) = +2a。

6. Unknowns on Both Sides | 两边均有未知数的方程

When the variable appears on both sides of the equation, our goal is to collect all variable terms on one side and all number terms on the other. This usually involves adding or subtracting a multiple of x to both sides, then solving as usual.

当方程两边都出现变量时，我们的目标是把所有含变量的项集中到一边，把所有数字项集中到另一边。这通常涉及在两边同时加上或减去一个带有 x 的倍数，然后照常求解。

Solve 7x − 3 = 4x + 9. Subtract 4x from both sides → 3x − 3 = 9. Add 3 → 3x = 12. Divide by 3 → x = 4. Check: 7 × 4 − 3 = 28 − 3 = 25, and 4 × 4 + 9 = 16 + 9 = 25 ✓.

解 7x − 3 = 4x + 9。两边同减 4x → 3x − 3 = 9。加 3 → 3x = 12。除以 3 → x = 4。验算：7 × 4 − 3 = 28 − 3 = 25，而 4 × 4 + 9 = 16 + 9 = 25 ✓。

Solve 2(3a + 2) = 4a + 18. Expand: 6a + 4 = 4a + 18. Subtract 4a → 2a + 4 = 18. Subtract 4 → 2a = 14. Divide by 2 → a = 7. Always check your solution in the original equation before declaring it final.

解 2(3a + 2) = 4a + 18。展开：6a + 4 = 4a + 18。减 4a → 2a + 4 = 18。减 4 → 2a = 14。除以 2 → a = 7。务必在宣布答案前，将解代回原方程进行验算。

A special case occurs when simplifying leads to a statement like 0 = 0 (infinitely many solutions) or something false like 0 = 5 (no solution). For example, 4x + 6 = 4x + 10 simplifies to 6 = 10, which is impossible, so there is no solution.

一种特殊情况是化简后得到像 0 = 0 这样的恒等式（无穷多解），或者得出谬误如 0 = 5（无解）。例如，4x + 6 = 4x + 10 化简后得到 6 = 10，这是不可能的，因此方程无解。

7. Equations Involving Fractions | 含分数的方程

Fractions in equations often look intimidating, but you can clear them by multiplying every term by the lowest common denominator (LCD) of all the fractions. This transforms the equation into a more familiar linear form.

方程中的分数常常看起来令人生畏，但你可以通过将所有项乘以所有分母的最小公分母 (LCD) 来消去分数，将方程转化为更熟悉的一次形式。

Solve x/3 + 2 = 5. Multiply every term (including the 2 and 5) by 3: (x/3)×3 + 2×3 = 5×3 → x + 6 = 15 → x = 9. You could also subtract 2 first and then multiply by 3; the answer remains the same.

解 x/3 + 2 = 5。将每一项（包括 2 和 5）都乘以 3：(x/3)×3 + 2×3 = 5×3 → x + 6 = 15 → x = 9。你也可以先减 2 再乘 3，答案保持不变。

Solve (2x)/5 − 1 = x/3. LCD of 5 and 3 is 15. Multiply every term by 15: 15×(2x/5) − 15×1 = 15×(x/3) → 6x − 15 = 5x. Subtract 5x → x − 15 = 0 → x = 15. Check: (2×15)/5 − 1 = 30/5 − 1 = 6 − 1 = 5, and 15/3 = 5 ✓.

解 (2x)/5 − 1 = x/3。5 和 3 的最小公分母是 15。将每一项乘以 15：15×(2x/5) − 15×1 = 15×(x/3) → 6x − 15 = 5x。减去 5x → x − 15 = 0 → x = 15。验算：(2×15)/5 − 1 = 30/5 − 1 = 6 − 1 = 5，而 15/3 = 5 ✓。

Be careful when a fraction has a binomial numerator like (x+1)/2. Still multiply each term by the LCD: 3 × (x+1)/2 = (3x+3)/2. The brackets must be retained to ensure the entire numerator is multiplied.

当分数的分子是二项式如 (x+1)/2 时要小心。仍需将每一项乘以 LCD：3 × (x+1)/2 = (3x+3)/2。括号必须保留，以确保整个分子都被乘到。

8. Applying Equations to Word Problems | 方程在实际问题中的应用

Many real-life scenarios can be modelled by linear equations. The key is to translate words into algebraic expressions and then form an equation based on equalities described in the problem.

许多现实情境都可以用一元一次方程来建模。关键在于将文字转化为代数表达式，然后根据问题中描述的等量关系建立方程。

Example: “Three times a number increased by seven equals forty.” Let the number be n → 3n + 7 = 40. Subtract 7 → 3n = 33 → n = 11.

示例：”一个数的三倍加七等于四十。” 设这个数为 n → 3n + 7 = 40。减 7 → 3n = 33 → n = 11。

Work systematically: (1) read carefully and identify the unknown; (2) assign a variable (often x or n); (3) build expressions for each part of the story; (4) write the equation using the equals sign where the problem states “is”, “gives”, “equals”, etc.; (5) solve; (6) check your answer in the context of the original wording.

按部就班进行：(1) 仔细阅读并确定未知量；(2) 设定变量（常用 x 或 n）；(3) 为每个情节部分构建表达式；(4) 在问题出现”是”、”得出”、”等于”等字眼的地方用等号写出方程；(5) 求解；(6) 在原文字语境中检查你的答案。

Perimeter problems are common: “A rectangle has a length that is twice its width. The perimeter is 54 cm. Find the width.” Let width = w, length = 2w. Perimeter = 2(l + w) = 2(2w + w) = 6w. Equation: 6w = 54 → w = 9 cm. Width is 9 cm, length 18 cm.

周长问题很常见：”一个矩形的长是宽的两倍，周长为 54 cm。求宽。” 设宽 = w，长 = 2w。周长 = 2(l + w) = 2(2w + w) = 6w。方程：6w = 54 → w = 9 cm。宽为 9 cm，长为 18 cm。

9. Common Mistakes and How to Avoid Them | 常见错误与如何避免

Even confident students can slip up. One frequent mistake is forgetting to apply an operation to every term on both sides. For instance, when solving 4x − 3 = x + 6, a student might subtract x from 4x but forget to subtract x from the −3, leading to 3x − 3 = 6. That step is actually correct if only subtracting x from the 4x and x terms. The real error is forgetting to carry the operation across all terms, like dividing only part of an expression.

即使是自信的学生也可能会出错。一个常见错误是忘记对两边所有项都施加操作。例如，解 4x − 3 = x + 6 时，学生可能从 4x 中减去 x，却忘记从 −3 中减去 x，导致 3x − 3 = 6。这一步其实是对的，如果只是从 4x 和 x 项中减 x。真正的错误在于操作时漏掉了某些项，比如只对整个表达式的一部分进行除法。

Another error: mishandling negative signs, especially when expanding brackets like −(2x − 5) becoming −2x − 5 instead of −2x + 5. Always remember that a minus outside a bracket flips the signs of all terms inside.

另一个错误：处理负号不当，尤其是展开括号时，比如 −(2x − 5) 变成了 −2x − 5 而不是 −2x + 5。务必记住，括号外的减号会翻转里面所有项的符号。

Careless arithmetic while adding or subtracting decimals and fractions is also a trap. Write each step clearly and avoid skipping stages. Use the check method: substitute your solution back into the original equation – if the two sides do not match, retrace your steps.

在加减小数和分数时的粗心计算也是一个陷阱。清晰地写下每一步，避免跳步。使用验算方法：将你的解代回原方程——如果两边不相等，就重新追溯你的步骤。

10. Checking Solutions and Building Good Habits | 检验解与养成好习惯

Checking your answer is not an optional extra; it is an essential part of solving equations. After finding a value, plug it into the left-hand side, evaluate, and then plug it into the right-hand side. If both yield the same number, your solution is almost certainly correct.

检验答案并不是可选的额外步骤，而是解方程的必要环节。得出一个值之后，把它代入左边计算，再代入右边计算。如果两边得到相同的数，那么你的解几乎肯定正确。

Form the habit of writing ‘LHS =’ and ‘RHS =’ on your working paper. For the equation 5(2x − 3) = 4x + 4, and solution x = 9.5 or 19/2, LHS = 5(2×9.5 − 3) = 5(19 − 3) = 5×16 = 80; RHS = 4×9.5 + 4 = 38 + 4 = 42. Wait – they are not equal, so the solution is wrong. This reveals an error and prompts you to re-solve. (The correct solution should be x = 19/6 or around 3.167, which gives LHS = RHS = 50/3? Let’s quickly check: 5(2×(19/6) − 3) = 5(38/6 − 18/6) = 5(20/6) = 100/6 = 50/3; RHS = 4×(19/6) + 4 = 76/6 + 24/6 = 100/6 = 50/3). So indeed checking caught the mistake.)

养成在草稿纸上写”左边 =”和”右边 =”的习惯。对于方程 5(2x − 3) = 4x + 4，假设你得出解 x = 9.5 或 19/2，左边 = 5(2×9.5 − 3) = 5(19 − 3) = 5×16 = 80；右边 = 4×9.5 + 4 = 38 + 4 = 42。等等——它们不相等，所以解是错的。这就揭示了一个错误，促使你重新求解。（正确的解应该是 x = 19/6 或约 3.167，代入后左边 = 右边 = 50/3）。因此检验确实能发现错误。

Neat presentation reduces mistakes. Align the equals signs vertically, show each inverse operation explicitly, and always use a new line for each logical step. Over time, these habits will make solving linear equations second nature.

整齐的书写能减少错误。将等号垂直对齐，明确写出每个逆运算，并且每个逻辑步骤都用新的一行。久而久之，这些习惯会让解一元一次方程变得像第二天性一样自然。

11. Extension: Literal Linear Equations and Formulae | 拓展：字母系数方程与公式

Sometimes you need to solve for a variable in a formula where there are other letters. The same balance principles apply. For example, make t the subject of v = u + at. Subtract u: v − u = at. Divide by a: t = (v − u)/a. Every operation is still a balancing act.

有时你需要在含有其他字母的公式中解出某个变量。同样的平衡原理依然适用。例如，把 t 变为公式 v = u + at 的主项。减去 u：v − u = at。除以 a：t = (v − u)/a。每一步操作仍然是平衡操作。

Treat the other letters as if they were numbers. This is excellent practice for rearranging formulae in science and for understanding functions. You also learn to be comfortable with fractions and negative terms in solutions.

把其他字母当作数字来处理。这可以很好地为科学课中的公式变形打下基础，也有助于理解函数。你还会学会从容地处理解中的分数和负项。

12. Summary and Further Practice | 总结与进一步练习

Solving linear equations is a skill that grows with methodical practice. Start with simple one-step cases, build up to two-step, then include brackets, unknowns on both sides, and fractions. Always apply the balance rule, use inverse operations in reverse order, check your answers, and present your work clearly.

解一元一次方程是一项通过有条理的练习而不断精进的技能。从简单的一步方程开始，逐步进阶到两步方程，然后纳入括号、两边均有未知数以及分数的情况。始终运用平衡法则，按逆顺序使用逆运算，检验答案，并清晰地展示你的演算过程。

Work through past papers, textbook exercises, and online interactive quizzes. The more you practise, the quicker you will recognise patterns and the less likely you are to make small slips. Remember, every complex algebra topic in Key Stage 4 and IGCSE builds on this fundamental skill.

多加练习过去的真题、课本习题以及在线互动测验。练得越多，你就越能迅速识别模式，并且越不容易犯小错。请记住，Key Stage 4 和 IGCSE 中的每一个复杂代数主题都建立在这个基本技能之上。

Always isolate the variable using opposite operations.
Always keep the equation balanced – what you do to one side, do to the other.
Always write the equals sign on a new line for each transformation.
Always check your final answer by substitution.

始终使用逆运算来隔离变量。
始终保持方程平衡——你对一边所做的，对另一边也要做。
每次变形都另起一行书写等号。
始终通过代入来检验最终答案。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Linear Equations | 解一元一次方程