Solving Linear Equations | 解线性方程

📚 Solving Linear Equations | 解线性方程

Linear equations are one of the most fundamental topics in Key Stage 3 mathematics. They describe relationships where the unknown appears only in simple forms, without powers or roots. Mastering how to solve these equations builds a strong foundation for algebra, problem-solving, and more advanced maths. This article will guide you step by step through the key methods, from one-step equations to more complex ones involving brackets, fractions, and variables on both sides, with plenty of examples and practice tips.

线性方程是 KS3 数学中最基础的主题之一。它们描述了未知数仅以简单形式出现、不含幂或根的等量关系。掌握如何解这些方程能为代数、问题解决和更高阶的数学打下坚实基础。本文将从一步方程开始，逐步带你学习更复杂的类型，包括含括号、分数和两边都有变量的方程，并提供大量示例与练习技巧。

---

1. Understanding Equations | 理解方程

An equation is a mathematical statement that shows two expressions are equal. It always contains an equals sign (=). The left-hand side (LHS) and the right-hand side (RHS) must have the same value. Our goal when solving an equation is to find the value of the unknown (often written as x) that makes the statement true.

方程是表明两个表达式相等的数学语句，它总是包含一个等号（=）。左边（LHS）和右边（RHS）的值必须相同。解方程的目标是找到使等式成立的未知数（常写作 x）的值。

Think of an equation as a balanced scale. 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.

可以把方程想象成一个平衡的天平。你对一边做的任何操作，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。

---

2. One-Step Equations: Addition & Subtraction | 一步方程：加法与减法

For equations like x + 5 = 12, we need to get x by itself. Because 5 is added to x, we do the inverse operation — subtract 5 — on both sides. This gives x + 5 – 5 = 12 – 5, so x = 7.

对于像 x + 5 = 12 这样的方程，我们需要将 x 单独留在一边。由于 x 加了 5，我们在两边同时进行逆运算——减去 5，得到 x + 5 – 5 = 12 – 5，所以 x = 7。

Similarly, for x – 3 = 9, the inverse of subtracting 3 is adding 3. Add 3 to both sides: x – 3 + 3 = 9 + 3, giving x = 12.

类似地，对于 x – 3 = 9，减去 3 的逆运算是加 3。两边同时加 3：x – 3 + 3 = 9 + 3，得到 x = 12。

x + a = b → x = b – a
x – a = b → x = b + a

---

3. One-Step Equations: Multiplication & Division | 一步方程：乘法与除法

When the unknown is multiplied by a number, we divide both sides by that number. For example, 4x = 20 means 4 × x = 20. Dividing both sides by 4 gives x = 20 ÷ 4 = 5.

当未知数乘以一个数时，我们将两边同时除以那个数。例如，4x = 20 意味着 4 × x = 20。两边同时除以 4 得到 x = 20 ÷ 4 = 5。

If an equation has a division, such as x ÷ 6 = 3 (or x/6 = 3), we do the inverse: multiply both sides by 6. So x = 3 × 6 = 18.

如果方程含有除法，如 x ÷ 6 = 3（或 x/6 = 3），我们进行逆运算：两边同时乘以 6。所以 x = 3 × 6 = 18。

a × x = b → x = b ÷ a
x ÷ a = b → x = b × a

---

4. Two-Step Equations | 两步方程

Two-step equations involve two operations applied to the variable. The general approach is to undo the addition or subtraction first, then undo the multiplication or division. Consider 2x + 3 = 11. Step 1: subtract 3 from both sides → 2x = 8. Step 2: divide both sides by 2 → x = 4.

两步方程涉及对变量施加的两种运算。一般方法是先撤销加法或减法，再撤销乘法或除法。以 2x + 3 = 11 为例。第 1 步：两边同时减去 3 → 2x = 8。第 2 步：两边同时除以 2 → x = 4。

Always reverse the order of operations: start with addition/subtraction, then deal with multiplication/division. This mirrors the order of operations in reverse (BIDMAS/BODMAS backwards).

永远要逆推运算顺序：先处理加减，再处理乘除。这相当于逆向的运算顺序（BIDMAS/BODMAS 的逆推）。

For equations like x/5 – 2 = 3, first add 2 to both sides: x/5 = 5. Then multiply both sides by 5: x = 25.

对于像 x/5 – 2 = 3 这样的方程，先两边加 2：x/5 = 5。然后两边乘以 5：x = 25。

---

5. Equations with Brackets | 带括号的方程

When an equation contains brackets, the first step is to expand them using the distributive law. For example, 3(y + 4) = 27 expands to 3y + 12 = 27. Then solve the two-step equation: subtract 12 → 3y = 15, divide by 3 → y = 5.

当方程含有括号时，第一步是用分配律展开括号。例如，3(y + 4) = 27 展开后得到 3y + 12 = 27。接着解两步方程：减去 12 → 3y = 15，除以 3 → y = 5。

Sometimes the coefficient of the bracket is negative, such as -2(x – 5) = 6. Expand carefully: -2 × x = -2x and -2 × (-5) = +10, giving -2x + 10 = 6. Then subtract 10 from both sides → -2x = -4, divide by -2 → x = 2.

有时括号前的系数为负，例如 -2(x – 5) = 6。展开时需小心：-2 × x = -2x，-2 × (-5) = +10，得到 -2x + 10 = 6。然后两边减 10 → -2x = -4，除以 -2 → x = 2。

---

6. Equations with Variables on Both Sides | 变量在两侧的方程

If you see an equation like 5x + 2 = 3x + 10, the variable appears on both sides. The strategy is to collect all variable terms on one side and constant terms on the other. Usually, subtract the smaller variable term from both sides. Here, subtract 3x from both sides → 2x + 2 = 10. Then subtract 2 → 2x = 8, divide by 2 → x = 4.

如果遇到像 5x + 2 = 3x + 10 这样的方程，变量出现在两边。策略是将所有含变量的项移到一边，常数项移到另一边。通常从两边减去较小的变量项。此处两边同时减去 3x → 2x + 2 = 10。然后减 2 → 2x = 8，除以 2 → x = 4。

Always check your solution by substituting back into the original equation: LHS = 5(4)+2=22, RHS = 3(4)+10=22. It balances.

始终将解代回原方程检验：左边 = 5(4)+2=22，右边 = 3(4)+10=22。平衡。

---

7. Solving Equations with Fractions | 含分数的方程

When an equation involves fractions, a powerful technique is to multiply every term by the lowest common denominator (LCD) to clear the fractions. For example, solve x/2 + 3 = x/4 + 5. The LCD of 2 and 4 is 4. Multiply everything by 4: 4×(x/2) + 4×3 = 4×(x/4) + 4×5 → 2x + 12 = x + 20. Now it’s a simple equation: subtract x from both sides → x + 12 = 20, subtract 12 → x = 8.

当方程包含分数时，一个有效的方法是将每一项乘以最小公分母（LCD）来去分母。例如，解 x/2 + 3 = x/4 + 5。2 和 4 的 LCD 是 4。每项乘以 4：4×(x/2) + 4×3 = 4×(x/4) + 4×5 → 2x + 12 = x + 20。现在就变成了简单方程：两边减 x → x + 12 = 20，减 12 → x = 8。

Remember: multiplying a fraction like (x+1)/3 by 3 gives just x+1. Multiply correctly to avoid errors.

记住：像 (x+1)/3 这样的分数乘以 3 后直接得到 x+1。正确进行乘法以避免错误。

---

8. Forming Equations from Word Problems | 从文字题建立方程

Many KS3 exam questions ask you to construct an equation from a real-life situation. The key is to define the unknown clearly, translate words into mathematical operations, and set up an equation.

许多 KS3 考试题目要求你根据实际情况列出方程。关键是要清晰地定义未知数，将文字转化为数学运算，并建立等式。

Example: “I think of a number, multiply it by 3, subtract 7 and the result is 20. Find the number.” Let the number be x. The statement becomes: 3x – 7 = 20. Solving: add 7 → 3x = 27, divide by 3 → x = 9.

示例： “我想了一个数，将它乘以 3，减去 7，结果是 20。求这个数。” 设这个数为 x。陈述变为：3x – 7 = 20。求解：加 7 → 3x = 27，除以 3 → x = 9。

Look for keywords: ‘sum’ means addition, ‘difference’ means subtraction, ‘product’ means multiplication, ‘quotient’ means division. Writing an equation step by step helps you keep track.

留意关键词：’和’意味着加法，’差’意味着减法，’积’意味着乘法，’商’意味着除法。一步步写出方程有助于理清思路。

---

9. Checking Your Solution | 检验你的解

Checking a solution is not just about spotting mistakes — it is a habit that builds confidence. Substitute the value you found back into the original equation and verify that both sides give the same result. If they do, you are correct. If not, retrace your steps.

检验解不仅是为了找出错误——它是一种增强信心的习惯。将你找到的值代回原方程，验证两边是否得到相同的结果。如果相同，你的答案就是正确的。如果不同，重新检查你的步骤。

For example, solve 2(3x – 4) = 10. We expand to 6x – 8 = 10, add 8 → 6x = 18, x = 3. Check: LHS = 2(3×3 – 4) = 2(9 – 4) = 2×5 = 10 = RHS. Correct.

例如，解 2(3x – 4) = 10。展开得 6x – 8 = 10，加 8 → 6x = 18，x = 3。检验：左边 = 2(3×3 – 4) = 2(9 – 4) = 2×5 = 10 = 右边。正确。

This step is especially useful when you suspect you might have made a sign error or an arithmetic slip during expansion.

当你怀疑自己在展开时可能犯了符号错误或算术失误时，这一步尤其有用。

---

10. Common Mistakes to Avoid | 常见错误及避免

Even simple equations can trip you up if you are not careful. Here are some frequent pitfalls and how to avoid them.

即使简单的方程，如果不小心，也可能会绊倒你。以下是一些常见的陷阱以及如何避免。

Mistake 错误	Correct approach 正确方法
Forgetting to keep the balance by performing operation on both sides. 忘记在两边进行相同操作以保持平衡。	Always write the operation on both sides of the equation. 始终在方程的两边同时写出运算。
Incorrectly applying inverse operations, e.g., adding instead of subtracting. 错误地应用逆运算，例如该减时却加了。	Ask: what is the opposite of this operation? Undo addition with subtraction, etc. 问自己：这个运算的反操作是什么？用减法撤销加法，等等。
Distributing a negative sign incorrectly, e.g., -2(x – 3) becomes -2x – 6 instead of -2x + 6. 负号分配错误，例如 -2(x – 3) 写成 -2x – 6 而非 -2x + 6。	Multiply each term inside the bracket by the coefficient, including the sign. 用系数乘以括号内的每一项，包括符号。
Dividing by the coefficient before dealing with addition/subtraction in two-step equations. 在两步方程中先除以系数再处理加减。	Reverse BIDMAS: undo addition/subtraction first. 逆向 BIDMAS：先撤销加减。
Forgetting to multiply every term by the LCD when clearing fractions. 去分母时忘记将每一项乘以 LCD。	Multiply the whole equation by the LCD, every single term. 将整个方程乘以 LCD，每一个单项。

---

11. Practice Questions Walkthrough | 练习题讲解

Let’s work through three typical KS3 exam-style questions, applying the methods we have discussed.

让我们来解决三道典型的 KS3 考试风格题目，运用我们讨论过的方法。

Q1: Solve 4(2x – 1) = 28.
Expand: 8x – 4 = 28. Add 4: 8x = 32. Divide by 8: x = 4. Check: 4(2×4 – 1) = 4(8-1) = 4×7 = 28.

题 1： 解 4(2x – 1) = 28。
展开：8x – 4 = 28。加 4：8x = 32。除以 8：x = 4。检验：4(2×4 – 1) = 4(8-1) = 4×7 = 28。

Q2: Solve 5x + 7 = 3x + 19.
Subtract 3x from both sides: 2x + 7 = 19. Subtract 7: 2x = 12. Divide by 2: x = 6. Check: LHS = 5×6 +7 = 37, RHS = 3×6 +19 = 37.

题 2： 解 5x + 7 = 3x + 19。
两边同时减去 3x：2x + 7 = 19。减 7：2x = 12。除以 2：x = 6。检验：左边 = 5×6 +7 = 37，右边 = 3×6 +19 = 37。

Q3: The perimeter of a rectangle is 38 cm. Its length is 3 cm more than twice its width. Find the dimensions.
Let width = w, then length = 2w + 3. Perimeter = 2(length + width) = 2(2w + 3 + w) = 2(3w + 3) = 6w + 6. Set equal to 38: 6w + 6 = 38. Subtract 6: 6w = 32. Divide by 6: w = 32/6 = 16/3 = 5⅓ cm. Length = 2(5⅓) + 3 = 10⅔ + 3 = 13⅔ cm. Check perim: 2(5⅓ + 13⅔) = 2×19 = 38.

题 3： 一个长方形的周长是 38 cm，其长度比宽度的两倍多 3 cm。求其尺寸。
设宽 = w，则长度 = 2w + 3。周长 = 2(长+宽) = 2(2w + 3 + w) = 2(3w + 3) = 6w + 6。令其等于 38：6w + 6 = 38。减 6：6w = 32。除以 6：w = 32/6 = 16/3 = 5⅓ cm。长度 = 2(5⅓) + 3 = 10⅔ + 3 = 13⅔ cm。检验周长：2(5⅓ + 13⅔) = 2×19 = 38。

---

12. Summary and Key Takeaways | 总结与关键要点

Solving linear equations is like following a clear recipe: isolate the variable by performing inverse operations, maintain balance, and always check your answer. Start with one-step equations to build fluency, then progress to two-step, brackets, variables on both sides, and fractions. Translating word problems into algebraic sentences is a skill that improves with practice. Keep this guide handy when you revise, and test yourself with a variety of questions to become confident and accurate.

解线性方程就像遵循一个清晰的食谱：通过执行逆运算来分离变量，保持平衡，并始终检查你的答案。从一步方程开始建立熟练度，然后发展到两步方程、带括号的方程、两边有变量的方程和含分数的方程。将文字题转化为代数语句是一项通过练习可以提升的技能。在复习时随身携带这份指南，并通过各种题目进行自测，以变得自信和准确。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）