Solving Linear Equations | 解一元一次方程

📚 Solving Linear Equations | 解一元一次方程

Linear equations are the foundation of algebra. In Key Stage 3 Mathematics, mastering the art of finding the unknown value – often represented by a letter like x – is essential for progressing to more advanced topics. This article breaks down the key methods, step by step, to help you solve linear equations confidently.

一元一次方程是代数的基石。在 KS3 数学中，掌握求未知数（通常用 x 这样的字母表示）的技巧，是进阶到更复杂内容的关键。本文将逐步拆解核心方法，帮助你自信地解出方程，把未知数变成已知。

1. What is a Linear Equation? | 什么是一元一次方程？

A linear equation is a mathematical sentence that shows two expressions are equal, and it contains only one variable raised to the power of 1. There are no squares, cubes, or roots of the variable. The goal is to find the value of the variable that makes the equation true.

一元一次方程是一个数学语句，它表明两个表达式相等，并且只包含一个次数为 1 的变量。方程中不出现变量的平方、立方或根号。我们的目标就是找到使方程成立的变量值。

For example, x + 5 = 12 is a linear equation. The expression on the left (x + 5) balances the right side (12). We solve it to find that x = 7. In Cambridge KS3, you will encounter equations with one operation, two operations, brackets, and variables on both sides.

例如，x + 5 = 12 就是一个一元一次方程。左边表达式 (x + 5) 与右边 12 保持平衡。我们解出 x = 7。在剑桥 KS3 中，你会遇到含一个运算、两个运算、含括号以及变量在等式两边等不同类型的方程。

2. The Balancing Method | 平衡法

Think of an equation as a perfectly 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 algebra: ‘Do unto one side as you do unto the other.’

把方程想象成一个完全平衡的跷跷板。你对一边做的任何操作，都必须对另一边做完全一样的操作，才能保持平衡。这是代数的黄金法则：“一边施之，另一边亦施之”。

If you add 3 to the left, you add 3 to the right. If you divide the left side by 5, you must divide the right side by 5. This principle allows us to isolate the variable on one side of the equation.

如果你在左边加 3，右边也必须加 3。如果你把左边除以 5，右边也必须除以 5。这个原理让我们能够将变量单独分离在等式的一边。

The balancing method is often shown with a scale. Using inverse operations (the opposite operation) helps to ‘undo’ whatever is attached to the variable. Addition undoes subtraction, multiplication undoes division, and vice versa.

平衡法常用天平来表示。使用逆运算（相反的运算）可以帮助“撤销”附加在变量上的运算。加法撤销减法，乘法撤销除法，反之亦然。

3. Solving Equations by Adding or Subtracting | 通过加减解方程

When a number is added to or subtracted from the variable, use the opposite operation to remove it. For x + 8 = 15, subtract 8 from both sides: x + 8 – 8 = 15 – 8, giving x = 7.

当一个数与变量相加或相减时，使用相反的运算将其移除。对于 x + 8 = 15，两边同时减去 8：x + 8 – 8 = 15 – 8，得到 x = 7。

For x – 4 = 11, add 4 to both sides: x – 4 + 4 = 11 + 4, so x = 15. Always write the operation on both sides clearly to avoid mistakes.

对于 x – 4 = 11，两边同时加上 4：x – 4 + 4 = 11 + 4，所以 x = 15。每次都要在两边清晰地写出运算步骤，避免出错。

A quick check: substitute your answer back into the original equation. 15 – 4 = 11, which is true. This habit builds confidence and accuracy.

快速检验：将答案代回原方程。15 – 4 = 11，成立。这个习惯能增强信心并提高准确度。

Equation	Inverse Operation	Solution
x + 9 = 20	Subtract 9	x = 11
x – 6 = 3	Add 6	x = 9
13 = x + 5	Subtract 5	x = 8

4. Solving Equations by Multiplying or Dividing | 通过乘除解方程

If the variable is multiplied by a number, divide both sides by that number. For 3x = 18, divide by 3: 3x ÷ 3 = 18 ÷ 3, hence x = 6. The coefficient of x tells you how many times x has been multiplied.

如果变量乘以了一个数，就将两边除以这个数。对于 3x = 18，两边除以 3：3x ÷ 3 = 18 ÷ 3，因此 x = 6。x 的系数就表示 x 被乘了多少倍。

If the variable is divided by a number, multiply both sides by the denominator. For x / 4 = 7, multiply both sides by 4: (x / 4) × 4 = 7 × 4, so x = 28. Remember that x / 4 is the same as ¼ x.

如果变量除以了一个数，就将两边乘以分母。对于 x / 4 = 7，两边乘以 4：(x / 4) × 4 = 7 × 4，所以 x = 28。记住，x / 4 等同于 ¼ x。

Watch out for negative coefficients. For –y = 10, multiply or divide by –1 to get y = –10. Treat the negative sign as a coefficient of –1.

注意负系数的情况。对于 –y = 10，两边乘以或除以 –1，得到 y = –10。将负号视为系数 –1 来处理。

5. Two-Step Equations | 两步方程

Many equations require two inverse operations. The general order is to undo addition or subtraction first, then handle multiplication or division. For example, 2x + 3 = 11.

许多方程需要两步逆运算。通常的顺序是，先处理加减，再处理乘除。例如，2x + 3 = 11。

Step 1: Subtract 3 from both sides: 2x = 8. Step 2: Divide both sides by 2: x = 4. Always undo the constant term before the coefficient.

第一步：两边减 3，得 2x = 8。第二步：两边除以 2，得 x = 4。一定要先处理常数项，再处理系数。

Another example: x / 5 – 2 = 6. First add 2 to both sides: x / 5 = 8. Then multiply both sides by 5: x = 40. The order matters: imagine the operations that were applied to x and reverse them in the opposite order (like unzipping a jacket).

另一个例子：x / 5 – 2 = 6。首先两边加 2：x / 5 = 8。然后两边乘以 5：x = 40。顺序很重要：想象一下对 x 施加运算的先后顺序，然后以相反的顺序逆向操作（就像拉开夹克拉链一样）。

6. Equations with Brackets | 含有括号的方程

When an equation contains brackets, you have two reliable methods. The first is to expand the brackets using the distributive law. For example, 3(a + 4) = 27.

当方程含有括号时，你有两种可靠的方法。第一种是利用分配律展开括号。例如，3(a + 4) = 27。

Expand: 3 × a + 3 × 4 = 3a + 12 = 27. Then subtract 12 from both sides: 3a = 15. Finally divide by 3: a = 5.

展开：3 × a + 3 × 4 = 3a + 12 = 27。然后两边减 12：3a = 15。最后除以 3：a = 5。

The second method is to treat the bracket as a single package. Divide both sides by the coefficient outside the bracket first. Using the same equation: 3(a + 4) = 27, divide both sides by 3 to get a + 4 = 9. Then subtract 4: a = 5. This approach is often quicker if the division results in a nice integer.

第二种方法是将括号整体当作一个包裹。先两边除以括号外的系数。用同一个方程：3(a + 4) = 27，两边除以 3 得到 a + 4 = 9。然后减 4：a = 5。如果除法能得到一个不错的整数，这种方法通常更快。

For equations like 2(3x – 1) = 5x + 4, expansion is usually the best first step because the bracket contains a variable term.

对于像 2(3x – 1) = 5x + 4 这样的方程，通常最好第一步先展开，因为括号内含有变量项。

7. Equations with Unknowns on Both Sides | 未知数在等号两边的方程

When the variable appears on both sides of the equals sign, the target is to collect all variable terms on one side and all constant terms on the other. For instance, 5x + 2 = 2x + 17.

当变量同时出现在等号两边时，目标是将所有含变量的项移到一边，所有常数项移到另一边。例如，5x + 2 = 2x + 17。

Subtract the smaller variable term from both sides to keep the coefficient positive. Here, subtract 2x from both sides: 5x – 2x + 2 = 2x – 2x + 17, which simplifies to 3x + 2 = 17. Then proceed as a two-step equation: subtract 2, divide by 3, giving x = 5.

先减去较小的变量项，以保持系数为正。这里从两边减去 2x：5x – 2x + 2 = 2x – 2x + 17，化简得 3x + 2 = 17。然后按两步方程处理：减 2，除以 3，得到 x = 5。

If you prefer, you can bring all terms to one side and set the expression equal to zero: 5x + 2 – 2x – 17 = 0 giving 3x – 15 = 0, then x = 5. Both methods are valid.

如果你愿意，也可以将所有项移到一边，使表达式等于零：5x + 2 – 2x – 17 = 0，得 3x – 15 = 0，然后 x = 5。两种方法都有效。

Avoid the common mistake of only moving the variable term without adjusting constants. Always perform the same operation on both sides.

避免只移变量项而不调整常数项这类常见错误。始终对两边执行相同操作。

8. Checking Your Solution | 检验你的解

Substituting the solution back into the original equation is a powerful habit that verifies your work. For the equation 4(x – 3) = 2x + 2, you found x = 7.

将解代回原方程是一个能验证解题过程的强大习惯。对于方程 4(x – 3) = 2x + 2，你求得 x = 7。

Left side: 4(7 – 3) = 4 × 4 = 16. Right side: 2 × 7 + 2 = 14 + 2 = 16. Both sides match, so x = 7 is correct.

左边：4(7 – 3) = 4 × 4 = 16。右边：2 × 7 + 2 = 14 + 2 = 16。两边匹配，所以 x = 7 正确。

This substitution check takes only a few seconds and catches sign errors, arithmetic mistakes, or distributing errors. In exams, it can save marks and reveal when a solution is extraneous (not common for linear, but still good practice).

这种代入检验只需几秒钟，就能发现符号错误、算术错误或分配律错误。在考试中，它能帮你保住分数，也能揭示不成立的解（虽然线性方程少见，但依然是良好习惯）。

9. Forming Equations from Word Problems | 根据应用题列方程

Translating a real-world situation into an equation is a key skill. Read the problem carefully, define your variable (let the unknown be x), and build an equation from the relationships described.

将现实情境转化为方程是一项关键技能。仔细读题，定义变量（设未知数为 x），然后根据描述的关系建立方程。

Example: ‘I think of a number, multiply it by 3, then add 7. The result is 31.’ Let the number be x. The expression is 3x + 7 = 31. Solve to find x = 8.

例题：“我想一个数，把它乘以 3，再加 7，结果是 31。” 设这个数为 x。表达式为 3x + 7 = 31。解得 x = 8。

For perimeter or geometry problems, use known formulas. For instance, a rectangular garden has length twice its width, and its perimeter is 48 m. Let width = w, then length = 2w. Perimeter = 2(w + 2w) = 6w = 48, so w = 8 m, length = 16 m.

对于周长或几何问题，使用已知公式。例如，一个矩形花园的长是宽的两倍，周长为 48 米。设宽 = w，则长 = 2w。周长 = 2(w + 2w) = 6w = 48，所以 w = 8 米，长 = 16 米。

Always check if the answer makes sense in context. A negative length would be a red flag!

始终检查答案在上下文中是否合理。负数长度就是一个警告信号！

10. Common Mistakes to Avoid | 常见错误避免

Mistake 1: Forgetting to balance both sides. When you subtract 4 from one side, you must subtract 4 from the other. Write every operation.

错误 1：忘记对两边保持平衡。当你从一边减 4 时，必须从另一边也减 4。要把每一步运算都写下来。

Mistake 2: Incorrectly applying the distributive property. –2(x – 3) becomes –2x + 6, not –2x – 6. Pay attention to the sign.

错误 2：错误地运用分配律。–2(x – 3) 应该等于 –2x + 6，而不是 –2x – 6。注意符号。

Mistake 3: Dividing incorrectly. In 4x = 2, x = 2 / 4 = ½ or 0.5, not 2. Remember to divide the right side by the coefficient.

错误 3：除法错误。4x = 2 中，x = 2 / 4 = ½ 或 0.5，而不是 2。记住右边也要除以系数。

Mistake 4: Mixing up the order of inverse operations. For 2x – 3 = 7, add 3 first, then divide by 2. Doing it the other way gives a wrong answer.

错误 4：逆运算的顺序颠倒。对于 2x – 3 = 7，先加 3，再除以 2。反过来做会得到错误答案。

Mistake 5: Losing a negative sign. –x = 8 means x = –8. Don’t drop the negative sign without multiplying by –1.

错误 5：弄丢负号。–x = 8 意味着 x = –8。不要在不乘以 –1 的情况下丢掉负号。

11. Practice Examples with Worked Solutions | 练习题与详解

Example 1: Solve 4(x – 2) + 7 = 2x + 9.

例题 1：解 4(x – 2) + 7 = 2x + 9。

Expand: 4x – 8 + 7 = 2x + 9 → 4x – 1 = 2x + 9. Subtract 2x: 2x – 1 = 9. Add 1: 2x = 10. Divide by 2: x = 5. Check: Left = 4(5–2)+7 = 4×3+7 = 19. Right = 2×5+9 = 19. Correct.

展开：4x – 8 + 7 = 2x + 9 → 4x – 1 = 2x + 9。减 2x：2x – 1 = 9。加 1：2x = 10。除以 2：x = 5。检验：左边 = 4(5–2)+7 = 4×3+7 = 19。右边 = 2×5+9 = 19。正确。

Example 2: A farmer has a total of 45 animals, some chickens (c) and some cows. There are 3 times as many chickens as cows. Set up and solve an equation.

例题 2：一个农场主共有 45 只动物，有些鸡 (c)，有些牛。鸡的数量是牛的 3 倍。建立并解方程。

Let cows = x, then chickens = 3x. Total: x + 3x = 4x = 45, so x = 11.25. Since we can’t have a fraction of an animal, adjust: likely the numbers were 11 cows and 33 chickens giving 44, but the question data might be 44 total. Check context—if the total was 45, it produces a non-integer, which suggests the farmer must have miscounted or the problem expects rounding, but typically KS3 problems yield integer solutions. So the original problem would use a multiple of 4 total. Here the algebra is sound: 4x = 44 would give x = 11 cows.

设牛 = x，则鸡 = 3x。总数：x + 3x = 4x = 45，所以 x = 11.25。因为动物不能是分数，需调整：很可能数量是 11 头牛和 33 只鸡，总数为 44，但题目数据可能为 45，产生非整数解，表明要么数据有误，要么题目预期你意识到不能有分数。不过 KS3 题目通常给出整数解。所以原题应用 4 的倍数总数。这里代数过程正确：4x = 44 得 x = 11 头牛。

Example 3: (x + 3)/2 = 5. Multiply both sides by 2: x + 3 = 10. Then x = 7.

例题 3：(x + 3)/2 = 5。两边乘以 2：x + 3 = 10。所以 x = 7。

12. Summary and Key Takeaways | 总结与核心要点

Solving linear equations is all about isolating the variable using inverse operations while keeping the equation balanced. Start with simpler one-step equations and build up to two-step, brackets, and unknowns on both sides.

解一元一次方程的核心就是运用逆运算、保持方程两边平衡，从而分离出变量。从简单的一步方程入手，逐步进阶到两步方程、含括号方程以及未知数在等号两边的方程。

Remember the golden rule: whatever you do to one side, do to the other. Practice forming your own equations from word problems, and always, always check your solution by substituting it back. Master these and you’ll have a solid algebraic foundation for years to come.

记住黄金法则：对一边做什么，对另一边也做同样的操作。多加练习从应用题中列出方程，并且一定要将解代回原式检验。掌握这些内容，你就为今后的代数学习打下了坚实的基础。

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Solving Linear Equations | 解一元一次方程