KS3 Cambridge Maths: Solving Linear Equations with Brackets and Fractions | KS3 剑桥数学：解含括号和分数的线性方程

📚 KS3 Cambridge Maths: Solving Linear Equations with Brackets and Fractions | KS3 剑桥数学：解含括号和分数的线性方程

Welcome to this Cambridge KS3 Maths revision guide on solving linear equations. This article focuses on the skills you need when an equation includes brackets, fractions, or unknowns on both sides – exactly the type of question often found in the p247_2.pdf worksheet. Work through the methods step by step, then test yourself with the practice questions.

欢迎阅读本剑桥 KS3 数学复习指南，主题是解线性方程。本文重点讲解方程中带括号、分数或两边都有未知数时所需的解题技巧，这正是 p247_2.pdf 练习中常见的题型。请逐步学习解题方法，然后用练习题进行自测。

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1. What Is a Linear Equation? | 什么是线性方程？

A linear equation is a statement that two expressions are equal, and the unknown variable is only raised to the power 1. You will see equations such as x + 7 = 12, 3x – 4 = 11, 2(x + 5) = 18, and x/3 + 1 = 6. The main task is to find the value of the unknown that makes the statement true.

线性方程是指两个表达式相等，且未知数的次数仅为 1 的等式。你会遇到如 x + 7 = 12、3x – 4 = 11、2(x + 5) = 18 和 x/3 + 1 = 6 这样的方程。主要任务是求出使等式成立的未知数的值。

3x + 5 = 20

2(x + 4) = 18

x/5 + 2 = 6

In Cambridge KS3 Mathematics, linear equations often appear in non-calculator papers, in word problems, and in sequences. You must be able to solve them accurately and show clear working.

在剑桥 KS3 数学中，线性方程经常出现在非计算器试卷、应用题和数列题中。你必须能够准确求解，并写出清晰的解题过程。

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2. The Balance Method and Inverse Operations | 天平法与逆运算

Think of an equation as a balance scale. To keep the equation true, you must do exactly the same operation to both sides. The opposite operations used to undo each other are addition and subtraction, and multiplication and division.

把方程想象成一台天平。为了保持方程成立，你必须对方程两边做完全相同的运算。互为逆运算的操作包括：加法和减法、乘法和除法。

• Inverse operations: + and – undo each other; × and ÷ undo each other. — 逆运算：加和减互相抵消；乘和除互相抵消。

For example, to solve x + 8 = 15, subtract 8 from both sides to get x = 7. To solve 4x = 20, divide both sides by 4 to get x = 5. Always apply the inverse operation in the reverse order of the order of operations.

例如，解 x + 8 = 15 时，两边同时减去 8，得到 x = 7。解 4x = 20 时，两边同时除以 4，得到 x = 5。始终按照运算顺序的相反顺序来使用逆运算。

x + 8 = 15 → x = 15 – 8 → x = 7

4x = 20 → x = 20 ÷ 4 → x = 5

Operation | 运算	Inverse operation | 逆运算
+ 6	– 6
– 9	+ 9
× 3	÷ 3
÷ 5	× 5

Keep your work tidy by writing each new equation directly below the previous one. This makes checking much easier.

书写要整齐，把每个新的方程直接写在上一个方程下面。这样检查起来会容易得多。

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3. Solving Equations with Brackets | 解含括号的方程

When an equation has brackets, you can expand the brackets first using the distributive law. For example, 3(x + 4) means 3 × x + 3 × 4, which equals 3x + 12. Then solve the resulting equation.

当方程中含有括号时，你可以先用分配律展开括号。例如，3(x + 4) 表示 3 × x + 3 × 4，等于 3x + 12。然后求解所得的方程。

3(x + 4) = 27

3x + 12 = 27

3x = 27 – 12

3x = 15 → x = 5

Another example: solve 4(x – 2) = 20. Expand to get 4x – 8 = 20. Add 8 to both sides: 4x = 28. Divide by 4: x = 7. You can also divide first if the number outside the bracket divides the right side evenly: x – 2 = 5, so x = 7.

另一个例子：解 4(x – 2) = 20。展开得到 4x – 8 = 20。两边加 8：4x = 28。除以 4：x = 7。你也可以先除以括号外的数，如果它能整除右边：x – 2 = 5，所以 x = 7。

When a bracket is multiplied by a negative number, be careful with signs. For example, -2(x + 5) = -2x – 10, not -2x + 10.

当括号乘以负数时，要注意符号。例如，-2(x + 5) = -2x – 10，而不是 -2x + 10。

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4. Solving Equations with Fractions | 解含分数的方程

Equations with fractions usually come in two forms. The first form has a coefficient that is a fraction, such as x/5 or 2x/3. The second form has an expression as the numerator over a denominator, such as (2x + 4)/3.

含分数的方程通常有两种形式。第一种形式是系数为分数，如 x/5 或 2x/3。第二种形式是分子为表达式、分母为数字，如 (2x + 4)/3。

For the first form, use multiplication as the inverse operation. To solve x/5 + 2 = 6, subtract 2 first: x/5 = 4. Then multiply both sides by 5: x = 20.

对于第一种形式，用乘法作为逆运算。解 x/5 + 2 = 6 时，先减 2：x/5 = 4。然后两边同时乘以 5：x = 20。

x/5 + 2 = 6 → x/5 = 4 → x = 20

For the second form, clear the denominator first by multiplying both sides by the denominator. Solve (2x + 4)/3 = 6 by multiplying both sides by 3: 2x + 4 = 18. Then subtract 4: 2x = 14, and divide by 2: x = 7.

对于第二种形式，先通过两边乘以分母来去掉分母。解 (2x + 4)/3 = 6 时，两边同时乘以 3：2x + 4 = 18。然后减 4：2x = 14，再除以 2：x = 7。

(2x + 4)/3 = 6 → 2x + 4 = 18 → 2x = 14 → x = 7

When the coefficient is a fraction such as 2x/3, think of it as 2x divided by 3. To solve 2x/3 = 8, multiply both sides by 3 to get 2x = 24, then divide by 2 to get x = 12.

当系数是分数如 2x/3 时，可以把它看作 2x 除以 3。解 2x/3 = 8 时，两边同时乘以 3，得到 2x = 24，再除以 2，得到 x = 12。

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5. Unknowns on Both Sides | 两边都有未知数

When an equation has the unknown on both sides, collect the variable terms on one side first. A good habit is to remove the smaller unknown term so that the coefficient of the variable stays positive.

当方程两边都有未知数时，首先把含未知数的项移到同一边。一个良好的习惯是消去较小的未知数项，这样未知数的系数会保持为正。

Solve 7x + 2 = 3x + 18. Subtract 3x from both sides to get 4x + 2 = 18. Subtract 2: 4x = 16. Divide by 4: x = 4.

解 7x + 2 = 3x + 18。两边同时减去 3x，得到 4x + 2 = 18。再减 2：4x = 16。除以 4：x = 4。

7x + 2 = 3x + 18 → 4x + 2 = 18 → 4x = 16 → x = 4

Solve 2x + 9 = 5x – 6. Subtract 2x from both sides: 9 = 3x – 6. Add 6 to both sides: 15 = 3x. Divide by 3: x = 5.

解 2x + 9 = 5x – 6。两边同时减去 2x：9 = 3x – 6。两边同时加 6：15 = 3x。除以 3：x = 5。

2x + 9 = 5x – 6 → 9 = 3x – 6 → 15 = 3x → x = 5

Always check that you have not reversed a sign when moving terms. Write each step clearly to avoid losing marks.

移项时始终要注意没有搞错符号。把每一步写清楚，以免丢分。

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6. Combining Brackets and Fractions | 括号与分数综合

Some Cambridge questions combine brackets and fractions in one equation. For example, solve (2(x + 3))/5 = 4. Clear the denominator first by multiplying both sides by 5: 2(x + 3) = 20. Divide by 2: x + 3 = 10. Subtract 3: x = 7.

一些剑桥考试题会把括号和分数结合在同一个方程中。例如，解 (2(x + 3))/5 = 4。先通过两边乘以 5 去分母：2(x + 3) = 20。再除以 2：x + 3 = 10。减 3：x = 7。

(2(x + 3))/5 = 4 → 2(x + 3) = 20 → x + 3 = 10 → x = 7

You can also expand first: 2x + 6 = 20, then solve to get 2x = 14, x = 7. Both methods are valid. Choose the one that gives easier numbers.

你也可以先展开：2x + 6 = 20，然后求解得到 2x = 14，x = 7。两种方法都有效。选择数字更容易处理的那一种。

Another example: solve 3(x – 2)/4 = 6. Multiply by 4: 3(x – 2) = 24. Divide by 3: x – 2 = 8. Add 2: x = 10.

另一个例子：解 3(x – 2)/4 = 6。两边乘以 4：3(x – 2) = 24。除以 3：x – 2 = 8。加 2：x = 10。

3(x – 2)/4 = 6 → 3(x – 2) = 24 → x – 2 = 8 → x = 10

When you see a negative fraction in front of a bracket, such as -(x + 5)/2 = 3, remember that the negative sign applies to the whole numerator after you multiply.

当你看到括号前有负分数时，如 -(x + 5)/2 = 3，请记住，乘以分母后，

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