Solving Linear Equations with Brackets and Fractions | 解含括号和分数的线性方程

📚 Solving Linear Equations with Brackets and Fractions | 解含括号和分数的线性方程

Linear equations are the foundation of algebra at KS3. Being able to solve equations with brackets and fractions helps you model real-world situations and prepares you for harder topics in Cambridge IGCSE and beyond.

线性方程是 KS3 代数的基石。能够解含括号和分数的方程可以帮助你建立现实情境的数学模型，并为剑桥 IGCSE 及更高阶段的学习做好准备。

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

A linear equation is a statement that two expressions are equal, where the unknown appears only to the power of 1. For example, 2x + 3 = 11 is linear because x is not squared or cubed.

线性方程是指两个表达式相等且未知数只出现 1 次方的等式。例如 2x + 3 = 11 就是线性方程，因为 x 没有平方或立方。

2x + 3 = 11

The goal is to find the value of the unknown that makes the equation true. We call this the solution or root of the equation. Examples of non-linear equations include x² = 9 and 1/x = 5, which are not covered at this stage.

我们的目标是求出使等式成立的未知数的值，这个值称为方程的解或根。非线性的例子包括 x² = 9 和 1/x = 5，这些不属于本阶段内容。

2. The Balance Method | 天平法

Think of an equation as a balanced scale. Whatever you do to one side, you must do to the other side to keep the balance.

把方程想象成一个平衡的天平。无论你对一边做了什么，必须对另一边做同样的操作才能保持平衡。

To solve x + 5 = 12, subtract 5 from both sides. This gives x = 7. In a balanced equation, adding, subtracting, multiplying, or dividing both sides by the same non-zero number keeps it true.

解 x + 5 = 12 时，两边同时减去 5，得到 x = 7。在平衡等式中，两边同时加、减、乘或除以同一个非零数，等式仍然成立。

x + 5 − 5 = 12 − 5 → x = 7

The balance method works because equality is preserved under the same operation. This is central to algebraic proof and to solving more difficult equations later.

天平法有效是因为等式在相同运算下保持不变，这正是代数证明以及日后解更复杂方程的核心。

3. Expanding Brackets First | 先去括号

When a bracket appears, expand it before applying the balance method. Use the distributive law: a(b + c) = ab + ac.

当方程中出现括号时，先用分配律 a(b + c) = ab + ac 展开括号，再使用天平法。

For 3(x + 2) = 21, first expand to get 3x + 6 = 21. Then subtract 6 from both sides to get 3x = 15. Finally divide both sides by 3 to get x = 5.

对于 3(x + 2) = 21，先展开得 3x + 6 = 21。然后两边减 6 得 3x = 15。最后两边除以 3 得 x = 5。

3(x + 2) = 21 → 3x + 6 = 21 → x = 5

If the bracket has a negative coefficient, expand carefully: -2(x − 3) = -2x + 6. Many errors happen because the negative sign is not distributed to every term inside the bracket.

如果括号前是负系数，展开时要小心：-2(x − 3) = -2x + 6。很多错误是因为负号没有分配到括号内的每一项。

4. Equations with Unknowns on Both Sides | 未知数在等式两边

If the unknown appears on both sides, collect like terms. First remove the smaller unknown term from both sides, then solve as usual.

如果未知数出现在等式两边，要先合并同类项。通常先消去两边中较小的未知数项，再照常求解。

Solve 5x + 2 = 3x + 10. Subtract 3x from both sides to get 2x + 2 = 10. Then subtract 2 from both sides to get 2x = 8, so x = 4.

解 5x + 2 = 3x + 10。两边同时减去 3x 得 2x + 2 = 10。再两边减 2 得 2x = 8，因此 x = 4。

5x − 3x + 2 = 3x − 3x + 10 → 2x = 8 → x = 4

If you have trouble choosing which side to collect on, collect the variable on the side with the larger coefficient to keep it positive. This reduces sign errors.

如果不确定把未知数合并到哪一边，可以把它移到系数较大的一边，使未知数项保持为正，这样可以减少符号错误。

5. Clearing Fractions | 去分母

To solve equations with fractions, multiply every term by the lowest common denominator (LCD). This clears the fractions and gives a simpler equation.

解含分数的方程时，将每一项乘以最小公分母 (LCD)，这样可以去掉分母，得到更简单的方程。

For x/2 + 3 = 7, the denominator is 2. Multiply every term by 2: 2(x/2) + 2(3) = 2(7), giving x + 6 = 14. Then x = 8.

对于 x/2 + 3 = 7，分母是 2。将每一项乘以 2：2(x/2) + 2(3) = 2(7)，得到 x + 6 = 14，所以 x = 8。

When there are two or more denominators, find the LCD. For example, in x/3 + x/4 = 7, the LCD is 12. Multiply through by 12 to obtain 4x + 3x = 84, so 7x = 84 and x = 12.

当有两个或多个分母时，要找到最小公分母。例如 x/3 + x/4 = 7，LCD 是 12。全式乘以 12 得 4x + 3x = 84，所以 7x = 84，x = 12。

x/3 + x/4 = 7 → ×12 → 4x + 3x = 84 → x = 12

Always check that you multiply both sides and every term by the LCD. For equations like x/2 + x/3 = 5, multiply by 6 to get 3x + 2x = 30.

始终确认将两边及每一项都乘以 LCD。对于 x/2 + x/3 = 5，乘以 6 得 3x + 2x = 30。

6. Common Mistakes to Avoid | 常见错误

A very common error is forgetting to multiply every term when clearing fractions. For example, in x/2 + 1 = 5, do not write x + 1 = 5; you must multiply the 1 as well.

一个常见错误是在去分母时忘记乘以每一个项。例如在 x/2 + 1 = 5 中，不能写成 x + 1 = 5；常数 1 也必须乘以 2。

Another mistake is to add or subtract before expanding brackets. Always deal with brackets first unless the equation is already in a form where brackets can be divided out.

另一个错误是在展开括号前就进行加减。除非方程已经可以约去括号外系数，否则应先处理括号。

Do multiply every term by the LCD.
确实要把每一个项都乘以 LCD。
Do not forget to change signs when moving terms across the equals sign.
移项时不要忘记变号。
Do check your final answer in the original equation.
务必把最终答案代回原方程检验。

7. Word Problems and Real-Life Contexts | 应用题与生活情境

Many KS3 questions ask you to form a linear equation from a word problem. Read the problem carefully, define the unknown with a letter, and write each relationship as an expression.

许多 KS3 题目要求你根据文字题列出线性方程。仔细读题，用字母设未知数，并把每个关系写成表达式。

Example: A rectangle has length 3 cm more than its width. The perimeter is 22 cm. Let the width be w. Then length is w + 3, and perimeter is 2(w + w + 3) = 22.

例题：一个矩形的长比宽多 3 cm，周长是 22 cm。设宽为 w，则长为 w + 3，周长为 2(w + w + 3) = 22。

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Solving Linear Equations with Brackets and Fractions | 解含括号和分数的线性方程