📚 Cambridge KS3 Mathematics: Solving Linear Equations with Brackets and Unknowns on Both Sides | 剑桥KS3数学：解含括号及两边含未知数的一元一次方程
Linear equations are the foundation of algebra at Key Stage 3. In this article, you will learn how to solve equations that contain brackets, fractions, and unknown terms on both sides, using the balancing method.
一元一次方程是 KS3 代数的基础。本文将介绍如何运用天平法解含有括号、分数以及两边都含未知数的方程。
1. What Is a Linear Equation? | 什么是一元一次方程？
of the unknown is 1. For example, 3x + 2 = 14 and 2(x − 5) = 3x + 1 are linear equations.
一元一次方程只有一个未知数，且未知数的最高次数为 1。例如，3x + 2 = 14 和 2(x − 5) = 3x + 1 都是一元一次方程。
The solution is the value of x that makes the equation true.
方程的解就是使等式成立的 x 值。
2. The Balancing Method | 天平法
Think of an equation as a balance. Whatever you do to one side, you must also do to the other side.
把方程想象成天平。你对一边做的运算，另一边也必须做同样的运算。
Key operations:
关键运算：
- Add or subtract the same amount from both sides.
- Multiply or divide both sides by the same non-zero number.
两边同时加上或减去同一个数；两边同时乘或除以同一个非零数。
Example: Solve x + 5 = 12.
例题：解 x + 5 = 12。
x = 12 − 5 = 7
2x = 14, so x = 7
Use inverse operations in reverse order to isolate x.
按相反顺序使用逆运算，把 x 单独留在等号一边。
3. Expanding Brackets First | 先去括号
s, expand them before collecting like terms. Use the distributive law: a(b + c) = ab + ac.
如果方程含有括号，先展开再合并同类项。使用分配律：a(b + c) = ab + ac。
Example: Solve 3(x + 2) = 21.
例题：解 3(x + 2) = 21。
3x + 6 = 21
3x = 15
x = 5
Be careful with negative signs: −2(x − 3) = −2x + 6.
注意负号：−2(x − 3) = −2x + 6。
4. Unknowns on Both Sides | 两边含未知数
When x appears on both sides, collect the x terms on one side and the numbers on the other.
当 x 出现在等式两边时，把含 x 的项移到一边，数字移到另一边。
Example: Solve 5x − 2 = 3x + 8.
例题：解 5x − 2 = 3x + 8。
5x − 3x = 8 + 2
2x = 10
x = 5
Moving a term across the equals sign changes its sign.
项移过等号时要变号。
5. Solving Equations with Fractions | 解含分数的方程
Multiply every term by the lowest common denominator to clear fractions.
将每一项乘以最小公分母，消去分母。
Example: Solve (x + 4) ÷ 3 = 6.
例题：解 (x + 4) ÷ 3 = 6。
x + 4 = 18
x = 14
If the equation is x/2 + 3 = 7, multiply all terms by 2 first: x + 6 = 14, so x = 8.
如果方程是 x/2 + 3 = 7，先每一项乘 2：x + 6 = 14，所以 x = 8。
6. Checking Your Answer by Substitution | 代入检验答案
Always substitute your answer back into the original equation to check.
一定要把答案代回原方程检验。
if x = 5, left side = 5 × 5 − 2 = 23 and right side = 3 × 5 + 8 = 23, so both sides match.
对于 5x − 2 = 3x + 8，若 x = 5，左边 = 5 × 5 − 2 = 23，右边 = 3 × 5 + 8 = 23，两边相等。
7. Common Mistakes to Avoid | 常见错误及避免方法
- Forgetting to multiply every term when clearing fractions.
- Expanding brackets with a negative multiplier incorrectly.
- Changing the sign incorrectly when moving terms across the equals sign.
常见错误包括：去分母时漏乘某项；括号外是负数时展开出错；移项时忘记变号。
Take one step at a time and write each stage clearly.
一步一步来，每一步都写清楚。
8. Word Problems Leading to Equations | 由文字题建立方程
Translate the words into an equation by letting x represent the unknown quantity.
设未知量为 x，把文字翻译成方程。
Example: Three times a number, increased by 4, equals 19. Let the number be x.
例题：一个数的 3 倍再加 4 等于 19。设这个数为 x。
3x + 4 = 19
3x = 15
x = 5
Check: 3 × 5 + 4 = 19.
检验：3 × 5 + 4 = 19。
9. Practice Questions | 练习题
Try these, then check your answers.
尝试以下题目，然后核对答案。
| 1) 4(x − 2) = 16 | Answer: x = 6 |
| 2) 7x + 3 = 4x + 15 | Answer: x = 4 |
| 3) (x + 5) ÷ 2 = 9 | Answer: x = 13 |
| 4) 2(x − 3) = x + 5 | Answer: x = 11 |
10. Summary and Key Takeaways | 小结与要点
Linear equations are solved by keeping the equation balanced, expanding brackets, clearing fractions, and collecting like terms.
解一元一次方程的关键是保持等式平衡、去括号、去分母并合并同类项。
- Use inverse operations in order.
- Always check your solution in the original equation.
按顺序使用逆运算；一定要把解代回原方程检验。
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