📚 Solving Linear Equations with Unknowns on Both Sides | 解未知数在两侧的一元一次方程
Welcome to this KS3 Cambridge Mathematics revision guide. In this article we focus on a key algebra skill: solving linear equations where the unknown appears on both sides, exactly the type of question often found in Exercise 2 on page 196 of the Cambridge Lower Secondary Mathematics course. You will learn how to balance equations, expand brackets, clear fractions, and check your solutions.
欢迎阅读这篇 KS3 剑桥数学复习指南。本文聚焦核心代数技能：解未知数同时出现在等号两侧的一元一次方程，这正是剑桥初中数学教材第 196 页练习 2 中常见的题型。你将学会如何平衡方程、展开括号、去分母，并检验所得解。
1. Understanding Linear Equations | 认识一元一次方程
= 14 is linear because the unknown x is not squared, cubed, or raised to any higher power.
一元一次方程是两边的表达式相等且未知数最高次数为 1 的代数等式。例如，3x + 5 = 14 就是一元一次方程，因为未知数 x 没有被平方、立方或更高次乘方。
Linear equation: ax + b = c
一元一次方程：ax + b = c
In KS3 Cambridge Mathematics, you are expected to solve equations that may require several steps: simplifying each side, expanding brackets, and moving terms between sides while keeping the equation balanced.
在 KS3 剑桥数学中，你应当能够求解需要多个步骤的方程：化简方程两边、展开括号，以及在保持等式平衡的前提下移动项。
2. The Balancing Method | 天平平衡法
Think of an equation as a balance scale. The equals sign is the pivot, and both sides must always have the same value. Whatever operation you perform on one side, you must also perform on the other side.
把方程想象成一个天平。等号是支点，两边必须始终保持相等的值。无论你对一边执行什么运算，也必须对另一边执行相同的运算。
= 5. The equation remains balanced because the same subtraction was applied to both sides.
例如，解 x + 7 = 12 时，两边同时减去 7。这样得到 x + 7 − 7 = 12 − 7，所以 x = 5。方程保持平衡，因为两边都执行了相同的减法。
3. Collecting Like Terms First | 先合并同类项
Before moving terms across the equals sign, simplify each side by collecting like terms. Like terms have exactly the same variable raised to the same power, such as 4x and −x, or 2 and 9.
在把项移到等号另一边之前，先合并同类项来化简两边。同类项是指变量和指数完全相同的项，例如 4x 和 −x，或 2 和 9。
becomes 3x + 2 = 13. Then subtract 2 from both sides: 3x = 11, and divide by 3: x = 11/3.
例如，解 4x + 2 − x = 13 时，先将 4x 和 −x 合并为 3x。方程变为 3x + 2 = 13。然后两边减 2：3x = 11，再除以 3：x = 11/3。
Always check each side before starting to move terms. A common mistake is to move a term before simplifying its own side, which can lead to sign errors or missed like terms.
在开始移项之前，一定要先检查两边。一个常见错误是在化简自己所在的一边之前就移动某项，这可能导致符号错误或漏掉同类项。
4. Expanding Brackets Before Solving | 先展开括号再求解
ac to expand them before collecting like terms. This is often essential when the unknown appears inside brackets on either or both sides.
当方程含有括号时，先用分配律 a(b + c) = ab + ac 展开括号，再合并同类项。当未知数出现在一侧或两侧的括号内时，这一步通常必不可少。
side: 3x + 12 = 27. Then subtract 12 from both sides: 3x = 15. Finally divide by 3: x = 5.
例如，解 3(x + 4) = 27。展开左边：3x + 12 = 27。然后两边减去 12：3x = 15。最后除以 3：x = 5。
2(x − 1) = 3 becomes 5 − 2x + 2 = 3 after expanding, because −2 multiplied by −1 gives +2.
当括号前出现负号时要特别小心。例如，5 − 2(x − 1) = 3 在展开后变为 5 − 2x + 2 = 3，因为 −2 乘以 −1 得到 +2。
5. Working with Unknowns on Both Sides | 处理未知数在两侧的方程
When the unknown appears on both sides of the equals sign, your first goal is to collect all variable terms on one side and all constant terms on the other side. Choose the side that keeps the coefficient of the variable positive if possible.
当未知数同时出现在等号两边时，你的首要目标是把所有含未知数的项集中到一边，把所有常数项集中到另一边。如果可能，尽量选择能让未知数系数保持为正的一边。
11, which gives 3x + 2 = 11. Then subtract 2 from both sides: 3x = 9. Divide by 3: x = 3.
例如，解 5x + 2 = 2x + 11。两边同时减去 2x：5x − 2x + 2 = 2x − 2x + 11，得到 3x + 2 = 11。然后两边减去 2：3x = 9。除以 3：x = 3。
. For example, 8 − x = 2x + 1 can be changed to 8 − 1 = 2x + x, giving 7 = 3x, so x = 7/3.
如果未知数最终带负系数，你可以将整个方程乘以 −1，或者干脆把含未知数的项移到另一边。例如，8 − x = 2x + 1 可变为 8 − 1 = 2x + x，得到 7 = 3x，所以 x = 7/3。
6. Dealing with Fractional Coefficients | 处理分数系数
Equations with fractions can be simplified by multiplying every term on both sides by the common denominator. This removes the fractions and makes the equation easier to solve.
含有分数的方程可以通过将两边每一项都乘以公分母来化简。这样就能消去分数，使方程更容易求解。
vely, multiply the original equation by 2 first: x + 6 = 14, then subtract 6 to get x = 8.
例如，解 x/2 + 3 = 7。先两边减去 3：x/2 = 4。然后两边乘以 2：x = 8。也可以先将原方程乘以 2：x + 6 = 14，再减去 6 得到 x = 8。
iply both sides by 3 first: 2x + 1 = 15. Then subtract 1: 2x = 14, and divide by 2: x = 7.
如果分数的分子是二项式，例如 (2x + 1)/3 = 5，先两边乘以 3：2x + 1 = 15。然后减去 1：2x = 14，再除以 2：x = 7。
7. Checking Solutions by Substitution | 代入检验解
After solving any equation, always substitute your answer back into the original equation to verify that the left side equals the right side. This is a quick way to catch arithmetic mistakes.
解出任何方程后，始终将答案代回原方程，检验左边是否等于右边。这是快速发现算术错误的好方法。
he right side is 2(3) + 11 = 6 + 11 = 17. Both sides equal 17, so the solution is correct.
例如，如果你解 5x + 2 = 2x + 11 得到 x = 3，检验时计算左边：5(3) + 2 = 15 + 2 = 17。右边为 2(3) + 11 = 6 + 11 = 17。两边都等于 17，因此解是正确的。
Substitution is especially important when the equation involves brackets or fractions, where mistakes in signs or distribution are easy to make.
当方程含有括号或分数时，代入检验尤其重要，因为符号或分配律很容易出错。
8. Common Errors and How to Avoid Them | 常见错误及避免方法
Below is a table of common errors students make when solving linear equations. Read each one carefully and try to avoid them in your own work.
下面是学生在解一元一次方程时常犯的错误表。请仔细阅读每一项，并在自己的练习中尽量避免。
| Mistake | 错误 | Why it is wrong | 为什么错 | Correction | 正确做法 |
| Forgetting to change sign when moving a term | 移项时忘记变号 | Moving a term across the equals sign is equivalent to adding or subtracting it on both sides, so its sign changes. | Treat every move as the same operation on both sides; write the sign change clearly. |
| Incorrectly expanding a negative bracket | 展开负括号时出错 | A negative sign outside a bracket multiplies every term inside, including constants. | Expand carefully: −2(x − 5) = −2x + 10. |
| Multiplying only one side by the denominator | 只把一边乘以分母 | The balance is broken unless every term on both sides is multiplied. | Multiply the entire equation by the common denominator. |
| Not checking the final answer | 没有检验最终答案 | Small arithmetic errors can pass unnoticed. | Always substitute the solution back into the original equation. |
9. Worked Example: Page 196 Exercise 2 Style Problem | 页码196练习2类型例题
Let us work through a question similar to those found on page 196, Exercise 2 of the Cambridge Lower Secondary Mathematics course.
让我们完整解答一道与剑桥初中数学教材第 196 页练习 2 类似的题目。
Solve 4(2x − 1) = 3(x + 5) + 2
解方程 4(2x − 1) = 3(x + 5) + 2
2x − 1) = 8x − 4. On the right, 3(x + 5) = 3x + 15, so the right side becomes 3x + 15 + 2.
步骤 1：展开两边。左边 4(2x − 1) = 8x − 4。右边 3(x + 5) = 3x + 15，所以右边变为 3x + 15 + 2。
by collecting the constant terms 15 and 2 to get 17. The equation is now 8x − 4 = 3x + 17.
步骤 2：化简右边，合并常数项 15 和 2 得到 17。方程现在为 8x − 4 = 3x + 17。
on the left. Subtract 3x from both sides: 8x − 3x − 4 = 3x − 3x + 17, giving 5x − 4 = 17.
步骤 3：将含未知数的项集中到左边。两边减去 3x：8x − 3x − 4 = 3x − 3x + 17，得到 5x − 4 = 17。
Step 4: Add 4 to both sides: 5x = 21. Then divide by 5: x = 21/5 or 4.2.
步骤 4：两边加 4：5x = 21。然后除以 5：x = 21/5 或 4.2。
46/5) + 2 = 138/5 + 10/5 = 148/5 = 29.6. Both sides are equal, so the solution is correct.
步骤 5：将 x = 21/5 代入原方程进行检验。左边：4(2(21/5) − 1) = 4(42/5 − 5/5) = 4(37/5) = 148/5 = 29.6。右边：3(21/5 + 5) + 2 = 3(21/5 + 25/5) + 2 = 3(46/5) + 2 = 138/5 + 10/5 = 148/5 = 29.6。两边相等，因此解正确。
10. Practice Questions and Answers | 练习题与答案
Try these problems before looking at the answers. Write each step clearly and check your final solution by substitution.
在查看答案之前先尝试做这些题。请清晰写出每一步，并通过代入检验最终解。
- 1. 2x + 3 = x + 7
- 2. 5(x − 2) = 3x + 4
- 3. (3x + 1)/2 = 8
- 4. 7 − 3(x + 1) = 2x − 5
- 5. 4x − 5 = 2(x + 3) + 1
Answers: 1. x = 4 | 2. x = 7 | 3. x = 5 | 4. x = 9/5 | 5. x = 6
答案：1. x = 4 | 2. x = 7 | 3. x = 5 | 4. x = 9/5 | 5. x = 6
3 = 2x − 5, so 4 − 3x = 2x − 5. Adding 3x and 5 to both sides gives 9 = 5x, hence x = 9/5.
例如第 4 题，展开得 7 − 3x − 3 = 2x − 5，所以 4 − 3x = 2x − 5。两边同时加 3x 和 5 得 9 = 5x，因此 x = 9/5。
11. Key Takeaways | 重点总结
Solving linear equations with unknowns on both sides is a core KS3 Cambridge algebra skill. Always simplify each side first, expand brackets carefully, collect variable terms on one side, and carry out the same operation on both sides to keep the equation balanced.
解未知数在两侧的一元一次方程是 KS3 剑桥代数的核心技能。务必先化简两边，仔细展开括号，将含未知数的项集中到一边，并对两边执行相同运算以保持方程平衡。
Remember to clear fractions by multiplying every term by the common denominator, watch for negative signs before brackets, and always check your solution by substitution. Regular practice of these steps will build strong algebraic fluency for Cambridge Checkpoint and beyond.
记住通过将每一项乘以公分母来消去分数，注意括号前的负号，并始终通过代入检验解。坚持练习这些步骤，将为剑桥 Checkpoint
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