📚 Solving Equations with Unknowns on Both Sides | 解含两侧未知数的方程
This article explores the key concepts and skills needed to solve linear equations where the unknown variable appears on both sides of the equals sign. Such equations are a core part of the Cambridge KS3 Mathematics curriculum, appearing frequently from Stage 8 onwards. Mastering them strengthens algebraic manipulation and prepares students for more advanced problem‑solving in later years.
本文探讨解一元一次方程所需的核心理念与技巧，这类方程在等号两侧均含有未知数。它是剑桥 KS3 数学课程的重要组成部分，从 Stage 8 起频繁出现。熟练掌握这类方程能够强化代数操作能力，为今后更高阶的问题求解打下基础。
1. Understanding the Equation | 理解方程
An equation is a statement that two mathematical expressions have the same value. When an unknown, often represented by a letter such as x, appears on both the left‑hand side (LHS) and the right‑hand side (RHS), the equation takes the general form ax + b = cx + d. The letters a, b, c and d stand for known numbers. Our task is to find the value of x that makes the equality true.
方程是表明两个数学表达式值相等的陈述。当未知数（通常用字母 x 表示）同时出现在等号左边和右边时，方程的一般形式为 ax + b = cx + d。字母 a、b、c 和 d 代表已知数。我们的任务就是找出使等式成立的 x 的值。
Until this point, you have probably solved equations where the unknown only sits on one side. The new challenge is to collect together all the x‑terms on one side and all the constant (number) terms on the other side. This process relies heavily on the concept of maintaining balance.
在此之前，你可能解过的方程中未知数都只位于一侧。新的挑战是将所有含 x 的项集中到一边，所有常数项集中到另一边。此过程极大地依赖于“保持平衡”的概念。
2. The Balance Method Revisited | 重温天平法
Think of an equation like a perfectly balanced set of scales. Whatever operation you perform on one side, you must perform exactly the same operation on the other side to keep the scales balanced. For example, adding 5 to the LHS and the RHS preserves equality; multiplying both sides by 3 also maintains equality.
想象方程就像一架完美平衡的天平。无论你对一边做了什么运算，对另一边也必须做完全相同的运算，才能保持天平平衡。比如，两边同时加 5 保持等价；两边同时乘 3 也保持等价。
When an unknown appears on both sides, we use the balance method to remove the variable from one side. If we have 5x + 2 = 3x + 10, we can subtract 3x from both sides. This eliminates the x‑term on the RHS and collects all the x‑terms on the LHS, leaving 2x + 2 = 10. The equation now looks like a simpler one‑sided unknown equation, which you already know how to solve.
当未知数出现在两边时，我们运用天平法将变量从其中一边消除。比如 5x + 2 = 3x + 10，可以从两边同时减去 3x。这样右边含 x 的项消失了，所有 x 项都集中到左边，得到 2x + 2 = 10。现在方程看起来就像只有一个未知数的简易方程，你已经知道如何求解。
Always ask yourself: ‘Am I keeping the scales balanced?’ when performing each step. This mindset prevents common errors such as adding on one side while subtracting from the other.
每一步运算时，都要问自己：“我是否保持天平平衡了？”这种思维习惯能够防止一边加、另一边减等常见错误。
3. A Step-by-Step Strategy | 分步求解策略
Adopting a clear sequence of operations will help you tackle any equation with unknowns on both sides. The recommended approach is to simplify each side first if there are parentheses, then bring the variable terms together on whichever side makes the coefficient positive, and finally isolate the unknown.
采用清晰的操作顺序，可以帮助你解决任何两侧含未知数的方程。推荐的做法是：如果有括号先化简每一边，然后把变量项集中到能使系数为正的一边，最后单独分离出未知数。
The standard steps can be summarised in the following table:
标准步骤可归纳如下表格：
| Step | Action | Example for 4x + 7 = 2x – 3 |
| 1 | Expand brackets if present | (none here) |
| 2 | Collect x‑terms on one side (subtract 2x) | 4x – 2x + 7 = -3 → 2x + 7 = -3 |
| 3 | Move constant to the other side (subtract 7) | 2x = -3 – 7 = -10 |
| 4 | Divide by the coefficient of x | x = -10 ÷ 2 = -5 |
| 5 | Check your solution | LHS: 4(-5)+7=-13; RHS: 2(-5)-3=-13 ✔ |
By following these five stages in order, you turn a tricky‑looking equation into a straightforward calculation. Practice with discipline until the process becomes automatic.
按顺序遵循这五个阶段，你可以把看似棘手的方程变成直截了当的计算。坚持练习，直到这一过程变成条件反射。
4. Worked Example 1 – Positive Coefficients | 样例 1 – 正系数
Let’s solve 7x – 1 = 3x + 11. Step 1: there are no brackets, so we skip that. Step 2: collect x‑terms. Since 7x > 3x, we choose to keep x on the LHS to keep its coefficient positive. Subtract 3x from both sides: 7x – 3x – 1 = 3x – 3x + 11, giving 4x – 1 = 11.
我们来解 7x – 1 = 3x + 11。第 1 步：没有括号，跳过。第 2 步：集中 x 项。因为 7x > 3x，我们选择把 x 保留在左边，以使系数为正。两边同时减 3x：7x – 3x – 1 = 3x – 3x + 11，得到 4x – 1 = 11。
Step 3: move the constant term -1 to the RHS by adding 1 to both sides: 4x – 1 + 1 = 11 + 1, so 4x = 12. Step 4: divide both sides by 4: x = 12 ÷ 4 = 3. Step 5: check: LHS: 7(3) – 1 = 20; RHS: 3(3) + 11 = 20. Both sides match, so x = 3 is correct.
第 3 步：把常数项 -1 移到右边，两边同时加 1：4x – 1 + 1 = 11 + 1，即 4x = 12。第 4 步：两边除以 4：x = 12 ÷ 4 = 3。第 5 步：检验：左边 7(3)-1=20，右边 3(3)+11=20。两边相等，所以 x = 3 正确。
Remember that you could have subtracted 7x instead, but that would give -4x = 12, leading to the same answer after dividing by -4. Choosing to make the coefficient positive at the start reduces sign errors.
要记得，你也可以选择减去 7x，这样会得到 -4x = 12，除以 -4 后答案相同。一开始就选择使系数为正，可以减少符号错误。
5. Worked Example 2 – Negative Coefficients and Terms | 样例 2 – 负系数与负项
Solve 2x – 9 = 5x + 6. This time, the coefficient on the RHS (5) is larger than on the LHS (2). To keep the coefficient positive, it is better to bring the x‑terms to the RHS. Subtract 2x from both sides: 2x – 2x – 9 = 5x – 2x + 6, leaving -9 = 3x + 6.
解方程 2x – 9 = 5x + 6。这一次，右边系数（5）比左边（2）大。为了让系数保持正数，最好把 x 项集中到右边。两边同时减 2x：2x – 2x – 9 = 5x – 2x + 6，得到 -9 = 3x + 6。
Now subtract 6 from both sides to move the constant: -9 – 6 = 3x + 6 – 6, giving -15 = 3x. Divide both sides by 3: -15 ÷ 3 = 3x ÷ 3, so -5 = x, or x = -5. Check: LHS: 2(-5) – 9 = -19; RHS: 5(-5) + 6 = -19. Correct.
现在两边减 6 来移动常数：-9 – 6 = 3x + 6 – 6，得到 -15 = 3x。两边除以 3：-15 ÷ 3 = 3x ÷ 3，所以 -5 = x，或 x = -5。检验：左边 2(-5)-9=-19，右边 5(-5)+6=-19。正确。
Notice how dealing with the multiplication at the end simplifies the process. Never guess the value of x; always use a step‑by‑step approach.
请注意，把乘除运算留到最后处理可以简化过程。千万不要靠猜 x 的值；始终坚持分步操作。
6. Dealing with Parentheses | 处理括号
Equations frequently include brackets that must be expanded before any terms are moved. For instance, 3(2x + 1) = 4(x – 2) + 5. The first action is to multiply out the brackets: on the LHS, 3 × 2x + 3 × 1 = 6x + 3. On the RHS, 4 × x – 4 × 2 = 4x – 8, then add 5: 4x – 8 + 5 = 4x – 3. The equation simplifies to 6x + 3 = 4x – 3.
方程中常包含括号，在移动任何项之前必须展开。例如，3(2x + 1) = 4(x – 2) + 5。第一步是去括号：左边 3×2x + 3×1 = 6x + 3。右边 4×x – 4×2 = 4x – 8，再加 5：4x – 8 + 5 = 4x – 3。方程简化为 6x + 3 = 4x – 3。
Now collect x‑terms: subtract 4x from both sides → 2x + 3 = -3. Subtract 3 from both sides → 2x = -6. Divide by 2 → x = -3. Always simplify each side fully before collecting like terms across the equals sign. A common mistake is to forget to multiply every term inside the bracket by the number outside.
现在集中 x 项：两边减 4x → 2x + 3 = -3。两边减 3 → 2x = -6。除以 2 → x = -3。一定要在合并等号两边的同类项之前，把每一边分别简化完毕。一个常见错误是忘记用括号外的数乘以括号内的每一项。
7. Equations Containing Fractions | 含分数的方程
When fractions appear, multiplying every term by the lowest common denominator (LCD) clears the fractions and turns the equation into an integer equation. For example, solve (x/3) + 1 = (x – 2)/4. The LCD of 3 and 4 is 12. Multiply every term by 12: 12×(x/3) + 12×1 = 12×[(x – 2)/4]. This gives 4x + 12 = 3(x – 2).
当方程含有分数时，将每一项乘以最简公分母（LCD）可以消去分母，将方程转化为整数方程。例如，解 (x/3) + 1 = (x – 2)/4。3 和 4 的 LCD 是 12。将每一项乘以 12：12×(x/3) + 12×1 = 12×[(x – 2)/4]。得到 4x + 12 = 3(x – 2)。
Now expand the RHS: 3x – 6. The equation is 4x + 12 = 3x – 6. Subtract 3x from both sides: x + 12 = -6. Subtract 12: x = -18. Check: LHS (-18/3)+1 = -6+1=-5; RHS (-18-2)/4 = -20/4=-5. Correct.
现在展开右边：3x – 6。方程为 4x + 12 = 3x – 6。两边减 3x：x + 12 = -6。减 12：x = -18。检验：左边 (-18/3)+1 = -6+1=-5；右边 (-18-2)/4 = -20/4=-5。正确。
Be careful when multiplying both sides by the LCD: if a term is not a fraction, still multiply it by the LCD to keep the equation balanced. Use brackets around numerators when needed to avoid sign errors.
两边乘以 LCD 时要小心：如果某一项不是分数，仍要乘以 LCD 以保持等式平衡。必要时用括号括住分子，避免符号错误。
8. Equations Involving Decimals and Percentages | 涉及小数和百分数的方程
Sometimes you will meet equations like 0.2x + 0.7 = 0.5x – 0.1. Although you can solve them directly, it is often easier to multiply through by a power of 10 to eliminate the decimals. Here, multiply each term by 10: 2x + 7 = 5x – 1. Now proceed as usual: subtract 2x from both sides → 7 = 3x – 1; add 1 → 8 = 3x; divide by 3 → x = 8/3.
你有时会遇到类似 0.2x + 0.7 = 0.5x – 0.1 的方程。尽管可以直接求解，但通常更简便的做法是两边同时乘以 10 的幂以消去小数。这里每一项乘以 10：2x + 7 = 5x – 1。现在按常规步骤求解：两边减 2x → 7 = 3x – 1；加 1 → 8 = 3x；除以 3 → x = 8/3。
The same technique applies to percentages. Rewrite a percentage such as 25% as a fraction (1/4) or a decimal (0.25), then clear fractions or decimals as preferred. Consistency is key; don’t multiply only some terms by the chosen number.
同样的技巧也适用于百分数。先将百分数如 25% 改写为分数 (1/4) 或小数 (0.25)，然后根据需要消去分数或小数。关键在于保持一致：不要只给部分项乘以所选的倍数。
9. Real‑Life Applications | 实际应用
These equations model many real‑world situations. For instance, Two companies offer different pricing plans: Alpha charges a £5 fixed fee plus £3 per item; Beta charges £1 per item plus a £9 fixed fee. For how many items will the total cost be the same? Let the number of items be x. Alpha cost = 3x + 5; Beta cost = x + 9. Equate them: 3x + 5 = x + 9. Solve: 2x = 4, so x = 2 items.
这类方程可以模拟许多实际情境。例如，两家公司提供不同的定价方案：Alpha 收 5 英镑固定费用加每件 3 镑；Beta 收每件 1 镑加 9 镑固定费用。购买多少件时总费用相同？设件数为 x。Alpha 费用 = 3x + 5；Beta 费用 = x + 9。列出方程：3x + 5 = x + 9。求解：2x = 4，所以 x = 2 件。
Another common problem involves ages. If John is three times as old as his son and in 12 years John will be twice as old as his son, find the son’s current age. Let the son’s age be x; John is 3x. In 12 years: John = 3x + 12, son = x + 12. Equation: 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12. The son is 12 years old.
另一个常见问题是年龄问题。若约翰的年龄是他儿子的三倍，12 年后他将是他儿子年龄的两倍，求儿子现在的年龄。设儿子年龄为 x，约翰为 3x。12 年后：约翰 = 3x + 12，儿子 = x + 12。方程：3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12。儿子现年 12 岁。
Recognising how to translate words into algebraic expressions is a vital skill. Look for phrases like ‘equal to’, ‘the same as’ and ‘twice as many’ to construct your equation.
学会如何将文字翻译为代数表达式是一项关键技能。留意“等于”“与……相同”“是……的两倍”等短语，以建立方程。
10. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
Even confident students can slip up. One frequent error is forgetting to change the sign when moving a term. For example, in 4x + 3 = x – 7, some learners incorrectly write 4x = x – 7 – 3 instead of subtracting 3 properly. Always remember to perform the same operation on both sides, rather than just ‘moving’ terms mentally.
即使自信的学生也可能出错。一个常见错误是移项时忘记变号。例如，在 4x + 3 = x – 7 中，有些同学错误地写成 4x = x – 7 – 3，而不是正确地减去 3。永远要记住对两边执行相同运算，而不仅仅在脑中“移项”。
Another pitfall is incomplete expansion of brackets, especially when a minus sign precedes the bracket: 5x – 2(x – 3) becomes 5x – 2x + 6, not 5x – 2x – 6. Multiplying the second term by -2 gives +6. Using brackets while working can prevent this. Also, after finding x, always substitute it back into the original equation to verify.
另一个陷阱是括号展开不全，尤其是括号前有负号时：5x – 2(x – 3) 应展开为 5x – 2x + 6，而不是 5x – 2x – 6。因为 -2 乘以 -3 得 +6。书写时使用括号可以避免此类错误。此外，求出 x 后，一定要代回原方程检验。
- Error: 2(x + 5) = 2x + 5 → Fix: 2(x + 5) = 2x + 10
- Error: 3x – 4 = x → subtract x gives 2x = -4 → Fix: 3x – 4 – x = 0 → 2x – 4 = 0 → 2x = 4
- Error: Dividing only one term by a number when clearing fractions → Fix: multiply every term.
中英对照避免错误清单：括号乘法漏乘；移项不变号；解方程后不检验；分数方程未每项乘分母。有意识地放慢速度，写出每步操作，能大幅提高正确率。
11. Practice Problems | 练习题
Try solving these equations on your own. The answers are provided below so you can check your work.
尝试独立解出下列方程。答案附后，以便核对。
- 6x – 5 = 2x + 7
- 4(3x – 1) = 5x + 10
- x/2 + 4 = x – 1
- 0.4x + 0.9 = 0.1x – 0.3
- 3x – 2(x + 1) = 8 – x
Answers: 1. x = 3; 2. x = 2; 3. x = 10; 4. x = -4; 5. x = 5.
答案：1. x = 3；2. x = 2；3. x = 10；4. x = -4；5. x = 5。
If you made any mistakes, go back and identify which step failed. Often, re‑checking expansion and sign changes reveals the issue. Keep a notebook of common slip‑ups to review before assessments.
如果你出了错，请回到原题并找出哪一步出了问题。通常，重新检查展开与符号变化就能发现问题所在。准备一本错题本，在测评前翻阅易犯错误，是很好的方法。
12. Summary and Key Takeaways | 总结与要点
Solving equations with unknowns on both sides is a logical extension of simpler one‑sided equations. The core principle remains that of keeping the balance and systematically isolating the variable. Always simplify each side first, collect the variable on the side that gives a positive coefficient, and check your answer by substitution.
解含两侧未知数的方程是简单一侧方程的逻辑延伸。核心原则依然是保持平衡并有步骤地分离变量。始终先化简每一边，将变量集中在能使系数为正的一边，并通过代入检验答案。
Regular practice with brackets, fractions and negative coefficients builds fluency. As you progress through KS3, you will meet ever more complex algebraic challenges, but the balanced, step‑by‑step method will remain your most trusted tool. Keep practising and remember that every equation can be tamed with patience and clear reasoning.
经常练习带括号、分数和负系数的题目，能够提升熟练度。在 KS3 的学习进程中，你会遇到更复杂的代数挑战，但保持平衡、分步求解的方法永远是你最可靠的利器。坚持练习，并记住：只要有耐心和清晰的推理，任何方程都能被驯服。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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