Decomposing Complex Problems: Problem-Solving Strategies | 分解复杂问题的解题策略

📚 Decomposing Complex Problems: Problem-Solving Strategies | 分解复杂问题的解题策略

Complex mathematical problems often look intimidating because they combine several ideas at once. The key strategy is to decompose the problem into smaller, manageable sub-problems and then assemble their results. This article presents a structured approach suitable for A-level and equivalent examinations.

复杂的数学问题常常因为同时结合多个知识点而显得难以入手。核心策略是：把大问题分解成若干可管理的小问题，再将结果整合起来。本文介绍一套适合 A-level 及同等水平考试的结构化方法。

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1. Understanding the Problem | 理解题意

Before any calculation is attempted, you must know exactly what the question is asking. Many errors occur because students solve a similar but different problem instead of the one written on the paper.

在任何计算开始之前，你必须准确知道题目在问什么。许多错误发生是因为学生解决的是“看起来相似”的另一个问题，而不是试卷上的原题。

• Read the problem at least twice and underline every numerical fact and condition.

  至少读题两遍，并划出每一个数值事实与条件。

• Restate the question in your own words: “What must I find?”

  用自己的话复述题目：“我需要求出什么？”

• Identify the domain of the answer: a length cannot be negative, a count must be an integer, and a probability must lie between 0 and 1.

  明确答案的范围：长度不能为负，数量必须是整数，概率必须介于 0 和 1 之间。

For example, the question “Solve x² = 4” asks for all real solutions, so x = 2 and x = −2 are both needed. A student who stops at x = 2 has decomposed the problem incompletely.

例如，题目“解方程 x² = 4”要求的是全部实数解，因此 x = 2 和 x = −2 都需要写出。只写到 x = 2 的学生并没有完整地分解这个问题。

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2. Decomposing into Smaller Parts | 将问题分解为更小的部分

The central technique of this entire strategy is division of a large problem into smaller tasks. Each small task can then be solved with a known method. This is often called the divide-and-conquer approach.

整套策略的核心技术是把一个大问题拆分成若干小任务。每个小任务都可以用已知方法解决。这种做法常被称为“分而治之”。

• Split by operation: first set up the equation, then solve the equation, then interpret the result.

  按步骤拆分：先建立方程，再解方程，最后解释结果。

• Split by case: if a question involves absolute value, inequalities, or different intervals, break it into separate cases.

  按情形拆分：如果题目涉及绝对值、不等式或不同区间，就分成独立的情形来处理。

• Split by component: in a geometry problem, separate shape facts from algebra facts; in a rate problem, separate time, distance, and speed.

  按成分拆分：在几何问题中，把图形性质与代数性质分开；在速率问题中，把时间、距离和速度分开。

Consider a train journey that has three parts: travel, waiting, and travel again. The total time is simply the sum of the three smaller times. If each part can be found separately, the whole problem is solved.

考虑一段包含三段过程的火车行程：行驶、等待、再行驶。总时间就是三个小段时间之和。只要每个小段时间能分别求出，整个问题也就解决了。

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3. Identifying Known and Unknown Quantities | 识别已知量与未知量

A powerful way to begin decomposition is to make a clear list of what is known and what is unknown. This turns a word problem into a mathematical statement.

开始分解的一个有效方法是清楚地列出“已知量”和“未知量”。这能把文字题转变成数学表述。

For example, consider a rectangle with perimeter 28 cm and area 45 cm². The known quantities are the perimeter P = 28 and the area A = 45. The unknown quantities are the length l and the width w.

例如，一个长方形的周长为 28 cm，面积为 45 cm²。已知量是周长 P = 28 和面积 A = 45；未知量是长 l 和宽 w。

• Assign variables to every unknown quantity and state what each variable represents.

  为每个未知量设变量，并说明每个变量代表什么。

• Write the conditions as equations: 2(l + w) = 28 and lw = 45.

  把条件写成方程：2(l + w) = 28 和 lw = 45。

• Simplify the first equation to l + w = 14, then decompose further by expressing l = 14 − w.

  将第一个方程化简为 l + w = 14，再进一步分解，写成 l = 14 − w。

The sub-problems now become: substitute l = 14 − w into lw = 45; solve the resulting quadratic; then state both dimensions.

现在的子问题变为：把 l = 14 − w 代入 lw = 45；解所得的一元二次方程；最后写出长和宽。

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4. Translating Words into Mathematical Language | 将文字转译为数学语言

Most real examination questions use words that must be converted into symbols. A set of keywords can help make this translation automatic.

大多数真实考试题目都使用文字，需要将其转换为符号。掌握关键词汇可以帮助你自动完成这种转译。

English phrase	Mathematical meaning
is, equals	=
more than, increased by	+
less than, decreased by	−
of, times, product of	×
per, ratio	÷

For example, “The product of a number and 5 is 20” becomes 5x = 20. The word “is” gives the equals sign, and “product” gives multiplication.

例如，“一个数与 5 的乘积是 20”应写成 5x = 20。单词“is”对应等号，“product”对应乘法。

In higher-level questions, phrases such as “rate of change” are translated as a derivative, and “sum to infinity” signals the use of a geometric series formula. Recognising these phrases is part of decomposition.

在更高层次的问题中，“rate of change”应转译为导数，“sum to infinity”则提示使用无穷等比数列求和公式。识别这些短语本身就是一种分解。

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5. Using a Visual Representation | 使用可视化表示

Drawing a diagram or graph can expose relationships that are hidden in words. Decomposition becomes easier when separate facts are visible at the same time.

画图或作图可以揭示文字背后隐藏的关系。当不同事实同时可见时，分解就会变得更容易。

• Geometry: mark all known lengths and unknown lengths on the diagram.

  几何题：在图上标出所有已知长度和未知长度。

• Probability: draw a tree diagram to separate sequential events or a Venn diagram to separate overlapping categories.

  概率题：画树形图分离先后事件，或用韦恩图分离重叠类别。

• Algebra: sketch a curve to identify roots, turning points, and asymptotes before solving analytically.

  代数题：先大致画出曲线，找出零点、极值点和渐近线，再做代数求解。

For instance, the equation x² − 4x + 3 = 0 can be understood through its graph. The curve y = x² − 4x + 3 crosses the x-axis at x = 1 and x = 3, which immediately suggests the factor form (x − 1)(x − 3) = 0.

例如，方程 x² − 4x + 3 = 0 可以通过图像来理解。曲线 y = x² − 4x + 3 与 x 轴相交于 x = 1 和 x = 3，这立刻提示因式分解形式 (x − 1)(x − 3) = 0。

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6. Solving Subproblems and Combining Results | 解决子问题并合并结果

Once a problem is decomposed, each part should be solved independently. The final step is to combine the results in a logical way, not simply to add them together blindly.

一旦问题被分解，每个部分都应独立求解。最后要做的不是盲目相加，而是以逻辑方式合并结果。

Consider this standard examination problem: a rectangular garden is 3 m longer than it is wide, and its perimeter is 30 m. Find the area.

看一道标准考试题：一个矩形花园比宽度长 3 m，周长为 30 m，求面积。

The problem can be decomposed as follows. First define the width w. Then the length is w + 3. The perimeter equation gives 2(w + (w + 3)) = 30.

这个问题可以这样分解：首先设宽为 w。那么长就是 w + 3。周长方程为 2(w + (w + 3)) = 30。

2(2w + 3) = 30 ⇒ 2w + 3 = 15 ⇒ 2w = 12 ⇒ w = 6

The width is 6 m, so the length is 9 m. The final sub-problem is to find the area: 6 × 9 = 54 m².

宽为 6 m，所以长为 9 m。最后一个子问题是求面积：6 × 9 = 54 m²。

When combining results, check whether the question requires a sum, a difference, a product, or a quotient. In this example, the final combination is a product; in other problems it may be a sum of separate cases.

合并结果时，要检查题目要求的是和、差、积还是商。在这个例子中，最终合并是乘法；在其他问题中，可能要求各情形的和。

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7. Working Backwards from the Answer | 从答案倒推

Some problems become much simpler when solved in reverse. This is especially true for questions involving repeated operations or reversal of a process.

有些问题倒着解会变得简单得多。尤其是涉及重复操作或恢复原过程的问题。

Suppose a price after a 20% discount is £80. To find the original price, work backwards: the final price equals original × 0.8, so original = 80 ÷ 0.8 = 100.

假设一件商品打八折后价格为 £80。要求原价，就应倒推：折扣后价格等于原价 × 0.8，所以原价 = 80 ÷ 0.8 = 100。

Working backwards also applies to sequences and transformations. If a point is translated by vector (3, −2) to reach (5, 1), the original point is found by subtracting the vector: (5 − 3, 1 + 2) = (2, 3).

倒推法也适用于数列和几何变换。如果一个点按向量 (3, −2) 平移后到达 (5, 1)，那么原来点坐标要通过减去向量求得：(5 − 3, 1 + 2) = (2, 3)。

• List the forward steps in order.

  按顺序列出正向步骤。

• Reverse each step in the opposite order.

  按相反顺序逆转每一步。

• Check that the result satisfies the original forward process.

  检查结果是否满足原正向过程。

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8. Looking for Invariants and Symmetry | 寻找不变量与对称性

Not every quantity changes during a problem. Some quantities remain constant, and these invariants can provide a shortcut. Symmetry can also reduce the amount of work needed.

并非每个量在问题中都会改变。有些量保持不变，这些“不变量”可以提供捷径。对称性也能减少计算工作量。

For example, the sum of the exterior angles of any polygon is always 360°. This invariant is independent of the number of sides and can be used to find a missing angle quickly.

例如，任何多边形的外角之和都是 360°。这个不变量与边数无关，可用来快速求出一个未知角。

Symmetry is useful in simultaneous equations. Consider the system:

对称性在方程组中非常有用。考虑以下方程组：

a + b = 10, a² − b² = 40

We decompose the second equation as (a − b)(a + b) = 40. Since a + b = 10, it follows that a − b = 4. Now the problem is simplified to two linear equations:

我们把第二个方程分解为 (a − b)(a + b) = 40。因为 a + b = 10，所以 a − b = 4。现在问题被化简为两个一次方程：

a + b = 10, a − b = 4 ⇒ a = 7, b = 3

The symmetry of the two equations allowed a direct decomposition without expanding the square term.

两个方程的对称性让我们无需展开平方项就能直接完成分解。

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9. Checking and Generalising | 检查与推广

After obtaining a solution, it must be checked against the original problem. This is the final and essential sub-problem: verifying consistency.

得到解答后，必须将其放回原题进行检验。这是最后一个必不可少的子问题：验证一致性。

• Substitute the answer back into every original equation.

  把答案代回每个原始方程。

• Check units: if the question gives metres per second and seconds, the distance must be in metres.

  检查单位：如果题目给的单位是米每秒和秒，那么距离的单位必须是米。

• Check restrictions: denominators cannot be zero, logarithm inputs must be positive, and square roots must be non-negative.

  检查限制条件：分母不能为零，对数的真数必须为正，根号内必须非负。

For example, solving 1/(x − 2) = 3 gives x − 2 = 1/3, so x = 7/3. The value x = 2 would make the denominator zero and must be rejected.

例如，解方程 1/(x − 2) = 3 得到 x − 2 = 1/3，因此 x = 7/3。若 x = 2 会使分母为零，必须舍去。

A completed solution should also be generalised when possible. If you have proved a pattern for three consecutive numbers, try the same reasoning for n consecutive numbers. This practice builds a deeper understanding and prepares you for unfamiliar questions.

当可能时，还应将解答推广。如果你证明了三个连续整数具有某个规律，就试着用同样的推理处理 n 个连续整数。这种练习能加深理解，并帮助你应对陌生题目。

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10. Practice Questions and Final Summary | 练习题与总结

The following questions should be solved by applying the decomposition strategies described above. Do not skip the checking stage.

以下练习题应运用上面介绍的分解答题策略来完成。不要跳过检验步骤。

• 1. Solve 7x − 3 = 2x + 12.

  1. 解方程 7x − 3 = 2x + 12。

• 2. The sum of two numbers is 28 and their difference is 12. Find the two numbers.

  2. 两个数之和为 28，差为 12。求这两个数。

• 3. A rectangle has area 60 cm² and its length is 4 cm more than its width. Find the perimeter.

  3. 一个长方形面积为 60 cm²，长比宽多 4 cm。求周长。

Question 1: subtract 2x from both sides, then add 3, then divide by 5 to obtain x = 3.

第 1 题：两边减去 2x，再加 3，最后除以 5，得到 x = 3。

Question 2: x + y = 28 and x − y = 12. Adding gives 2x = 40, so x = 20, and therefore y = 8.

第 2 题：x + y = 28，x − y = 12。两式相加得 2x = 40，所以 x = 20，进而 y = 8。

Question 3: let width be w. Then w(w + 4) = 60, so w² + 4w − 60 = 0, which factors as (w + 10)(w − 6) = 0. Since w > 0, w = 6 and length = 10, so perimeter = 32 cm.

第 3 题：设宽为 w，则 w(w + 4) = 60，即 w² + 4w − 60 = 0，因式分解为 (w + 10)(w − 6) = 0。因为 w > 0，所以 w = 6，长为 10，周长为 32 cm。

In summary, the universal problem-solving strategy is to understand the problem, decompose it into smaller parts, translate each part into mathematics, solve each part, combine the results, and finally check the answer. Mastering this cycle will allow you to face complex examination questions with confidence.

Published by TutorHao | Mathematics Revision Series | aleveler.com

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