Solving Equations with Algebraic Fractions | 解含代数分数的方程

📚 Solving Equations with Algebraic Fractions | 解含代数分数的方程

Algebraic fractions appear frequently in IGCSE Mathematics examinations. Learning to solve equations that contain algebraic fractions is a fundamental skill that connects your knowledge of fractions, factorisation, and linear or quadratic equations. This revision guide will walk you through every essential method step by step, with worked examples tailored to the Edexcel IGCSE syllabus.

含代数分数的方程在IGCSE数学考试中频繁出现。学会解含有代数分数的方程，是连接分数运算、因式分解以及一次或二次方程知识的关键技能。本复习指南将针对爱德思IGCSE考纲，一步步带你掌握每一个核心方法，并配有标准例题。

1. What Are Algebraic Fractions? | 什么是代数分数？

An algebraic fraction is a fraction whose numerator or denominator contains an algebraic expression, such as (x + 1)/2, 3/(x – 2), or (2x + 1)/(x + 3). When such fractions appear inside an equation, the unknown variable x may appear in the numerator, the denominator, or both.

代数分数是指分子或分母中含有代数表达式的分数，例如 (x + 1)/2、3/(x – 2) 或 (2x + 1)/(x + 3)。当这类分数出现在方程中时，未知数 x 可能出现在分子、分母或同时出现在两者之中。

A key rule to remember is that the denominator of any fraction can never equal zero. If your solution makes a denominator zero, that value must be rejected.

需要牢记的一条关键规则是：任何分数的分母都不能等于零。如果你的解使某个分母为零，那么这个解必须被舍去。

2. The Golden Rule: Multiply by the Denominator(s) | 黄金法则：乘以分母

The most powerful strategy for solving equations involving algebraic fractions is to eliminate the fractions by multiplying both sides of the equation by the common denominator. This transforms the equation into a simpler form that contains no fractions.

解含代数分数方程最强大的策略是通过在方程两边同时乘以公分母来消去分数。这样可以将方程转化为不含分数的更简单形式。

For example, consider the equation:

例如，考虑方程：

x/2 + 3 = 5

Notice that the denominator of the only fraction is 2. Multiplying every term on both sides by 2 gives:

注意到唯一分数的分母是2。在两边每一项都乘以2得到：

2 × (x/2) + 2 × 3 = 2 × 5

x + 6 = 10

x = 4

Always multiply every term on both sides of the equation, not just the fractional terms, to keep the equation balanced.

切记要在方程两边的每一项上都乘以该数，而不仅仅是含有分数的项，这样才能保持方程平衡。

3. Single Fraction Equals a Number | 单个分数等于一个数

When an equation has a single algebraic fraction on one side and a number on the other, multiply both sides by the denominator of the fraction.

当方程的一边只有一个代数分数，而另一边是一个数时，在方程两边乘以该分数的分母即可。

Worked Example 1 | 例题1：

7/(x – 1) = 2

Multiply both sides by (x – 1):

两边同乘 (x – 1)：

7 = 2(x – 1)

7 = 2x – 2

9 = 2x

x = 9/2 = 4.5

Check: when x = 4.5, the denominator x – 1 = 3.5, which is not zero, so this solution is valid.

验算：当 x = 4.5 时，分母 x – 1 = 3.5，不为零，因此解有效。

4. Fraction Equals Fraction (Cross-Multiplication) | 分数等于分数（交叉相乘）

When the equation has exactly one fraction on each side, the fastest method is cross-multiplication. If a/b = c/d, then a × d = b × c.

当方程两边各只有一个分数时，最快的方法是交叉相乘。若 a/b = c/d，则 a × d = b × c。

Worked Example 2 | 例题2：

3/(x + 2) = 5/(2x – 1)

Cross-multiplying:

交叉相乘：

3(2x – 1) = 5(x + 2)

6x – 3 = 5x + 10

6x – 5x = 10 + 3

x = 13

Check: neither denominator equals zero when x = 13, since x + 2 = 15 and 2x – 1 = 25. The solution is x = 13.

验算：当 x = 13 时，两个分母都不为零，因为 x + 2 = 15，2x – 1 = 25。所以解为 x = 13。

5. Fractions with Linear Denominators | 含一次线性分母的分数

When an equation contains more than two terms with different linear denominators, you need to find the lowest common multiple (LCM) of all the denominators and multiply every term by it.

当方程包含两个以上带有不同一次分母的项时，你需要找出所有分母的最小公倍数 (LCM)，并将每一项都乘以这个最小公倍数。

Worked Example 3 | 例题3：

4/x + 3/(x + 1) = 2

The denominators are x and (x + 1). Their LCM is x(x + 1). Multiply each term by x(x + 1):

分母为 x 和 (x + 1)。它们的最小公倍数为 x(x + 1)。将每一项乘以 x(x + 1)：

x(x + 1) × 4/x + x(x + 1) × 3/(x + 1) = x(x + 1) × 2

4(x + 1) + 3x = 2x(x + 1)

Expanding and simplifying:

展开并化简：

4x + 4 + 3x = 2x² + 2x

7x + 4 = 2x² + 2x

0 = 2x² – 5x – 4

Using the quadratic formula x = (-b ± √(b² – 4ac))/2a with a = 2, b = -5, c = -4:

使用二次方程求根公式 x = (-b ± √(b² – 4ac))/2a，其中 a = 2，b = -5，c = -4：

x = (5 ± √(25 + 32))/4 = (5 ± √57)/4

Neither solution makes a denominator zero, so both are valid.

两个解都不会使分母为零，因此都有效。

6. Quadratic Denominators and Factorisation | 二次分母与因式分解

Sometimes the denominator is a quadratic expression, such as x² – 1 or x² – 3x + 2. Before finding the LCM, factorise every quadratic denominator fully. This often reveals that one denominator is actually a multiple of another.

有时分母是二次表达式，例如 x² – 1 或 x² – 3x + 2。在求最小公倍数之前，先将每个二次分母完全因式分解。这通常会揭示出某个分母实际上是另一个分母的倍数。

Worked Example 4 | 例题4：

2/(x² – 1) + 1/(x + 1) = 1

Factorise x² – 1 = (x + 1)(x – 1). So the denominators are (x + 1)(x – 1) and (x + 1). The LCM is (x + 1)(x – 1). Multiplying every term by this LCM:

因式分解 x² – 1 = (x + 1)(x – 1)。因此分母分别为 (x + 1)(x – 1) 和 (x + 1)。最小公倍数为 (x + 1)(x – 1)。将每一项乘以这个最小公倍数：

2 + (x – 1) = (x + 1)(x – 1)

2 + x – 1 = x² – 1

x + 1 = x² – 1

x² – x – 2 = 0

Factorising gives (x – 2)(x + 1) = 0, so x = 2 or x = -1. However, substituting x = -1 into the original equation creates a denominator of zero in both fractions. Therefore x = -1 must be rejected, and the only valid solution is x = 2.

因式分解得 (x – 2)(x + 1) = 0，所以 x = 2 或 x = -1。然而，将 x = -1 代入原方程会使两个分数的分母均为零。因此 x = -1 必须舍去，唯一有效解为 x = 2。

7. Recognising Excluded Values | 识别排除值

Before solving any equation containing algebraic fractions, it is wise to identify the excluded values of x — the values that make any denominator equal to zero. These values cannot be part of the solution set.

在解任何含代数分数的方程之前，明智的做法是先识别 x 的排除值——即会使任一分母为零的值。这些值不能出现在解集之中。

If a denominator is (x – 3), then x = 3 is excluded.
如果分母为 (x – 3)，则 x = 3 被排除。
If a denominator is (2x + 5), then x = -5/2 is excluded.
如果分母为 (2x + 5)，则 x = -5/2 被排除。
If a denominator is x² – 4 = (x – 2)(x + 2), then x = 2 and x = -2 are excluded.
如果分母为 x² – 4 = (x – 2)(x + 2)，则 x = 2 和 x = -2 被排除。

After solving the equation, always compare your answers with the excluded values. If any answer coincides with an excluded value, delete it from the final solution set.

解出方程后，务必把你的答案与排除值进行比较。如果某个答案与排除值重合，就要从最终解集中删除它。

8. Solving When x Appears in the Numerator | 当 x 出现在分子中时

If x appears only in the numerators, the equation behaves much like an ordinary linear or quadratic equation. The fractions are simply numerical coefficients that can be cleared by multiplying by their common denominator.

如果 x 只出现在分子中，这个方程的行为就与普通的一次或二次方程非常相似。这些分数只是数值系数，可以通过乘以它们的公分母来消去。

Worked Example 5 | 例题5：

(x + 1)/3 + (x – 1)/4 = 5

The LCM of 3 and 4 is 12. Multiply every term by 12:

3 和 4 的最小公倍数为 12。将每一项乘以 12：

4(x + 1) + 3(x – 1) = 60

4x + 4 + 3x – 3 = 60

7x + 1 = 60

7x = 59

x = 59/7

When x is only in the numerator, denominator restrictions are usually not an issue, since ordinary numbers never equal zero. Still, always check your final answer.

当 x 只出现在分子中时，分母限制通常不是问题，因为普通数值永远不会等于零。不过，仍然要检查最终答案。

9. Simplifying Fractions Before Solving | 解题前先化简分数

Before applying any method, look carefully at each algebraic fraction. If the numerator and denominator share a common factor, simplify the fraction first. This can turn a complicated quadratic into a simple linear expression.

在运用任何方法之前，仔细审视每一个代数分数。如果分子与分母含有公因式，先化简该分数。这可以把一个复杂的二次式化成简单的一次式。

Worked Example 6 | 例题6：

(x² – 9)/(x – 3) = 7

Notice that x² – 9 = (x + 3)(x – 3). Therefore:

注意 x² – 9 = (x + 3)(x – 3)。因此：

((x + 3)(x – 3))/(x – 3) = x + 3

The original equation simplifies to x + 3 = 7, giving x = 4.

原方程化简为 x + 3 = 7，得到 x = 4。

Important: we cancelled (x – 3), which is only valid when x ≠ 3. In the original equation, x = 3 is indeed excluded, so the cancellation is legitimate here.

重要提示：我们约去了 (x – 3)，这仅在 x ≠ 3 时成立。在原方程中，x = 3 确实是被排除的，所以此处的约分是合理的。

10. Forming Quadratic Equations from Fractions | 由分数构建二次方程

Many IGCSE examination questions are designed so that clearing the denominators produces a quadratic equation. You must then solve it either by factorisation, completing the square, or the quadratic formula.

许多IGCSE考题的设计意图是：去分母后会得到一个二次方程。此时你必须通过因式分解、配方法或求根公式来解这个二次方程。

Worked Example 7 | 例题7：

(x + 1)/x + x/2 = 3

The denominators are x and 2. The LCM is 2x. Multiply every term by 2x:

分母为 x 和 2。最小公倍数为 2x。将每一项乘以 2x：

2(x + 1) + x² = 6x

2x + 2 + x² = 6x

x² – 4x + 2 = 0

Since this does not factorise nicely, use the quadratic formula:

由于该二次式不易因式分解，使用求根公式：

x = (4 ± √(16 – 8))/2 = (4 ± √8)/2 = 2 ± √2

The excluded value is x = 0. Neither 2 + √2 nor 2 – √2 equals zero, so both are valid solutions.

排除值为 x = 0。2 + √2 和 2 – √2 都不等于零，因此两者均为有效解。

11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students frequently lose marks on algebraic fraction questions due to a small number of repeated errors. Being aware of these can greatly improve your accuracy.

学生在含代数分数题目上丢分，通常是由于少数几个反复出现的错误。了解这些错误可以极大提升你的准确率。

Mistake / 错误	Example / 示例	Correct Approach / 正确做法
Forgetting to multiply non-fraction terms by the LCM	解 x/3 + 2 = 5 时只把 x/3 乘以3	Multiply every term by the LCM: x + 6 = 15
Forgetting to multiply non-fraction terms by the LCM（中文对照）	Solving x/3 + 2 = 5 by multiplying only the fraction by 3	每一项都乘以最小公倍数：x + 6 = 15
Ignoring excluded values	Accepting x = -1 in Example 4	Reject x = -1 because it makes denominators zero
Ignoring excluded values（中文对照）	在例题4中接受了 x = -1	舍去 x = -1，因为它使分母为零
Incorrect cross-multiplication when a side has more than one term	Treating 1/x + 1/2 = 1/3 as if only 1/x and 1/3 exist	First combine terms on each side, then cross-multiply
Incorrect cross-multiplication when a side has more than one term（中文对照）	把 1/x + 1/2 = 1/3 当作只有 1/x 和 1/3 进行交叉相乘	先将每一边的项合并，再进行交叉相乘
Sign errors when expanding negative brackets	Expanding -(x + 2) as -x + 2	Use -x – 2
Sign errors when expanding negative brackets（中文对照）	将 -(x + 2) 展开为 -x + 2	应写成 -x – 2

Always write down the excluded values before you begin solving, and always substitute your final answers back into the original equation to check them.

在开始解题前一定要先写下排除值，并且一定要把最终答案代回原方程进行校验。

12. Exam-Style Practice Questions | 考试风格练习

Below are four practice questions that mirror the style and difficulty of Edexcel IGCSE questions. Try each one before reading the solution.

以下是四道模拟爱德思IGCSE题型与难度的练习题。请先尝试作答，再阅读解答。

Question 1 | 练习1：

5/(2x – 1) = 3/x

Question 2 | 练习2：

(x + 2)/5 – (x – 1)/2 = 1

Question 3 | 练习3：

2/(x + 1) + 3/(x – 2) = 1

Question 4 | 练习4：

4/(x² – 4) + 1/(x – 2) = 2/(x + 2)

Solutions | 解答：

Question 1: Cross-multiply: 5x = 3(2x – 1) → 5x = 6x – 3 → x = 3. Valid since neither denominator is zero.

练习1：交叉相乘：5x = 3(2x – 1) → 5x = 6x – 3 → x = 3。两个分母均不为零，解有效。

Question 2: LCM of 5 and 2 is 10. Multiply: 2(x + 2) – 5(x – 1) = 10 → 2x + 4 – 5x + 5 = 10 → -3x + 9 = 10 → x = -1/3.

练习2：5 和 2 的最小公倍数为 10。乘以10：2(x + 2) – 5(x – 1) = 10 → 2x + 4 – 5x + 5 = 10 → -3x + 9 = 10 → x = -1/3。

Question 3: LCM is (x + 1)(x – 2). Multiply: 2(x – 2) + 3(x + 1) = (x + 1)(x – 2). Then 2x – 4 + 3x + 3 = x² – x – 2 → 5x – 1 = x² – x – 2 → x² – 6x – 1 = 0 → x = 3 ± √10. Both valid.

练习3：最小公倍数为 (x + 1)(x – 2)。乘以：2(x – 2) + 3(x + 1) = (x + 1)(x – 2)。即 2x – 4 + 3x + 3 = x² – x – 2 → 5x – 1 = x² – x – 2 → x² – 6x – 1 = 0 → x = 3 ± √10。两者均有效。

Question 4: Factorise x² – 4 = (x + 2)(x – 2). LCM is (x + 2)(x – 2). Multiplying gives 4 + (x + 2) = 2(x – 2). Then x + 6 = 2x – 4 → x = 10. Valid since no denominator is zero.

练习4：因式分解 x² – 4 = (x + 2)(x – 2)。最小公倍数为 (x + 2)(x – 2)。乘以得 4 + (x + 2) = 2(x – 2)。即 x + 6 = 2x – 4 → x = 10。所有分母均不为零，解有效。

Solving equations with algebraic fractions becomes straightforward once you master three core steps: identify excluded values, eliminate denominators by multiplying by the LCM, and solve the resulting linear or quadratic equation. Always verify your answers in the original equation. With regular practice, you will find these questions to be some of the most predictable marks on the paper.

解含代数分数的方程，只要掌握三个核心步骤就会变得直截了当：识别排除值、乘以最小公倍数消去分母、解所得的一次或二次方程。始终将答案代入原方程验算。通过规律练习，你会发现这类题目是试卷中最容易稳定拿分的题型之一。

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Solving Equations with Algebraic Fractions | 解含代数分数的方程