📚 Solving Equations with Algebraic Fractions | 解含代数分数的方程
Algebraic fractions appear frequently in IGCSE Mathematics, especially in equations that require a structured approach to remove denominators. An algebraic fraction is simply a fraction in which the numerator, denominator, or both contain expressions with variables. Solving equations with algebraic fractions involves clearing denominators, simplifying, and solving the resulting polynomial—often linear or quadratic—while always checking for invalid solutions that make any denominator zero. This topic is fundamental for Edexcel IGCSE and builds directly on your skills with rational expressions, factorisation, and equation-solving strategies.
代数分数在IGCSE数学中频繁出现，尤其是在需要系统方法去分母的方程里。代数分数就是分子、分母或两者均含有变量表达式的分数。解含代数分数的方程需要去分母、化简并求解所得的多项式方程（通常是一次或二次方程），同时必须检查那些使分母为零的无效解。本主题是Edexcel IGCSE的基础内容，直接建立在有理表达式、因式分解和方程求解技巧之上。
1. Identifying Algebraic Fractions in Equations | 识别方程中的代数分数
g these forms is the first step because you cannot treat them as ordinary linear equations – ariable in the denominator means the equation is not linear until you clear the fractions.
含代数分数的方程至少有一项的分母中含有变量。例如 3/x = 2, (x+1)/(x-4) = 5 以及 1/(x+2) + 2/(x-3) = 1。识别出这些形式是第一步，因为你不能把它们当成普通的一次方程处理——分母中的变量意味着在去分母之前，该方程不是线性的。
2. The Core Strategy: Clearing Denominators | 核心策略：去分母
To remove denominators safely, we multiply every term on both sides of the equation by the lowest common denominator (LCD) of all the algebraic fractions. Because we are multiplying by an expression that contains the variable, the resulting equation may have extra solutions that make the original denominators zero. Therefore, it is essential to check all potential solutions at the end.
为了安全地消去分母，我们要用所有代数分数的共同分母（即最简公分母，LCD）乘以方程两边的每一项。由于乘上的式子可能含变量，得到的方程可能会引入使原分母为零的额外解。因此，最后必须检验所有可能的解。
The method works for any number of terms and for denominators that are monomials, binomials, or more complex polynomials. We always simplify numerators and factorise denominators where possible before finding the LCD.
该方法适用于任意多项，也适用于单项式、二项式或更复杂多项式构成的分母。我们在寻找最简公分母之前，只要有可能，就应先化简分子并对分母进行因式分解。
3. Finding the Lowest Common Denominator (LCD) | 寻找最简公分母
The LCD is the smallest expression that contains each denominator as a factor. For denominators like x, x+2, and x(x-1), you identify all distinct factors and take the highest power of each. For example, for 1/x and 1/(x+2), the LCD is x(x+2). For 2/(x²-1) and 3/(x+1), factorise the first denominator as (x-1)(x+1), so the LCD is (x-1)(x+1).
最简公分母（LCD）是能包含每一个分母作为其因式的表达式。对于像 x, x+2 和 x(x-1) 这样的分母，你需要找出所有不同的因式，并取每个因式的最高次幂。例如，对于 1/x 和 1/(x+2)，LCD 是 x(x+2)。对于 2/(x²-1) 和 3/(x+1)，将第一个分母因式分解为 (x-1)(x+1)，因此 LCD 为 (x-1)(x+1)。
Always write denominators in fully factorised form. This prevents missing common factors and keeps the multiplication as simple as possible. Do not cancel terms that are inside a sum; factorisation is safe only on the denominator level before multiplying.
始终把分母写成完全因式分解的形式。这样可以避免遗漏公因式，并使乘法尽可能简单。不要约去和式内部的项；只有在做乘法之前对方程的分母进行因式分解才是安全的。
4. Solving Linear Equations with Algebraic Fractions | 解含代数分数的一次方程
3x - 2x, giving x = -7. Check denominators: x+1= -6 ≠ 0, x-2 = -9 ≠ 0, so x = -7 is valid.
当去分母后得到的每项都是一次方程时，过程非常直接。解方程 2/(x+1) = 3/(x-2)。两边同乘 LCD (x+1)(x-2)：得 2(x-2) = 3(x+1)。展开：2x - 4 = 3x + 3。移项：-4 - 3 = 3x - 2x，得出 x = -7。检验分母：x+1 = -6 ≠ 0，x-2 = -9 ≠ 0，因此 x = -7 是有效解。
d simplify to a linear equation, then solve. Always check if any solution makes an original denominator zero.
在某些情况下，方程的一边可能有常数项。例如 4/x - 1 = 2/(x+3)。两边同乘 LCD x(x+3)：得 4(x+3) - x(x+3) = 2x。展开化简得到一次方程，然后求解。始终要检查是否有解使原分母为零。
5. Handling Binomial Denominators with Care | 小心处理二项式分母
3/(x+2) = 4/(x-3) leads to 3(x-3) = 4(x+2), which is linear: 3x - 9 = 4x + 8, so x = -17.
当分母是二项式，例如 (x+2) 和 (x-3)，LCD 就是它们的乘积 (x+2)(x-3)。两边相乘后通常会得到一个二次式，但经过抵消后可化简为一次方程。例如 3/(x+2) = 4/(x-3) 可导出 3(x-3) = 4(x+2)，这是一次方程：3x - 9 = 4x + 8，解得 x = -17。
This yields a quadratic equation, not linear, which we solve by expanding and rearranging.
有时一个分母是另一个分母的倍数。例如 (x-1) 和 (x-1)²，LCD 为 (x-1)²。在方程 2/(x-1) + 3/(x-1)² = 1 中，两边同乘 (x-1)² 得 2(x-1) + 3 = (x-1)²，这会产生一个二次方程而不是一次方程。我们通过展开和移项来求解它。
6. Equations That Lead to Quadratic Forms | 转化为二次形式的方程
rmula or completing the square, then check both solutions do not make any denominator zero.
许多IGCSE代数分数方程在去分母后会被设计成二次方程。考虑 1/(x+2) + 1/(x-2) = 1。LCD 是 (x+2)(x-2) = x² - 4。每一项相乘得：(x-2) + (x+2) = (x+2)(x-2)。化简左边：2x = x² - 4。移项：x² - 2x - 4 = 0。使用二次公式或配方法求解，然后检查两个解都不会使原分母为零。
quadratic. Always discard any root that equals ±1, as this would make a denominator zero.
另一种常见型式是单个分数等于一个常数加另一个分数，经过交叉相乘后得到二次方程。例如 2/(x-1) = 3/(x+1) + 1。同乘 LCD (x-1)(x+1) 后，要细心展开，把所有项移到一边，然后解二次方程。始终舍去等于 ±1 的根，因为这会使分母为零。
7. Factorising and Simplifying Before Solving | 求解前因式分解并化简
ply through: 2 + (x-2) = 3(x+2). This simplifies to 2 + x - 2 = 3x + 6, giving x = -3x - 6? 3x + 6 → -2x = 6 → x = -3. Check denominators: x-2 = -5, x+2 = -1, all non-zero, so valid.
对分母进行因式分解有助于快速找出 LCD 并减少不必要的运算。例如在方程 2/(x²-4) + 1/(x+2) = 3/(x-2) 中，注意 x²-4 可分解为 (x-2)(x+2)。那么 LCD 是 (x-2)(x+2)。两边同乘：2 + (x-2) = 3(x+2)。化简为 2 + x - 2 = 3x + 6，得 x = 3x + 6 → -2x = 6 → x = -3。检验分母：x-2 = -5, x+2 = -1，均不为零，因此有效。
Attempt to factorise numerators as well if you suspect common factors with denominators. However, be cautious: you cannot cancel across the equals sign or within a sum unless you take the factor outside a bracket that multiplies a whole term. The safest approach is to factorise denominators, find the LCD, multiply, then simplify.
如果你怀疑分子与分母有公因式，也可以尝试分解分子。但要小心：不能跨越等号约分，也不能在和式中任意约分，除非能把公因式提到括号外并与某一项整体相乘。最安全的方法是分解分母，找到 LCD，相乘后再化简。
8. Using Cross-Multiplication for Two Equal Fractions | 对两个相等分数使用交叉相乘
ut that works because it is equivalent to multiplying both sides by the product BD. It is especially efficient for linear denominators. Just remember that it is valid only if neither denominator is zero.
当方程由一个代数分数等于另一个代数分数构成，即 A/B = C/D 时，我们可以直接交叉相乘：A×D = B×C。这个技巧是一种快捷方式，等效于两边同乘分母的乘积 BD。对于线性分母尤其高效。但需记住，只有当分母都不为零时交叉相乘才成立。
+ 2 = 2x² - 6x. Rearrange into standard form: 0 = 2x² - 6x - x² - 3x - 2 → x² - 9x - 2 = 0? Let' - 9x - 2. Actually x² - 9x - 2 = 0. Solve by quadratic formula, then ensure x ≠ 3, x ≠ -1.
示例：(x+2)/(x-3) = (2x)/(x+1)。交叉相乘：(x+2)(x+1) = (2x)(x-3)。展开：x² + 3x + 2 = 2x² - 6x。化为标准形式：0 = 2x² - 6x - x² - 3x - 2 → x² - 9x - 2 = 0? 仔细计算：2x² - 6x - x² - 3x - 2 = x² - 9x - 2。实际上是 x² - 9x - 2 = 0。用二次公式求解，同时确保 x ≠ 3 且 x ≠ -1。
Cross-multiplication can also be extended to three or more fractions if you group them appropriately, but it is usually simpler to multiply by the LCD directly.
交叉相乘法也可推广到三个或更多个分数，只需适当地分组，但通常直接乘 LCD 更简单。
9. Checking for Extraneous Solutions | 检查增根
An extraneous solution arises when the value obtained from the cleared equation makes one or more original denominators zero. This happens because multiplying by the LCD can introduce solutions that were not allowed in the original equation. Every solution must be substituted back into the original equation's denominators to confirm they are non-zero.
当由去分母后的方程求出的值使一个或多个原分母为零时，就会产生增根。这是因为乘 LCD 可能会引入原方程不允许的解。每一个解都必须代回原方程的分母中，以确认它们不为零。
e would get numerators: 3 = 1 + 2, an identity, meaning all x except x=3 are solutions; x=3 is excluded. Always state domain restrictions: x ≠ values that make any denominator zero.
例如，解 1/(x-2) = 2/(x-2) + 1。去分母得 1 = 2 + (x-2) → 1 = x → x = 1。分母 x-2 = -1 ≠ 0，所以有效。但考虑 (x+3)/(x-1) = 2/(x-1) - 1。去分母得 x+3 = 2 - (x-1) → x+3 = 2 - x + 1 → 2x = 0 → x = 0，没问题。但如果方程是 3/(x-3) = 1/(x-3) + 2/(x-3)，会得到分子恒等式 3=1+2，这意味着除 x=3 外的所有 x 都是解；x=3 必须排除。始终说明定义域限制：x ≠ 使任何分母为零的值。
10. Common Errors and How to Avoid Them | 常见错误及避免方法
orrect step is [2/(x-1) + 3] × LCD = [4/x] × LCD, so 3 must also be multiplied by the LCD.
一个常见错误是忘记用 LCD 乘以每一项，尤其是常数项。当方程是 2/(x-1) + 3 = 4/x 时，学生有时只乘了分数项，而把 3 漏掉了。正确的步骤是 [2/(x-1) + 3] × LCD = [4/x] × LCD，因此 3 也必须乘以 LCD。
Another error is incorrectly expanding products when denominators involve binomials. Always use brackets. If you have 2(x+1)(x-2), expand systematically. Also, avoid premature cancellation: you cannot cancel x from (2x+3)/(x) inside an equation unless you factor and simplify after clearing denominators or handle with care.
另一个错误是在分母含有二项式时，错误地展开乘积。一定要使用括号。如果有 2(x+1)(x-2)，要系统地展开。此外，避免过早约分：在方程内部不能直接将 (2x+3)/x 中的 x 约掉，除非你去分母后进行因式分解和化简，或谨慎处理。
A third mistake is ignoring the need to check solutions. Even if your algebra is correct, a solution may be invalid if it makes any original denominator zero. In IGCSE, you may be asked explicitly to reject extraneous solutions.
第三个错误是忽略检验解的必要性。即便代数运算正确，如果解使原分母为零，它仍然是无效的。在 IGCSE 中，可能会明确要求你舍去增根。
11. Worked Examples with Detailed Steps | 详细解题示例
Example 1 (Linear): Solve 5/(x-2) = 3/(x+1).
示例1（一次）：解 5/(x-2) = 3/(x+1)。
erms: 5x - 3x = -6 - 5 → 2x = -11 → x = -11/2. Check: x-2 ≠ 0, x+1 ≠ 0, solution is valid.
LCD 为 (x-2)(x+1)。相乘：5(x+1) = 3(x-2)。展开：5x + 5 = 3x - 6。移项：5x - 3x = -6 - 5 → 2x = -11 → x = -11/2。检验：x-2 ≠ 0, x+1 ≠ 0，解有效。
Example 2 (Quadratic): Solve 2/(x+1) + 3/(x-2) = 1.
示例2（二次）：解 2/(x+1) + 3/(x-2) = 1。
- 4 + 3x + 3 = 5x - 1. Right: x² - x - 2. Equation: 5x - 1 = x² - x - 2 → 0 = x² - 6x - 1? = x² - x - 2 - 5x + 1 = x² - 6x - 1. Correct. So x² - 6x - 1 = 0 → using formula: x = [6 ± √36 + 4] / 2 = [6 ± √40]/2 = 3 ± √10. Check denominators: neither √10 nor -√10 makes x+1=0 or x-2=0. Both valid.
LCD: (x+1)(x-2)。两边相乘：2(x-2) + 3(x+1) = (x+1)(x-2)。展开左边：2x - 4 + 3x + 3 = 5x - 1。右边：x² - x - 2。得方程：5x - 1 = x² - x - 2 → 移项：0 = x² - x - 2 - 5x + 1 = x² - 6x - 1。确实 x² - 6x - 1 = 0。用公式解：x = [6 ± √36+4]/2 = [6 ± √40]/2 = 3 ± √10。检验分母：±√10 都不会使 x+1=0 或 x-2=0。两个解均有效。
Example 3 (Extraneous check): Solve x/(x-3) + 2/(x-3) = 3.
示例3（增根检查）：解 x/(x-3) + 2/(x-3) = 3。
= 3x - 9 → 2+9 = 2x → x = 11/2. Check denominator: 11/2 - 3 = 11/2 - 6/2 = 5/2 ≠ 0, fine.
由于分母相同，左边合并为 (x+2)/(x-3) = 3。两边同乘：x+2 = 3(x-3) → x+2 = 3x - 9 → 2+9 = 2x → x = 11/2。检验分母：11/2 - 3 = 5/2 ≠ 0，可以。
12. Exam Tips and Summary | 考试技巧与总结
In IGCSE exams, algebraic fraction questions often carry high marks and require clear logical steps. Remember to: factorise all denominators first; state the domain restrictions (values that make denominators zero) if required; find the LCD; multiply every term, using brackets; expand and simplify carefully; solve the resulting equation; and finally discard any extraneous roots by substituting into original denominators.
在IGCSE考试中，代数分数类题目常常分值较高，要求写出清晰的逻辑步骤。请记住：首先对所有分母进行因式分解；如有需要，注明定义域限制（使分母为零的值）；找到最简公分母；用括号将每一项乘以 LCD；仔细展开并化简；解出所得的方程；最后通过代回原分母舍去增根。
Practice varied problems including those with two fractions equal, those with sum or difference equal to a constant, and those requiring cross-multiplication. Always present your solution neatly, showing the multiplication step and the check. This systematic approach will minimise errors and maximise your marks.
练习各种题型，包括两个分数相等、和或差等于常数的情形，以及需要交叉相乘的题目。始终整洁地展示解答，写出乘法步骤和检验过程。这种系统的方法将最大限度地减少错误并帮你获得高分。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图