Equations and Identities | 方程与恒等式

📚 Equations and Identities | 方程与恒等式

In A-level mathematics, the distinction between an equation and an identity is fundamental. While both involve algebraic expressions joined by an equals sign, they behave in very different ways and demand different approaches. This article will clarify the difference, explore solving techniques, and show you how to prove and use identities confidently in exams.

在 A-level 数学中，区分方程与恒等式是基础中的基础。虽然两者都涉及由等号连接的代数表达式，但它们的行为方式截然不同，需要不同的处理方法。本文将厘清二者的区别，深入探讨求解技巧，并帮助你自信地证明和运用恒等式应对考试。

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1. What is an Equation? | 什么是方程？

An equation is a mathematical statement that two expressions are equal, but only for certain specific values of the variable(s). For example, the equation 2x + 3 = 7 is only true when x = 2. When we say we ‘solve’ an equation, we are finding the value(s) of the variable that make the statement true.

方程是表示两个表达式相等的数学陈述，但它仅对变量的某些特定值成立。例如，方程 2x + 3 = 7 仅在 x = 2 时成立。当我们说”解”一个方程时，我们实际上是在寻找使该陈述成立的变量的值。

Equations can be classified by their degree: linear equations (degree 1), quadratic equations (degree 2), cubic equations (degree 3), and so on. An equation involving multiple variables, such as x + y = 5, has infinitely many solution pairs — its solutions form a line or curve when plotted graphically.

方程可按次数分类：线性方程（一次）、二次方程、三次方程，以此类推。涉及多个变量的方程，如 x + y = 5，有无穷多组解——在图像上，它的解构成一条直线或曲线。

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2. What is an Identity? | 什么是恒等式？

An identity is a statement that two expressions are equal for all values of the variable(s). It is denoted using the symbol ≡ (read as ‘is identically equal to’). For instance, (x + 1)² ≡ x² + 2x + 1 is true for every real number x. Unlike an equation, an identity requires no ‘solving’ — it is always true.

恒等式是表示两个表达式对变量的所有取值都相等的陈述，用符号 ≡（读作”恒等于”）表示。例如，(x + 1)² ≡ x² + 2x + 1 对每个实数 x 都成立。与方程不同，恒等式无需”求解”——它永远成立。

Some common identities you should recognise instantly include: a² − b² ≡ (a + b)(a − b), (a ± b)² ≡ a² ± 2ab + b², and the difference of squares. These are algebraic tools that allow you to rewrite expressions without changing their value.

一些你应当一眼就能认出的常见恒等式包括：a² − b² ≡ (a + b)(a − b)、完全平方公式 (a ± b)² ≡ a² ± 2ab + b² 以及平方差公式。这些是允许你在不改变表达式值的情况下重写表达式的代数工具。

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3. The Key Difference | 关键区别

The crucial distinction lies in the scope of truth: an equation is true for specific values of the variable, whereas an identity is true for all values of the variable. Consider x² − 3x + 2 = 0: this is an equation, true only when x = 1 or x = 2. But x² − 3x + 2 ≡ (x − 1)(x − 2) is an identity, true for every x.

关键区别在于成立的范围：方程对变量的特定值成立，而恒等式对变量的所有值都成立。考虑 x² − 3x + 2 = 0：这是一个方程，仅在 x = 1 或 x = 2 时成立。但 x² − 3x + 2 ≡ (x − 1)(x − 2) 是一个恒等式，对所有 x 都成立。

In practice, the symbol ‘=’ is often used for both in textbooks, but in exam questions you must look for context. If a question says ‘solve’, it is an equation. If it says ‘prove’ or ‘show that’, it is an identity. Mixed questions may present an identity with unknown coefficients and ask you to find them.

在实际中，教科书里常常统一用”=”，但在考试题目中你必须根据语境判断。如果题目说”求解”，那就是方程；如果说”证明”或”说明”，那就是恒等式。混合题型可能给出带未知系数的恒等式，要求你求出这些系数。

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4. Solving Linear Equations | 解线性方程

Linear equations take the form ax + b = 0 (where a ≠ 0) and have exactly one solution. The method is straightforward: expand brackets, collect like terms, and isolate the variable. For example, to solve 3(x − 2) + 5 = 2x + 1, first expand: 3x − 6 + 5 = 2x + 1, then simplify to 3x − 1 = 2x + 1, giving x = 2.

线性方程的形式为 ax + b = 0（其中 a ≠ 0），恰有一个解。方法很直接：去括号、合并同类项、分离变量。例如，解 3(x − 2) + 5 = 2x + 1，先展开：3x − 6 + 5 = 2x + 1，化简得 3x − 1 = 2x + 1，即 x = 2。

When the unknown appears on both sides, gather all variable terms on one side and constants on the other. If the equation involves fractions, multiply through by the lowest common multiple of the denominators first. Always check your answer by substituting it back into the original equation.

当未知数出现在等号两边时，将所有含变量的项移到一边，常数移到另一边。如果方程含分数，先乘以分母的最小公倍数。始终将答案代回原方程验算。

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5. Solving Quadratic Equations | 解二次方程

A quadratic equation has the standard form ax² + bx + c = 0 (a ≠ 0). It can have two distinct real roots, one repeated root, or no real roots. There are three main methods of solution: factorisation, completing the square, and the quadratic formula.

二次方程的标准形式为 ax² + bx + c = 0（a ≠ 0）。它可能有两个不同的实根、一个重根，或无实根。求解方法主要有三种：因式分解、配方法，以及求根公式。

Factorisation: If ax² + bx + c can be written as (px + q)(rx + s), then set each factor to zero. For example, x² − 5x + 6 = 0 factors as (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

因式分解：如果 ax² + bx + c 可以写成 (px + q)(rx + s)，则令每个因子为零。例如，x² − 5x + 6 = 0 可以分解为 (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。

Completing the square: Rewrite x² + bx + c as (x + b/2)² − (b/2)² + c, then solve by taking square roots. This method also reveals the vertex of the parabola and is essential for finding maximum or minimum values.

配方法：将 x² + bx + c 改写为 (x + b/2)² − (b/2)² + c，然后通过开平方求解。这种方法还能揭示抛物线的顶点，是求最大值或最小值的必备工具。

Quadratic formula: For any quadratic equation, the roots are given by:

x = (−b ± √(b² − 4ac)) / 2a

The discriminant Δ = b² − 4ac determines the nature of the roots: if Δ > 0, two distinct real roots; if Δ = 0, one repeated root; if Δ < 0, no real roots (two complex roots).

判别式 Δ = b² − 4ac 决定根的性质：若 Δ > 0，有两个不同的实根；若 Δ = 0，有一个重根；若 Δ < 0，无实根（有两个复数根）。

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6. Simultaneous Equations | 联立方程

Simultaneous equations are a set of equations in multiple unknowns that must be satisfied at the same time. The most common cases at A-level are two linear equations in two unknowns, and one linear equation combined with one quadratic equation.

联立方程是一组必须同时满足的多个未知数的方程。A-level 中最常见的情况是二元一次方程组，以及一个一次方程与一个二次方程的组合。

Elimination method: Multiply one or both equations so that the coefficients of one variable match, then add or subtract to eliminate that variable. For example, solving 2x + y = 7 and x − y = 2: adding gives 3x = 9, so x = 3, then y = 1.

消元法：将一个或两个方程乘以适当的数，使某个变量的系数相同，然后相加或相减以消去该变量。例如，解 2x + y = 7 和 x − y = 2：两式相加得 3x = 9，所以 x = 3，继而 y = 1。

Substitution method: Rearrange one equation to express one variable in terms of the other, then substitute into the second equation. This is essential when one equation is quadratic — substitute the linear expression into the quadratic to obtain a single quadratic in one variable, then solve. Be sure to find both x and y values and check that each pair satisfies both original equations.

代入法：将一个方程变形，用另一个变量表示其中一个变量，然后代入第二个方程。当一个方程是二次方程时，这种方法必不可少——将一次表达式代入二次方程，得到关于一个变量的一元二次方程，然后求解。务必求出所有 x 和 y 的解对，并检验每组解是否同时满足两个原方程。

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7. Proving Identities | 证明恒等式

To prove an identity, you must show that the left-hand side (LHS) and right-hand side (RHS) are equal for all values of the variable. The most reliable strategy is to start with the more complicated side and manipulate it algebraically until it matches the other side exactly.

证明恒等式，你必须证明左边（LHS）和右边（RHS）对变量的所有取值都相等。最可靠的策略是从较复杂的一边入手，通过代数变形逐步将其化为另一边。

Example: Prove that (a + b)² − (a − b)² ≡ 4ab.

例题：证明 (a + b)² − (a − b)² ≡ 4ab。

Expanding the LHS: (a² + 2ab + b²) − (a² − 2ab + b²) = a² + 2ab + b² − a² + 2ab − b² = 4ab. Since the LHS simplifies exactly to the RHS, the identity is proved. Never assume what you are trying to prove — work from one side only.

展开左边：(a² + 2ab + b²) − (a² − 2ab + b²) = a² + 2ab + b² − a² + 2ab − b² = 4ab。由于左边恰好化简为右边，恒等式得证。永远不要直接假设要证明的结论——必须只从一边出发逐步变形。

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8. Finding Unknown Coefficients | 求未知系数

A classic exam question presents an identity containing unknown constants and asks you to determine their values. The key principle is the comparison of coefficients: if two polynomials are identically equal, then the coefficients of corresponding powers of x must be equal.

一类经典考题给出含有未知常数的恒等式，要求你确定这些常数的值。核心原理是比较系数法：如果两个多项式恒等，则 x 的各次幂对应的系数必须相等。

Example: Given that x² + 5x + 6 ≡ (x + a)(x + b), find the values of a and b.

例题：已知 x² + 5x + 6 ≡ (x + a)(x + b)，求 a 和 b 的值。

Expanding the RHS: x² + (a + b)x + ab. Comparing coefficients: a + b = 5 and ab = 6. Solving these simultaneously, the pairs (a, b) = (2, 3) or (3, 2) both satisfy the equations. Hence the factorisation is x² + 5x + 6 ≡ (x + 2)(x + 3).

展开右边：x² + (a + b)x + ab。比较系数：a + b = 5 且 ab = 6。联立求解，得 (a, b) = (2, 3) 或 (3, 2) 均满足方程。因此因式分解为 x² + 5x + 6 ≡ (x + 2)(x + 3)。

Alternatively, since the identity holds for all x, you can substitute cleverly chosen values of x to obtain equations for the unknowns. Substituting x = 0 often kills the variable terms and reveals the constant term directly.

另一种方法是：因为恒等式对所有x 都成立，你可以巧妙地选取 x 的特定值代入，得到关于未知数的方程。代入 x = 0 通常可以消去含变量的项，直接揭示常数项。

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9. Identities in Algebraic Fractions | 代数分式中的恒等式

Identities play a central role in simplifying algebraic fractions and in partial fractions. For example, the identity x² − 1 ≡ (x − 1)(x + 1) allows us to simplify the fraction (x² − 1)/(x − 1) to x + 1, provided x ≠ 1.

恒等式在简化代数分式和部分分式中扮演核心角色。例如，恒等式 x² − 1 ≡ (x − 1)(x + 1) 使我们能将分式 (x² − 1)/(x − 1) 简化为 x + 1（前提是 x ≠ 1）。

When decomposing a rational expression into partial fractions, we write the fraction as a sum of simpler fractions and then use the comparison-of-coefficients method or substitute strategic values to find the unknown numerators. This is a direct application of the identity principle.

当将有理式分解为部分分式时，我们把它写成若干个更简单分式之和，然后用比较系数法或代入特殊值来求未知分子。这正是恒等式原理的直接应用。

Example: Express 5/((x + 1)(x − 2)) in partial fractions.

例题：将 5/((x + 1)(x − 2)) 化为部分分式。

5/((x + 1)(x − 2)) ≡ A/(x + 1) + B/(x − 2)

Multiplying both sides by (x + 1)(x − 2): 5 ≡ A(x − 2) + B(x + 1). Substituting x = 2 gives 5 = 3B, so B = 5/3. Substituting x = −1 gives 5 = −3A, so A = −5/3. Thus the identity is 5/((x + 1)(x − 2)) ≡ −5/(3(x + 1)) + 5/(3(x − 2)).

两边同乘 (x + 1)(x − 2)：5 ≡ A(x − 2) + B(x + 1)。代入 x = 2 得 5 = 3B，所以 B = 5/3。代入 x = −1 得 5 = −3A，所以 A = −5/3。因此恒等式为 5/((x + 1)(x − 2)) ≡ −5/(3(x + 1)) + 5/(3(x − 2))。

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10. Common Mistakes and Pitfalls | 常见错误与陷阱

Mistake 1: Treating an identity as an equation. If you try to ‘solve’ an identity, you will either get no solution or every value as a solution — which is meaningless. Recognise the symbol ≡ and the phrase ‘prove’ or ‘show that’ as signals that you are dealing with an identity.

错误一：把恒等式当方程解。如果试图”解”恒等式，你会得到无解或所有值都是解——这毫无意义。看到符号 ≡ 以及”证明”或”说明”等表述，应意识到你面对的是恒等式。

Mistake 2: Dividing by a variable. Dividing both sides of an equation by a variable expression such as x risks losing a solution. For instance, in x² = 3x, dividing by x gives x = 3, but you also lose the solution x = 0. Always factorise instead of dividing by an unknown quantity.

错误二：除以变量。在方程两边同时除以含变量的表达式（如 x）会导致丢根。例如，在 x² = 3x 中，除以 x 得 x = 3，但丢失了解 x = 0。应始终用因式分解，而非除以未知量。

Mistake 3: Forgetting to check extraneous roots. When squaring both sides of an equation (common in solving equations with square roots), extraneous roots may be introduced. Always substitute your answers back into the original equation to verify them.

错误三：忘记检验增根。当方程两边同时平方时（常见于解含根号的方程），可能会引入增根。务必把答案代回原方程逐一检验。

Mistake 4: Sign errors in expansion. When expanding (a − b)², a frequent error is writing a² − b² instead of a² − 2ab + b². Carefully apply the identity (a ± b)² ≡ a² ± 2ab + b² and remember that (a − b)² is not equal to a² − b² in general.

错误四：展开时符号出错。展开 (a − b)² 时，常见错误是写成 a² − b²，而正确应为 a² − 2ab + b²。应仔细运用恒等式 (a ± b)² ≡ a² ± 2ab + b²，并记住 (a − b)² 一般不等于 a² − b²。

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11. Exam-Style Practice | 考试风格练习

Question 1: Determine whether each of the following is an equation or an identity: (a) 3x + 7 = 16; (b) (x − 3)(x + 3) ≡ x² − 9; (c) 2x + 5 = 2x − 1.

练习一：判断下列各题为方程还是恒等式：(a) 3x + 7 = 16；(b) (x − 3)(x + 3) ≡ x² − 9；(c) 2x + 5 = 2x − 1。

Answer: (a) Equation — true only when x = 3. (b) Identity — true for all x. (c) Equation with no solution — no value of x satisfies it.

解答：(a) 方程——仅在 x = 3 时成立。(b) 恒等式——对所有 x 成立。(c) 无解的方程——没有任何 x 值能满足它。

Question 2: Given that x³ + ax² + bx − 6 ≡ (x − 1)(x + 2)(x + c), find the values of a, b and c.

练习二：已知 x³ + ax² + bx − 6 ≡ (x − 1)(x + 2)(x + c)，求 a、b、c 的值。

Answer: First, substitute x = 1 into both sides: the LHS gives 1 + a + b − 6, and the RHS gives 0, so a + b = 5. Substituting x = −2: LHS gives −8 + 4a − 2b − 6 = 0, so 4a − 2b = 14, that is 2a − b = 7. Solving a + b = 5 and 2a − b = 7 simultaneously: adding gives 3a = 12, so a = 4 and b = 1. Now the constant term comparison: expanding the RHS, the constant is (−1)(2)(c) = −2c, which must equal −6, so c = 3. Hence a = 4, b = 1, c = 3.

解答：首先将 x = 1 代入两边：左边为 1 + a + b − 6，右边为 0，所以 a + b = 5。代入 x = −2：左边为 −8 + 4a − 2b − 6 = 0，即 4a − 2b = 14，也就是 2a − b = 7。联立 a + b = 5 和 2a − b = 7：两式相加得 3a = 12，所以 a = 4，b = 1。再比较常数项：展开右边，常数项为 (−1)(2)(c) = −2c，必须等于 −6，所以 c = 3。因此 a = 4，b = 1，c = 3。

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12. Summary of Key Points | 要点总结

An equation is true for specific values of the variable and is solved by algebraic manipulation; an identity is true for all values and is proved by rewriting one side into the other. The symbol ≡ signals an identity. Master the three methods for solving quadratics — factorisation, completing the square, and the quadratic formula — and use the discriminant to predict the nature of roots.

方程对变量的特定值成立，通过代数变形来求解；恒等式对所有值成立，通过将一边变形为另一边来证明。符号 ≡ 表示恒等式。掌握解二次方程的三种方法——因式分解、配方法、求根公式——并使用判别式预判根的性质。

For simultaneous equations, use elimination or substitution appropriately. To find unknown coefficients in an identity, compare coefficients or substitute strategic values. Avoid common pitfalls: never divide by a variable, always check for extraneous roots, and carefully expand squared brackets.

对于联立方程，灵活运用消元法或代入法。要在恒等式中求未知系数，可比较系数或代入特殊值。避免常见陷阱：不要除以变量，始终检验增根，并仔细展开平方项。

Above all, practise regularly. The more you work with equations and identities, the more natural the distinction becomes — and the faster you will recognise which technique a question demands.

最重要的是，定期练习。你接触的方程与恒等式越多，它们的区别就越自然——你也越能迅速识别题目需要哪种技巧。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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