Exercise 1E: Simplifying Algebraic Expressions and Substitution | 练习1E：代数表达式的化简与代入

📚 Exercise 1E: Simplifying Algebraic Expressions and Substitution | 练习1E：代数表达式的化简与代入

Mastering the manipulation of algebraic expressions is a cornerstone of the IB Mathematics curriculum. In this article, we break down the essential skills covered in a typical Exercise 1E — from collecting like terms and expanding brackets to substituting values and simplifying rational expressions. Each concept is presented with clear English explanations followed by their Chinese counterparts to reinforce bilingual understanding.

掌握代数表达式的运算是 IB 数学课程的核心基础。在这篇文章中，我们逐一拆解典型练习 1E 所涵盖的基本技能——从合并同类项、展开括号，到代入求值和分式化简。每个概念都先给出清晰的英文解释，再附上对应的中文译文，以强化双语理解。

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1. Understanding Algebraic Expressions | 理解代数表达式

An algebraic expression is a mathematical phrase that combines numbers, variables (letters representing unknown quantities), and operation symbols. It does not contain an equality sign, which distinguishes it from an equation.

代数表达式是由数字、变量（代表未知量的字母）和运算符号组成的数学短语。它不包含等号，这使它区别于方程。

In the expression 4x² – 3xy + 7, the parts separated by addition or subtraction are called terms. Here, 4x², –3xy, and 7 are three distinct terms. The numerical factor of a term is its coefficient, so the coefficient of x² is 4 and the coefficient of xy is –3.

在表达式 4x² – 3xy + 7 中，由加法或减法分隔开的各个部分称为项。此处 4x²、–3xy 和 7 是三个不同的项。项中的数字因数被称为系数，因此 x² 的系数是 4，xy 的系数是 –3。

Constant terms, like 7, have no variable part. Understanding the anatomy of an expression is the first step towards simplifying it efficiently.

像 7 这样的常数项没有变量部分。充分了解表达式的构成是高效化简它的第一步。

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2. Collecting Like Terms | 合并同类项

Like terms are terms that contain exactly the same variables raised to the same powers — only their coefficients may differ. For example, 5a and –2a are like terms, but 3a² and 3a are not.

同类项是指含有完全相同的变量且各变量的指数也相同的项——仅仅是系数可能不同。例如，5a 和 –2a 是同类项，但 3a² 和 3a 不是。

To simplify an expression, we group and combine the coefficients of like terms. Consider 2x + 5y – x + 3y. The x-terms can be combined: 2x – x = x; the y-terms: 5y + 3y = 8y. The simplified expression is x + 8y.

要化简一个表达式，我们需要将同类项的系数进行合并。例如 2x + 5y – x + 3y，x 项可合并为 2x – x = x；y 项为 5y + 3y = 8y。化简后的表达式是 x + 8y。

Always be careful with signs: a term like –4z is added to other z-terms by taking its negative coefficient into account, so 3z – 4z = –z.

务必注意符号：像 –4z 这样的项，在与其他 z 项相加时，要连同其负号系数一并计算，因此 3z – 4z = –z。

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3. Expanding Single Brackets | 展开单括号

Removing a single bracket involves multiplying each term inside the bracket by the factor outside. This is a direct application of the distributive law: a(b + c) = ab + ac.

去掉单个括号需要将括号外的因数与括号内的每一项相乘。这是乘法分配律的直接应用：a(b + c) = ab + ac。

For instance, expand 3(2x – 5). Multiply 3 by 2x to get 6x, then multiply 3 by –5 to get –15. The result is 6x – 15.

例如，展开 3(2x – 5)。用 3 乘以 2x 得到 6x，再用 3 乘以 –5 得到 –15。结果为 6x – 15。

When the factor includes a negative sign, distribute the negative carefully: –2(4 – y) = –8 + 2y, because –2 × 4 = –8 and –2 × (–y) = +2y.

当因数含有负号时，要仔细分配负号：–2(4 – y) = –8 + 2y，因为 –2 × 4 = –8，而 –2 × (–y) = +2y。

–x(3x – 2) = –3x² + 2x

–x(3x – 2) = –3x² + 2x

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4. Expanding Double Brackets | 展开双括号

To expand the product of two binomials, use the FOIL method — First, Outer, Inner, Last — or simply ensure every term in the first bracket multiplies every term in the second bracket.

要展开两个二项式的乘积，可以使用 FOIL 法则——先乘首项、外项、内项、尾项——或者简单地确保第一个括号中的每一项都与第二个括号中的每一项相乘。

Expand (x + 3)(x – 4): First terms x·x = x², Outer x·(–4) = –4x, Inner 3·x = 3x, Last 3·(–4) = –12. Combine the middle terms –4x + 3x = –x. The expansion is x² – x – 12.

展开 (x + 3)(x – 4)：首项 x·x = x²，外项 x·(–4) = –4x，内项 3·x = 3x，尾项 3·(–4) = –12。合并中间项 –4x + 3x = –x。展开结果为 x² – x – 12。

A special case is the perfect square: (a + b)² = a² + 2ab + b². Likewise, (a – b)² = a² – 2ab + b². Recognising these patterns speeds up simplification.

一个特殊情形是完全平方公式：(a + b)² = a² + 2ab + b²。类似地，(a – b)² = a² – 2ab + b²。识别这些模式能加快化简速度。

(2x + 1)² = 4x² + 4x + 1

(2x + 1)² = 4x² + 4x + 1

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5. Substitution – Replacing Variables | 代入求值

Substitution means replacing each variable in an expression with a given numerical value and then evaluating the result using the correct order of operations (BIDMAS/BODMAS).

代入是指将表达式中的每个变量替换为给定的数值，然后按照正确的运算顺序（括号、指数、乘除、加减）计算出结果。

If we are asked to evaluate 2x² – 3y for x = –1 and y = 4, we first replace: 2(–1)² – 3(4). Inside the exponent, (–1)² = 1, so we get 2(1) – 12 = 2 – 12 = –10.

若要求我们计算当 x = –1 和 y = 4 时 2x² – 3y 的值，先进行替换：2(–1)² – 3(4)。指数部分 (–1)² = 1，于是得到 2(1) – 12 = 2 – 12 = –10。

Be especially careful when the variable has a negative coefficient or is squared. Brackets around the substituted value help avoid sign errors: putting –1 inside ( ) ensures we square the negative value correctly.

当变量带有负号或有平方时尤其要小心。在代入值外加括号有助于避免符号错误：将 –1 放在括号内可确保正确地平方负数。

Expression	Substitution	Step-by-step	Value
3x – y	x=2, y=5	3(2) – 5	1
x² + 2x	x=–3	(–3)² + 2(–3) = 9 – 6	3

Practising substitution builds the foundation for evaluating functions, plotting graphs, and solving equations later in the IB course.

练习代入求值可为 IB 课程后续的函数求值、作图以及解方程打下基础。

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6. Simplifying with Fractions | 分式化简

Algebraic fractions follow the same rules as numerical fractions. When adding or subtracting, find a common denominator first. For instance, to simplify (x/2) + (x/3), the common denominator is 6, giving (3x/6) + (2x/6) = 5x/6.

代数分式遵循与数字分式相同的规则。加减运算时，首先要找到公分母。例如，要化简 (x/2) + (x/3)，公分母是 6，得到 (3x/6) + (2x/6) = 5x/6。

When multiplying fractions, multiply numerators together and denominators together, then simplify by cancelling common factors: (2x/3) × (6/x) = (12x)/(3x) = 4, provided x ≠ 0.

分式相乘时，分子乘分子，分母乘分母，然后约去公因数进行化简：(2x/3) × (6/x) = (12x)/(3x) = 4，前提是 x ≠ 0。

Always state restrictions on the variable that prevent division by zero. Simplifying (x² – 4)/(x – 2) factorises to ((x–2)(x+2))/(x–2) = x + 2, for x ≠ 2.

务必注明使分母为零的变量限制条件。化简 (x² – 4)/(x – 2) 时，因式分解得 ((x–2)(x+2))/(x–2) = x + 2，其中 x ≠ 2。

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7. Applying Exponent Rules | 运用指数法则

Exponentiation is a shorthand for repeated multiplication. The base tells us which number is being multiplied, and the exponent (or power) tells us how many times.

指数是重复乘法的简写形式。底数指示哪个数在相乘，指数（或幂）指示相乘的次数。

Key rules include: aᵐ × aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, and aᵐ ÷ aⁿ = aᵐ⁻ⁿ. Also, a⁰ = 1 for any non-zero a, and a⁻ⁿ = 1/aⁿ.

关键法则包括：aᵐ × aⁿ = aᵐ⁺ⁿ，(aᵐ)ⁿ = aᵐⁿ，以及 aᵐ ÷ aⁿ = aᵐ⁻ⁿ。此外，对于任何非零的 a，a⁰ = 1，且 a⁻ⁿ = 1/aⁿ。

Applying these rules correctly prevents errors when simplifying expressions like (2x³)² ÷ x⁴. First, (2x³)² = 4x⁶, then dividing by x⁴ gives 4x².

正确应用这些法则能在化简类似 (2x³)² ÷ x⁴ 的表达式时避免出错。首先，(2x³)² = 4x⁶，然后除以 x⁴ 得到 4x²。

(3a⁻²b³)² = 9a⁻⁴b⁶ = 9b⁶/a⁴

(3a⁻²b³)² = 9a⁻⁴b⁶ = 9b⁶/a⁴

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8. Common Mistakes to Avoid | 常见错误及避免方法

One frequent error is to incorrectly combine unlike terms, such as writing 2a + 3b as 5ab. These are not like terms and cannot be added in that way.

一个常见错误是错误地合并不同类项，比如将 2a + 3b 写成 5ab。它们不是同类项，不能这样相加。

Another pitfall is mishandling the minus sign during expansion: forgetting that –(x – 3) becomes –x + 3, not –x – 3.

另一个陷阱是在展开时错误处理负号：忘记了 –(x – 3) 会变成 –x + 3，而非 –x – 3。

In substitution, students often forget to apply exponents before multiplication, or they write (–2)² as –4. Remember the square of a negative is positive.

在代入时，学生常常忘记先算指数再算乘法，或将 (–2)² 写成 –4。请记住，负数的平方是正数。

Avoid dividing by zero without stating restrictions. In an expression like 5/(x – 2), you must note that x ≠ 2.

避免在未注明限制条件的情况下除以零。在诸如 5/(x – 2) 的表达式中，你必须注明 x ≠ 2。

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9. Worked Example (Step-by-Step) | 分步示例

Let’s simplify 2(3x – 1) – (x + 4)(x – 2) and then evaluate for x = –1.

我们来化简 2(3x – 1) – (x + 4)(x – 2) 并代入 x = –1 求值。

Step 1: Expand the single bracket: 2(3x – 1) = 6x – 2.

第1步：展开单括号：2(3x – 1) = 6x – 2。

Step 2: Expand the double brackets: (x + 4)(x – 2) = x² – 2x + 4x – 8 = x² + 2x – 8.

第2步：展开双括号：(x + 4)(x – 2) = x² – 2x + 4x – 8 = x² + 2x – 8。

Step 3: Combine the results, remembering the subtraction: 6x – 2 – (x² + 2x – 8) = 6x – 2 – x² – 2x + 8.

第3步：合并结果，注意括号前的减号：6x – 2 – (x² + 2x – 8) = 6x – 2 – x² – 2x + 8。

Step 4: Collect like terms: –x² + (6x – 2x) + (–2 + 8) = –x² + 4x + 6.

第4步：合并同类项：–x² + (6x – 2x) + (–2 + 8) = –x² + 4x + 6。

Step 5: Substitute x = –1: –(–1)² + 4(–1) + 6 = –1 – 4 + 6 = 1.

第5步：代入 x = –1：–(–1)² + 4(–1) + 6 = –1 – 4 + 6 = 1。

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10. Key Takeaways | 要点总结

Always check for like terms before attempting any other simplification — combining them reduces the expression to its simplest form.

在尝试其他任何化简之前，始终先检查同类项——合并它们可将表达式化为最简形式。

Expanding brackets correctly relies on distributing each term systematically and paying attention to negative signs. Use FOIL for double brackets to stay organised.

正确展开括号依靠系统地分配每一项并注意负号。对双括号使用 FOIL 法则可保持条理清晰。

When substituting values, enclose negative numbers in parentheses to avoid sign mistakes, and follow the order of operations strictly.

代入数值时，将负数放入括号内以避免符号错误，并严格遵循运算顺序。

Exponent rules and fraction simplification both demand careful factorisation and cancellation. Never cancel terms unless they are factors of the entire numerator and denominator.

指数法则和分式化简都需要仔细的因式分解与约分。除非是分子和分母的整体公因数，否则切勿随意约去项。

Regular practice with Exercise 1E-style problems builds the fluency needed for the algebra-heavy IB Mathematics exams, whether you are following the Analysis & Approaches or Applications & Interpretation route.

经常练习 1E 类型的题目有助于培养流利的代数处理能力，无论你选择的是 IB 数学分析与方法还是应用与解释路径，这都能为代数密集的考试做好准备。

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