Prelims pp01-02 Core Homework 9 Answers: Algebra Essentials | 预考第1-2页核心作业9答案：代数基础

📚 Prelims pp01-02 Core Homework 9 Answers: Algebra Essentials | 预考第1-2页核心作业9答案：代数基础

This revision article walks through the answers and methods for Core Homework 9 from Prelims Paper 1 and Paper 2. The focus is on algebra essentials that frequently appear in early high-stakes assessments, including linear equations, factorising quadratics, inequalities, simultaneous equations, and algebraic fractions.

本复习文章讲解预考试卷1和试卷2中核心作业9的答案与方法。重点放在高利害考试早期常见的代数基础知识，包括线性方程、二次式因式分解、不等式、联立方程和代数分式。

1. Understanding the Core Homework 9 Problems | 理解核心作业9的问题

Core Homework 9 typically asks students to solve a set of algebraic problems. The questions are designed to test fluency with symbolic manipulation rather than just memorising rules. In Papers 1 and 2, the same skills appear with small variations in wording.

核心作业9通常要求学生解决一组代数问题。这些题目旨在考查符号运算的熟练度，而不仅仅是记忆规则。在试卷1和试卷2中，同样的技能以略有不同的措辞出现。

The answer key below explains each step so that you can locate errors in your own working. Always show full method because marks are awarded for correct substitution and simplification, not only for the final answer.

下面的答案解析逐步说明，以便你可以在自己的计算过程中找出错误。始终展示完整方法，因为得分不仅给最终答案，还给正确的代入和化简步骤。

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2. Solving Linear Equations | 解线性方程

es to isolate the term containing x: 3x = 15. Then divide both sides by 3 to obtain x = 5.

Q9(a) 通常给出一个线性方程，如 3x + 5 = 20。第一步是从两边减去 5，以分离含 x 的项：3x = 15。然后两边除以 3，得到 x = 5。

comes 2x - 6 = 10, then 2x = 16, so x = 8. Checking by substitution confirms the solution.

如果方程含有括号，首先展开。例如，2(x - 3) = 10 变为 2x - 6 = 10，然后 2x = 16，所以 x = 8。通过代入检验可确认解。

3x + 5 = 20 → x = 5

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3. Expanding Brackets Accurately | 准确展开括号

Use the distributive law: x(x - 2) + 3(x - 2), which gives x² - 2x + 3x - 6 = x² + x - 6.

Q9(b) 的核心任务是展开两个二项式的乘积，例如 (x + 3)(x - 2)。使用分配律：x(x - 2) + 3(x - 2)，得到 x² - 2x + 3x - 6 = x² + x - 6。

acket: -(x - 4) = -x + 4. Many candidates lose a mark by writing -x - 4 instead of -x + 4.

当括号外有负号时要小心符号：-(x - 4) = -x + 4。许多考生因为写成 -x - 4 而不是 -x + 4 而丢分。

(x + 3)(x - 2) = x² + x - 6

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4. Factorising Quadratics | 因式分解二次式

: 2 and 3. Therefore x² + 5x + 6 = (x + 2)(x + 3). Always expand the answer back to check.

要对 x² + 5x + 6 进行因式分解，找到两个数，它们相乘得 6 且相加得 5：2 和 3。因此 x² + 5x + 6 = (x + 2)(x + 3)。始终将答案展开回去检查。

For a quadratic with a negative constant, such as x² - x - 12, the pair must multiply to -12 and add to -1. The numbers are 3 and -4, so the factorised form is (x + 3)(x - 4).

对于常数项为负的二次式，例如 x² - x - 12，这对数必须相乘得 -12 且相加得 -1。这两个数是 3 和 -4，所以因式分解形式为 (x + 3)(x - 4)。

x² + 5x + 6 = (x + 2)(x + 3)

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5. Solving Quadratic Equations by Factorising | 通过因式分解解二次方程

. Then set each bracket equal to zero: x + 2 = 0 gives x = -2, and x + 3 = 0 gives x = -3.

当 Q9 要求解 x² + 5x + 6 = 0 时，首先因式分解为 (x + 2)(x + 3) = 0。然后令每个括号等于零：x + 2 = 0 得 x = -2，x + 3 = 0 得 x = -3。

Do not divide both sides by x, because that can remove a valid solution. Instead, bring all terms to one side and factorise. The zero product property then guarantees at least one factor is zero.

不要两边除以 x，因为那样可能会丢掉一个有效解。相反，将所有项移到一边并进行因式分解。零积性质保证至少一个因式为零。

x² + 5x + 6 = 0 → x = -2 or x = -3

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6. Linear Inequalities | 线性不等式

de by 5: x ≤ 3. Since you divide by a positive number, the inequality sign stays the same.

在 Q9(c) 中，你可能需要解一个不等式，如 5x - 3 ≤ 12。两边加 3：5x ≤ 15。除以 5：x ≤ 3。因为除以正数，不等号方向保持不变。

If the coefficient of x is negative, reverse the sign. For example, -2x < 8 becomes x > -4. Show this reversal clearly in your working.

如果 x 的系数为负，则反转不等号。例如，-2x < 8 变为 x > -4。在计算过程中清楚展示这一反转。

5x - 3 ≤ 12 → x ≤ 3

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7. Simultaneous Equations by Elimination | 用消元法解联立方程

= 12, so x = 4. Substitute back into x - y = 2 to get y = 2. The solution is x = 4, y = 2.

Q9(d)

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