AQA A-Level Maths Miscellaneous Exercise 3: Equations and Inequalities | AQA A-Level 数学综合练习3：方程与不等式

📚 AQA A-Level Maths Miscellaneous Exercise 3: Equations and Inequalities | AQA A-Level 数学综合练习3：方程与不等式

This article covers the key techniques and worked examples you need for the AQA A-Level Mathematics Year 1 course, particularly the topic of equations and inequalities. We will walk through solving linear and quadratic equations, simultaneous equations, inequalities, and common examination pitfalls, using the style of the ‘Miscellaneous Exercise 3’ from the official textbook.

本文涵盖 AQA A-Level 数学一年级课程中方程与不等式板块的核心技巧和典型例题。我们将按照官方教材’综合练习3’的风格，系统讲解一次方程、二次方程、联立方程组、不等式以及常见的考试陷阱。

1. Core Linear Equations | 核心线性方程

Solving a linear equation requires isolating the variable using inverse operations. Always perform the same operation on both sides to keep the equation balanced. For example, solve 3x + 5 = 14: subtract 5 from both sides to get 3x = 9, then divide by 3 to obtain x = 3.

解一次方程需要通过逆运算隔离未知数。始终在等式两边进行相同操作以保持平衡。例如解 3x + 5 = 14：两边同减 5 得 3x = 9，再同除以 3 得到 x = 3。

Clear fractions by multiplying through by the denominator / 去分母：两边乘以分母。
Expand brackets before simplifying / 先去括号再合并。
Collect like terms on each side / 将同类项移到同一边。

2. Solving Quadratic Equations | 解二次方程

A quadratic equation takes the form ax² + bx + c = 0. You can solve it by factorising, completing the square, or using the quadratic formula. The method you choose depends on the question and the type of roots expected.

二次方程形如 ax² + bx + c = 0。解法包括因式分解、配方法或求根公式。具体选择取决于题目的形式与预期的根的类型。

x = (−b ± √(b² − 4ac)) / 2a

For example, solve x² − 5x + 6 = 0. Factorise to (x − 2)(x − 3) = 0, giving roots x = 2 or x = 3. Alternatively, you can complete the square: (x − 5/2)² = 1/4, leading to the same solutions.

例如解 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0，所以根为 x = 2 或 x = 3。也可以用配方法：(x − 5/2)² = 1/4，同样得到解。

3. Simultaneous Equations | 联立方程

When you have two equations involving two unknowns, you can solve them using substitution or elimination. For a linear and a quadratic pair, substitution is the standard method. After substitution, you may obtain a quadratic equation, which can have two, one, or zero real solutions.

当有两个方程且含两个未知数时，可以用代入法或消元法求解。若一个为线性、另一个为二次，通常使用代入法。代入后得到二次方程，其解可能有两个、一个或没有实数解。

Example: y = 2x + 1 and x² + y² = 10. Substitute y: x² + (2x + 1)² = 10 → 5x² + 4x − 9 = 0. Solve to get x = 1 or x = −9/5. Then find the corresponding y values: (1, 3) and (−9/5, −13/5).

例：y = 2x + 1 与 x² + y² = 10。代入 y：x² + (2x + 1)² = 10 → 5x² + 4x − 9 = 0。解得 x = 1 或 x = −9/5，再求对应 y：交点分别为 (1, 3) 和 (−9/5, −13/5)。

4. Linear Inequalities | 一次不等式

Linear inequalities are solved like equations, but multiplying or dividing by a negative number reverses the inequality sign. Remember: if you swap sides, the inequality also reverses.

解一次不等式与解方程类似，但两边同乘或同除负数时要反向不等号。还要注意：将左右两边互换时，不等号方向也必须改变。

4x − 7 > 9 → 4x > 16 → x > 4 / 所以解为 x > 4。
−3x ≤ 12 → x ≥ −4 (notice the sign flip) / 注意符号反转。
On a number line, an open circle indicates ‘strictly <' or '>‘, while a closed circle indicates ‘≤’ or ‘≥’. / 数轴上，空心圆圈表示严格小于或大于，实心圆圈表示小于等于或大于等于。

5. Quadratic Inequalities | 二次不等式

To solve a quadratic inequality, first rearrange to have zero on one side, then find the roots of the quadratic. Sketch the graph or test intervals to decide which region(s) satisfy the inequality.

解二次不等式时，先将一边化为零，再求二次方程的根。通过画草图或测试区间判断哪些区域满足不等式。

Example: solve x² − x − 12 > 0. Factorise: (x − 4)(x + 3) > 0. The roots are 4 and −3. The graph opens upwards, so the inequality holds for x < −3 or x > 4.

例：解 x² − x − 12 > 0。因式分解：(x − 4)(x + 3) > 0。根为 4 和 −3。抛物线开口向上，因此解为 x < −3 或 x > 4。

If a > 0, the solution to ax² + bx + c > 0 is x < α or x > β, while ax² + bx + c < 0 is α < x < β.

6. Using the Discriminant | 使用判别式

The discriminant b² − 4ac determines the nature of the roots of a quadratic equation. This knowledge is often tested in contexts such as finding when a line intersects, touches, or misses a curve.

判别式 b² − 4ac 决定了二次方程根的性质。这一知识点经常用于判断一条直线与曲线相交、相切或不相交的问题。

Discriminant value / 判别式值	Roots / 根	Graph interpretation / 图形意义
b² − 4ac > 0	Two distinct real roots / 两个不同实数根	Intersects x-axis twice / 与 x 轴交于两点
b² − 4ac = 0	One repeated real root / 一个重根	Touches x-axis / 与 x 轴相切
b² − 4ac < 0	No real roots / 无实数根	Does not meet x-axis / 不与 x 轴相交

For example, for the equation kx² + 2x + 1 = 0, find the value of k such that the equation has equal roots. Set the discriminant to zero: 2² − 4(k)(1) = 0 → 4 − 4k = 0 → k = 1.

例如，对于方程 kx² + 2x + 1 = 0，求使方程有等根的 k 值。令判别式为零：2² − 4(k)(1) = 0 → 4 − 4k = 0 → k = 1。

7. Set-Builder Notation and Number Lines | 集合描述法与数轴表示

Solutions to inequalities are often expressed in set-builder notation or interval notation. In AQA, you may be asked to represent the solution on a number line. For example, x ≤ 3 or x > 7 is written as x ∈ (−∞, 3] ∪ (7, ∞).

不等式的解常常用集合描述法或区间表示法。在 AQA 考试中，你可能需要把解表示在数轴上。例如 x ≤ 3 或 x > 7 可写成 x ∈ (−∞, 3] ∪ (7, ∞)。

Use square brackets [ ] for inclusive / 方括号表示包含端点。
Use round brackets ( ) for exclusive or infinite / 圆括号表示不包含端点或无穷。
A ‘union’ symbol ∪ joins separate regions / 符号 ∪ 表示并集联接不同区域。

8. Equations with Fractions and Decimals | 含分数和小数的方程

When equations contain fractions, multiply every term by the lowest common multiple (LCM) of denominators to simplify. This often removes the fractions entirely, making the equation easier to solve.

当方程含有分数时，将每一项乘以分母的最小公倍数（LCM）来简化，通常可以完全去掉分数，使方程更容易求解。

(x − 1)/2 + (2x + 3)/3 = 2

Multiply by 6: 3(x − 1) + 2(2x + 3) = 12 → 3x − 3 + 4x + 6 = 12 → 7x + 3 = 12 → x = 9/7.

两边同乘以6：3(x − 1) + 2(2x + 3) = 12 → 3x − 3 + 4x + 6 = 12 → 7x + 3 = 12 → x = 9/7。

9. Forming and Solving Equations from Word Problems | 根据应用题建立并求解方程

Many AQA questions in ‘Miscellaneous Exercises’ require you to translate a real-world scenario into an algebraic equation. Define your variable clearly, form the equation from the given conditions, then solve it. Check your answer against the original statement.

在’综合练习’中，许多 AQA 题目要求你把实际情境转化成代数方程。首先要定义未知数，然后根据条件构建方程并求解，最后用原题检查答案是否合理。

Example: The length of a rectangle is 4 cm more than its width. The perimeter is 36 cm. Let width = w, length = w + 4. Then 2(w + (w + 4)) = 36 → 4w + 8 = 36 → w = 7, length = 11 cm.

例：长方形的长比宽多 4 cm，周长为 36 cm。设宽 = w，长 = w + 4。则 2(w + (w + 4)) = 36 → 4w + 8 = 36 → w = 7，长 = 11 cm。

10. When a Line Meets a Quadratic | 直线与二次曲线的位置关系

Substituting the linear equation into the quadratic gives a quadratic equation in one variable. The discriminant of this resulting quadratic tells you how many intersection points exist. This is a favourite exam style question because it combines several skills at once.

将直线方程代入二次方程，就得到一个一元二次方程。此二次方程的判别式决定了交点的个数。这是考试中非常常见的题型，综合考查了多个技能。

Worked example: y = 2x + k and y = x² + 3x + 1. Find the range of k for which the line intersects the curve at two points. Substitute: 2x + k = x² + 3x + 1 → x² + x + (1 − k) = 0. Discriminant: 1² − 4(1)(1 − k) > 0 → 1 − 4 + 4k > 0 → 4k > 3 → k > 3/4.

例题：直线 y = 2x + k 与曲线 y = x² + 3x + 1 相交于两点，求 k 的范围。代入：2x + k = x² + 3x + 1 → x² + x + (1 − k) = 0。判别式：1² − 4(1)(1 − k) > 0 → 1 − 4 + 4k > 0 → 4k > 3 → k > 3/4。

11. Common Mistakes and Tips | 常见错误与技巧

Below are the most common errors students make with equations and inequalities, and how to avoid them.

以下列出学生在方程与不等式部分最容易犯的错误及避免方法。

Forgetting to reverse the inequality when multiplying/dividing by a negative / 乘除负数时忘记反转不等号。
Mis-expanding double brackets – always check with FOIL or an area method / 展开括号出错——用 FOIL 或面积法检查。
When solving simultaneous equations, substitute back to find the other variable – don’t stop halfway / 联立方程时，解出一个未知数后要代入求另一个，不能半途停下。
Ignoring impossible roots (e.g., negative lengths) / 忽略不合理的根（如负长度）。
Forgetting to set the equation to zero before factorising / 因式分解前忘记把方程化为零。

12. Practice Exercises for You | 为你的练习题

Use the following questions to test yourself. Answers are provided at the end of this section, but try them first without peeking.

用下列问题自测吧。答案在本节末尾，但请先尝试独立完成。

Solve the equation 5(2x − 3) = 4(x + 6). / 解方程 5(2x − 3) = 4(x + 6)。
Find the roots of x² − 7x + 10 = 0. / 求 x² − 7x + 10 = 0 的根。
Solve the simultaneous equations: y = x² − 2x + 1 and y = 3x − 5. / 解联立方程：y = x² − 2x + 1 与 y = 3x − 5。
Solve the inequality 2x² + 3x − 2 < 0. / 解不等式 2x² + 3x − 2 < 0。
Find the values of p such that the equation px² + 4x − 1 = 0 has two distinct real roots. / 求 p 的值，使得方程 px² + 4x − 1 = 0 有两个不同的实数根。

Answers / 答案: 1. x = 13/2 (or 6.5) 2. x = 2, 5 3. (3, 4) 4. −2 < x < 1/2 5. p > −4 (and p ≠ 0)

If you got any of these wrong, revisit the relevant sections above. Mastery comes from repeated practice, so attempt more questions from your textbook.

如果这些题目有任何做错了，请回看上文对应的小节。熟练掌握源于反复练习，请多做教材中的题目。

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AQA A-Level Maths Miscellaneous Exercise 3: Equations and Inequalities | AQA A-Level 数学综合练习3：方程与不等式