📚 Mixed Exercise 3: Equations and Inequalities | 混合练习3：方程与不等式
Mixed Exercise 3 from the Edexcel A-Level Mathematics Pure Year 1 textbook is a consolidation of all key concepts in Chapter 3: Equations and Inequalities. This exercise challenges students to apply methods for solving linear equations, quadratic equations, simultaneous equations, linear and quadratic inequalities, and to use the discriminant to determine the nature of roots. By working through these problems, learners build fluency in algebraic manipulation, graphical interpretation and logical reasoning. The mixed nature means you must quickly recognise which technique is suitable for each question — a vital skill in the exam.
Edexcel A-Level 数学纯数第一册教材中的混合练习3是对第三章“方程与不等式”全部核心概念的综合巩固。该练习要求学生运用解线性方程、二次方程、联立方程、线性不等式和二次不等式的方法，并利用判别式判断根的性质。通过完成这些题目，学生可以提升代数运算、图形解读和逻辑推理的熟练度。混合题型意味着你必须迅速识别每道题应使用哪种技巧——这在考试中至关重要。
1. Solving Linear Equations | 解线性方程
Linear equations in one variable are the simplest type, usually of the form ax + b = cx + d. The strategy is to collect like terms, isolate the variable on one side, and then divide by the coefficient. Be careful when negative signs appear before brackets, and always check your solution by substitution. In Mixed Exercise 3, you may encounter equations with fractions; the first step is often to multiply through by a common denominator to clear the fractions.
一元一次线性方程是最简单的类型，通常形式为 ax + b = cx + d。求解策略是合并同类项，将未知数移到一边，再除以系数。括号前出现负号时要特别小心，并且一定要通过代入检验解。在混合练习3中，你可能会遇到含有分数的方程；第一步通常是两边乘以公分母，消去分母。
Example: Solve 3(2x – 1) – 2(x + 4) = 7. First expand: 6x – 3 – 2x – 8 = 7 → 4x – 11 = 7 → 4x = 18 → x = 4.5. When fractions are involved, e.g. (x+1)/3 – (2x-1)/4 = 2, multiply by 12 to get 4(x+1) – 3(2x-1) = 24, then expand and simplify.
例如：解 3(2x – 1) – 2(x + 4) = 7。先去括号：6x – 3 – 2x – 8 = 7 → 4x – 11 = 7 → 4x = 18 → x = 4.5。含分数时，如 (x+1)/3 – (2x-1)/4 = 2，乘以12得 4(x+1) – 3(2x-1) = 24，然后去括号化简。
2. Quadratic Equations and Factorisation | 二次方程与因式分解
A quadratic equation is written as ax² + bx + c = 0, where a ≠ 0. The three main methods for solving are factorising, completing the square, and using the quadratic formula x = [-b ± √(b² – 4ac)] / (2a). In Mixed Exercise 3, many quadratics are factorisable, but some demand the formula. Always write the equation in standard form with zero on one side before attempting to factorise. Remember to look for a common factor first, as it simplifies the process.
二次方程的一般形式为 ax² + bx + c = 0，其中 a ≠ 0。三种主要求解方法是因式分解、配方法和使用求根公式 x = [-b ± √(b² – 4ac)] / (2a)。在混合练习3中，许多二次方程是可分解的，但有些需要用公式。在尝试因式分解之前，一定要将方程写成标准形式，即一边为零。记得先提取公因子，这样能简化过程。
For factorisation, find two numbers that multiply to ac and add to b. For example, 2x² + 7x – 15 = 0: ac = -30, numbers 10 and -3. Rewrite: 2x² + 10x – 3x – 15 = 0 → 2x(x+5) – 3(x+5) = 0 → (2x-3)(x+5)=0 → x = 3/2 or x = -5. When the quadratic does not factorise neatly, use the formula directly.
因式分解时，找到两个数，其积为 ac，和为 b。例如 2x² + 7x – 15 = 0: ac = -30，两数为 10 和 -3。重写：2x² + 10x – 3x – 15 = 0 → 2x(x+5) – 3(x+5) = 0 → (2x-3)(x+5)=0 → x = 3/2 或 x = -5。当二次式不能整洁分解时，直接使用公式。
3. Simultaneous Equations – One Linear and One Quadratic | 联立方程 – 一次与二次
Mixed Exercise 3 typically includes pairs where one equation is linear and the other is quadratic, e.g. y = 2x + 1 and y = x² + 3x – 2. The substitution method is most effective: replace y in the quadratic using the linear expression, then solve the resulting quadratic in one variable. You will often obtain two solutions for x, and each generates a corresponding y-value. Always write solutions as coordinate pairs (x, y).
混合练习3通常包含一个一次方程和一个二次方程的联立方程组，例如 y = 2x + 1 和 y = x² + 3x – 2。代入法最有效：用一次关系式替换二次方程中的 y，然后求解关于一个变量的二次方程。你通常会得到两个 x 的解，每个 x 产生相应的 y 值。务必把解写成坐标对 (x, y) 的形式。
Example: Solve y = x² – 4x + 3 and y = 2x – 2. Set x² – 4x + 3 = 2x – 2 → x² – 6x + 5 = 0 → (x-1)(x-5)=0 → x=1, x=5. Then y = 2(1)-2 = 0 and y = 2(5)-2 = 8. Solutions: (1, 0) and (5, 8). Also be aware that the line could be tangent to the curve, leading to a repeated root (one intersection point).
例如：解 y = x² – 4x + 3 和 y = 2x – 2。令 x² – 4x + 3 = 2x – 2 → x² – 6x + 5 = 0 → (x-1)(x-5)=0 → x=1, x=5。于是 y = 2(1)-2 = 0，y = 2(5)-2 = 8。解为 (1, 0) 和 (5, 8)。同时要注意，直线可能与曲线相切，此时得到重根（一个交点）。
4. The Discriminant and Nature of Roots | 判别式与根的性质
For a quadratic ax² + bx + c = 0, the discriminant Δ = b² – 4ac tells us the nature of the roots without solving the equation. If Δ > 0, there are two distinct real roots; if Δ = 0, there is one repeated real root; if Δ < 0, there are no real roots. Mixed Exercise 3 uses this concept to find unknown coefficients when given conditions on the number of solutions, such as a line intersecting a curve exactly once.
对于二次方程 ax² + bx + c = 0，判别式 Δ = b² – 4ac 可以在不解方程的情况下告诉我们根的性质。若 Δ > 0，有两个相异实根；若 Δ = 0，有一个重根；若 Δ < 0，无实根。混合练习3利用这一概念，在给定解的个数条件（如直线与曲线恰好相交一次）时，求出未知系数。
| Discriminant Δ | Nature of roots |
|---|---|
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root (tangent case) |
| Δ < 0 | No real roots |
For example, find k such that y = x² + kx + 9 has exactly one root. Set Δ = 0: k² – 4(1)(9) = 0 → k² = 36 → k = ±6. Always express the condition mathematically and solve the resulting equation. In simultaneous equations, you often eliminate one variable to form a quadratic in the other, then apply discriminant conditions.
例如，求 k 使得 y = x² + kx + 9 恰好有一个根。令 Δ = 0：k² – 4(1)(9) = 0 → k² = 36 → k = ±6。务必用数学式表达条件，然后解出方程。在联立方程中，常消去一个变量得到关于另一个变量的二次方程，再应用判别式条件。
5. Linear Inequalities | 线性不等式
Linear inequalities are solved like linear equations, but with one crucial rule: if you multiply or divide by a negative number, you must reverse the inequality sign. The solution is typically expressed as a range, e.g. x > 5, and can be shown on a number line. Compound inequalities such as -3 < 2x + 1 ≤ 7 are handled by isolating x step by step while keeping the inequality valid throughout.
线性不等式的解法与线性方程类似，但有一条重要规则：如果乘以或除以一个负数，必须把不等号方向反转。解通常表示为一个范围，如 x > 5，并可以在数轴上表示。对于复合不等式，如 -3 < 2x + 1 ≤ 7，需要逐步分离 x，同时确保整个过程中不等关系保持不变。
Solve: 4 – 3x > 10. Subtract 4: -3x > 6. Divide by -3 (reverse sign): x < -2. The solution set is all real numbers less than -2. In the exam, notation is important: you may use set notation {x : x < -2} or interval notation (-∞, -2). Mixed Exercise 3 includes both straightforward inequalities and those involving brackets and fractions.
解：4 – 3x > 10。减4：-3x > 6。除以 -3（变号）：x < -2。解集是所有小于 -2 的实数。考试中，符号使用很重要：你可以用集合符号 {x : x < -2} 或区间符号 (-∞, -2)。混合练习3既有简单的不等式，也包含带括号和分数的题目。
6. Quadratic Inequalities | 二次不等式
To solve a quadratic inequality such as x² – 5x + 6 > 0, first find the critical values by solving the corresponding equation x² – 5x + 6 = 0, giving x = 2 and x = 3. Then use a sign diagram or sketch the graph to determine where the quadratic is positive. Because the graph of y = x² – 5x + 6 is a U-shaped parabola, it is greater than zero outside the interval between the roots, i.e. x < 2 or x > 3. If the inequality were < 0, the solution would be between the roots.
解二次不等式如 x² – 5x + 6 > 0，首先通过求解对应方程 x² – 5x + 6 = 0 找到临界值，得 x = 2 和 x = 3。然后利用符号表或草图判断二次式在何处为正。因为 y = x² – 5x + 6 的图像是开口向上的抛物线，所以在两根之外的区间大于零，即 x < 2 或 x > 3。如果不等号是 < 0，解集则在两根之间。
Always bring the inequality to the form where one side is zero. For x² ≤ 4, rewrite as x² – 4 ≤ 0 → (x-2)(x+2) ≤ 0. Critical values -2 and 2; the quadratic is ≤ 0 between -2 and 2 inclusive. Mixed Exercise 3 reinforces the graphical approach: a quadratic with a positive coefficient of x² is ∪-shaped; negative coefficient is ∩-shaped. This mental picture is essential for writing correct intervals.
务必先将不等式化为一边为零的形式。对于 x² ≤ 4，改写为 x² – 4 ≤ 0 → (x-2)(x+2) ≤ 0。临界值 -2 和 2；二次式在 -2 和 2 之间（含端点）小于等于零。混合练习3强化了图像法：x² 的系数为正时，图像是 ∪ 形；系数为负时是 ∩ 形。这个脑内画面对于写出正确区间至关重要。
7. Regions Defined by Inequalities | 不等式所表示的区域
Inequalities in two variables describe regions in the coordinate plane. For example, y > 2x + 1 indicates the region above the line y = 2x + 1 (dashed line for strict inequality). A system of inequalities defines the intersection of several regions. It is common to shade the unwanted regions, leaving the desired region unshaded. In Mixed Exercise 3, you may be asked to label regions satisfying conditions like y ≤ x², y > x, and x < 3.
含有两个变量的不等式描述了坐标平面上的区域。例如，y > 2x + 1 表示直线 y = 2x + 1 上方的区域（严格不等式用虚线）。不等式组则定义了几个区域的交集。常见的作法是涂去不需要的区域，让所求区域留白。在混合练习3中，你可能需要标出满足诸如 y ≤ x²、y > x 和 x < 3 等条件的区域。
To sketch, first draw each boundary line or curve, using solid for ≤ or ≥ and dashed for < or >. Then test a point (e.g. the origin) to decide which side to shade. For quadratic boundaries like y ≥ x² – 4, the graph of y = x² – 4 is a parabola; y ≥ means the region above (including) the curve. Always check with a specific point not on the boundary if you are unsure. Label the feasible region clearly.
作图时，首先画出每条分界线或曲线，≤ 或 ≥ 用实线，< 或 > 用虚线。然后取一点（如原点）检验，确定涂哪一侧。对于如 y ≥ x² – 4 的二次边界，y = x² – 4 的图像是抛物线；y ≥ 表示曲线上方（含曲线）的区域。如果不确定，可以取一个不在边界上的特定点进行检验。清晰标出可行域。
8. Mixed Word Problems | 混合应用题
Many questions in Mixed Exercise 3 present real-world contexts requiring the formation of equations or inequalities. You might need to set up a quadratic equation from geometric conditions, such as the area of a rectangle or the trajectory of a projectile, then solve and interpret the answer in context. It is essential to define variables clearly and check whether all algebraic solutions make sense — for instance, negative lengths must be discarded.
混合练习3中有许多题目给出实际背景，需要建立方程或不等式。你可能要根据几何条件，如矩形面积或抛射物轨迹，列出二次方程，然后求解并在上下文中解释答案。明确设定变量并检查所有代数解是否合理至关重要——例如，负的长度必须舍去。
Example: A rectangular garden has length 3 m more than its width. The area is 40 m². Find the dimensions. Let width = w, length = w + 3. Area = w(w+3) = 40 → w² + 3w – 40 = 0 → (w+8)(w-5)=0 → w = 5 (reject w = -8). So width = 5 m, length = 8 m. Also, word problems may involve cost, revenue, or profit leading to inequalities like x² – 20x + 75 ≤ 0, where only positive integer solutions are admissible.
例如：一个矩形花园的长比宽多3米，面积为40平方米。求尺寸。设宽为 w，长为 w + 3。面积 = w(w+3) = 40 → w² + 3w – 40 = 0 → (w+8)(w-5)=0 → w = 5（舍去 w = -8）。因此宽为5米，长为8米。此外，应用题还可能涉及成本、收入或利润，得出如 x² – 20x + 75 ≤ 0 的不等式，此时只有正整数解可接受。
In all problems of Mixed Exercise 3, pay attention to the precise wording: ‘at least’, ‘exceeds’, ‘no more than’ translate into ≥, >, ≤ respectively. Practising these word problems strengthens your modelling skills, a key component of the Pure Mathematics assessment.
在混合练习3的所有习题中，注意精确定义：“at least” 译作 ≥，“exceeds” 译作 >，“no more than” 译作 ≤。练习这些应用题能增强建模能力，这是纯数评估中的关键部分。
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