Equations and Inequalities in IGCSE Mathematics | IGCSE 数学中的方程与不等式

📚 Equations and Inequalities in IGCSE Mathematics | IGCSE 数学中的方程与不等式

Equations and inequalities are central to the IGCSE Mathematics syllabus. They allow you to express relationships, find unknown values, and model real-life situations in a precise way. A strong command of solving equations and inequalities is essential for success in both Core and Extended papers.

方程与不等式是 IGCSE 数学大纲的核心内容。它们帮助你表达数量关系、求解未知数，并准确地模拟现实情境。熟练掌握解方程和不等式的方法，对于在 Core 和 Extended 试卷中取得好成绩至关重要。

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1. Understanding Linear Equations | 理解线性方程

ng a linear equation means finding the value of the variable that makes the equation true.

线性方程是指未知数的最高次数为 1 的方程。它通常可以写成 ax + b = 0 或 ax + b = cx + d 的形式。解线性方程就是求出使等式成立的未知数的值。

e unknown is x, and the highest power of x is 1. The solution is the number that can replace x so that both sides have the same value.

例如：在方程 3x + 5 = 20 中，未知数是 x，x 的最高次数为 1。方程的解就是能够代入 x，使等号两边数值相等的那个数。

Equation form	Example	Highest power
ax + b = 0	4x - 9 = 0	1
ax + b = cx + d	5x + 2 = 3x + 10	1

Recognising the type of equation helps you choose the correct solving strategy from the start.

从一开始就识别方程的类型，有助于你选择正确的求解策略。

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2. Solving Linear Equations Step by Step | 逐步解线性方程

To solve a linear equation, perform inverse operations in reverse order. First simplify both sides, then add or subtract terms to collect like terms, and finally multiply or divide to isolate the variable.

解线性方程时，需要按照逆运算的相反顺序进行。先化简两边，再通过加减移项合并同类项，最后通过乘除将未知数单独留在等号一边。

• Simplify both sides by expanding brackets and collecting like terms. / 先展开括号并合并同类项，化简两边。
• Move variable terms to one side and constant terms to the other. / 将含有未知数的项移到一边，常数项移到另一边。
• Divide by the coefficient of the variable. / 最后除以未知数的系数。

+ 6 = 10. Then subtract 6 from both sides: 2x = 4. Finally divide both sides by 2: x = 2.

示例：给定 2(x + 3) = 10，先展开括号得 2x + 6 = 10。然后两边减 6 得 2x = 4。最后两边除以 2 得 x = 2。

2(x + 3) = 10 → 2x + 6 = 10 → 2x = 4 → x = 2

Always check by substituting your solution back into the original equation. If the left side equals the right side, your answer is correct.

一定要把解代回原方程进行检验。如果左边等于右边，说明答案正确。

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3. Linear Inequalities on a Number Line | 数轴上的线性不等式

rcle for strict inequalities (<, >) and a closed circle for inclusive inequalities (≤, ≥).

不等式用 <、>、≤ 或 ≥ 来比较两个表达式。线性不等式的解通常是一个取值范围。在数轴上，严格不等式（<、>）用空心圆表示，包含等于的不等式（≤、≥）用实心圆表示。

eans all values less than or equal to 5, with a closed circle at 5 and shading to the left.

例如：x > 2 表示所有大于 2 的数，在 2 处画空心圆，并向右画阴影。相反，x ≤ 5 表示所有小于或等于 5 的数，在 5 处画实心圆，并向左画阴影。

Inequality	Circle type	Direction of shading
x < 3	Open	Left
x ≥ -1	Closed	Right

Reading and drawing number-line diagrams correctly is commonly tested in IGCSE papers.

正确地读图和画数轴图是 IGCSE 试卷中的常见考点。

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4. Solving Linear Inequalities | 解线性不等式

Solving linear inequalities is similar to solving linear equations, but there is one key difference: when you multiply or divide both sides by a negative number, you must reverse the inequality sign.

解线性不等式与解线性方程类似，但有一个关键区别：当两边同时乘以或除以一个负数时，必须改变不等号的方向。

s by 3: x ≤ 5. The inequality sign stays the same because we divided by a positive number.

示例：对于 3x + 4 ≤ 19，两边减 4 得 3x ≤ 15。然后两边除以 3 得 x ≤ 5。因为除以的是正数，不等号方向保持不变。

Example with reversal: Solve -2x > 8. Divide both sides by -2 and reverse the sign: x < -4. This step is easy to forget, so write a reminder next to your working.

变号示例：解 -2x > 8。两边除以 -2 并改变不等号方向：x < -4。这一步很容易忘记，建议在解题时写下提醒。

-2x > 8 → x < -4

In the final answer, you may be asked to show the solution on a number line or write it in set-builder notation such as {x : x < -4}.

最后答案可能要求你在数轴上表示解，或用集合表示法写出解，例如 {x : x < -4}。

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5. Compound Inequalities | 复合不等式

ns x is greater than 2 and less than or equal to 7. Solve each part carefully, and represent the final range on a number line or in set-builder notation.

复合不等式由两个不等式组合而成。例如，2 < x ≤ 7 表示 x 大于 2 且小于等于 7。需要仔细求解每个部分，并在数轴或集合表示法中表示最终范围。

Subtract 1 from all three parts: 2 ≤ 2x < 8. Then divide all three parts by 2: 1 ≤ x < 4.

示例：解 3 ≤ 2x + 1 < 9。三部分同时减 1 得 2 ≤ 2x < 8。再同时除以 2 得 1 ≤ x < 4。

3 ≤ 2x + 1 < 9 → 2 ≤ 2x < 8 → 1 ≤ x < 4

) in interval notation, although at IGCSE you often only need a number-line diagram or a clear inequality statement.

写最终答案时要使用正确的表示法。例如，解 1 ≤ x < 4 可以写成区间 [1, 4)，不过在 IGCSE 中通常只需要数轴图或清晰的不等式表达。

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6. Quadratic Equations: Factorisation | 二次方程：因式分解法

standard method is factorisation. Write the quadratic as a product of two brackets, then set each bracket equal to zero because if a product is zero, at least one factor must be zero.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。一种标准方法是用因式分解。将二次式写成两个括号的乘积，然后令每个括号分别等于零，因为如果乘积为零，至少有一个因式为零。

left side: (x + 2)(x + 3) = 0. Therefore x + 2 = 0 or x + 3 = 0, giving x = -2 or x = -3.

示例：解 x² + 5x + 6 = 0。将左边因式分解得 (x + 2)(x + 3) = 0。因此 x + 2 = 0 或 x + 3 = 0，得到 x = -2 或 x = -3。

x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = -2 or x = -3

Always check the signs inside the brackets. For a positive constant and a positive coefficient of x, both numbers are positive. If the constant is positive but the coefficient of x is negative, both numbers are negative.

检查括号内的符号非常重要。如果常数项为正且 x 的系数为正，那么两个数都为正。如果常数项为正但 x 的系数为负，那么两个数都为负。

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7. Quadratic Formula | 二次公式

ily, use the quadratic formula. For ax² + bx + c = 0, the solutions are given by x = [-b ± √b² - 4ac] / 2a. The expression b² - 4ac is called the discriminant.

当二次方程不容易因式分解时，可以使用二次公式。对于 ax² + bx + c = 0，解为 x = [-b ± √b² - 4ac] / 2a。表达式 b² - 4ac 称为判别式。

x = [-b ± √b² - 4ac] / 2a

Example: Solve x² + 4x + 1 = 0. Here a = 1, b = 4, c = 1. Then x = [-4 ± √16 - 4] / 2 = [-4 ± √12] / 2 = -2 ± √3.

示例：解 x² + 4x + 1 = 0。这里 a = 1，b = 4，c = 1。则 x = [-4 ± √16 - 4] / 2 = [-4 ± √12] / 2 = -2 ± √3。

The discriminant tells you how many real solutions there are. If b² - 4ac > 0, there are two distinct

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