Linear Equations and Inequalities for IGCSE Mathematics | IGCSE 数学：线性方程与不等式

📚 Linear Equations and Inequalities for IGCSE Mathematics | IGCSE 数学：线性方程与不等式

Linear equations and inequalities are foundational algebra topics in IGCSE Mathematics. They appear in both Core and Extended papers, often as standalone questions or embedded in word problems and graphs. A linear equation contains variables raised only to the power of 1, and its graph is a straight line. Linear inequalities extend equations by using symbols such as <, >, ≤ and ≥, and their solutions are ranges of values rather than single numbers. This article reviews the essential methods, common question types, and exam tips you need to master this topic.

线性方程和不等式是 IGCSE 数学中代数的基础主题。它们出现在核心和扩展试卷中，通常作为独立题目或嵌入应用题和图像题中。线性方程只含一次幂的变量，其图像是一条直线。线性不等式通过使用 <、>、≤ 和 ≥ 等符号扩展了方程，其解是值的范围而不是单个数字。本文回顾了掌握该主题所需的基本方法、常见题型和考试技巧。

1. What is a Linear Equation? | 什么是线性方程？

A linear equation is an algebraic statement in which the highest power of the variable is 1. For example, 2x + 3 = 11, 5y = 20, and 3(x + 1) = 2x + 7 are linear equations. Linear equations in one variable have exactly one solution, provided the coefficient of the variable is not zero. In two variables, such as y = 2x + 1, the equation describes a straight line on the coordinate plane.

线性方程是变量最高次数为 1 的代数表达式。例如，2x + 3 = 11、5y = 20 和 3(x + 1) = 2x + 7 都是线性方程。一元线性方程，只要变量系数不为零，就只有一个解。对于两变量线性方程，例如 y = 2x + 1，该方程在坐标平面上描述一条直线。

In IGCSE exams, you need to solve linear equations accurately and show clear working. Marks are often awarded for method even if the final answer is incorrect, so writing balanced operations is essential. The key principle is to keep the equation balanced: whatever operation you do to one side, you must do to the other. A common check is that any equation such as x² = 9 or xy = 12 is not linear because the variables are squared or multiplied together.

在 IGCSE 考试中，你需要准确解线性方程并展示清晰的步骤。即使最终答案错误，方法也常常能得分，因此写出平衡的运算是必要的。关键原则是保持方程平衡：你对一边进行什么运算，就必须对另一边进行相同运算。一个常见的判别是，诸如 x² = 9 或 xy = 12 的方程不是线性方程，因为变量被平方或相乘。

2. Solving Basic Linear Equations | 解基本线性方程

To solve a basic linear equation, isolate the variable by using inverse operations in the correct order. The standard order is to undo addition or subtraction first, then undo multiplication or division. For example, to solve 4x + 7 = 27, subtract 7 from both sides to get 4x = 20, then divide both sides by 4 to get x = 5.

解基本线性方程时，使用逆运算按正确顺序分离变量。标准顺序是先解除加法或减法，再解除乘法或除法。例如，解 4x + 7 = 27，先从两边减去 7 得到 4x = 20，然后两边除以 4 得到 x = 5。

The general steps for solving an equation like ax + b = c are:

• Subtract b from both sides: ax = c – b
• Divide both sides by a (if a ≠ 0): x = (c – b) / a

解 ax + b = c 这类方程的一般步骤是：

• 两边减去 b：ax = c – b
• 两边除以 a（若 a ≠ 0）：x = (c – b) / a

x = (c – b) / a

Always substitute your solution back into the original equation to check it. If the left-hand side equals the right-hand side, your answer is correct. For example, solve -3x + 5 = 20. Subtract 5: -3x = 15. Divide by -3: x = -5. Substitute back: -3(-5) + 5 = 15 + 5 = 20, so the solution works.

始终将解代回原方程进行检查。如果左边等于右边，那么你的答案就是正确的。例如，解 -3x + 5 = 20。两边减 5：-3x = 15。除以 -3：x = -5。代回：-3(-5) + 5 = 15 + 5 = 20，所以该解成立。

3. Equations with Brackets | 含括号的方程

When an equation contains brackets, expand them first before collecting like terms. Use the distributive law: a(b + c) = ab + ac. For example, solve 3(x + 2) = 21. Expanding gives 3x + 6 = 21. Then subtract 6 from both sides: 3x = 15, and divide by 3: x = 5.

当方程含有括号时，先展开括号再合并同类项。使用分配律：a(b + c) = ab + ac。例如，解 3(x + 2) = 21。展开得 3x + 6 = 21。然后两边减 6：3x = 15，再除以 3：x = 5。

If there is a negative sign before the bracket, be careful with signs. For example, solve 5 – 2(x – 1) = 9. Expand: 5 – 2x + 2 = 9, so 7 – 2x = 9. Subtract 7: -2x = 2, divide by -2: x = -1. The sign error of writing 5 – 2x – 2 instead of 5 – 2x + 2 is one of the most common mistakes in IGCSE algebra.

如果括号前有负号，要小心符号。例如，解 5 – 2(x – 1) = 9。展开：5 – 2x + 2 = 9，所以 7 – 2x = 9。两边减 7：-2x = 2，除以 -2：x = -1。把 5 – 2x + 2 写成 5 – 2x – 2 的符号错误是 IGCSE 代数中最常见的错误之一。

For a more complex example, solve 2(3x – 1) + 3(x + 2) = 20. Expand: 6x – 2 + 3x + 6 = 20. Collect like terms: 9x + 4 = 20. Subtract 4: 9x = 16, so x = 16/9. Expand carefully and check each term before collecting.

对于更复杂的例子，解 2(3x – 1) + 3(x + 2) = 20。展开：6x – 2 + 3x + 6 = 20。合并同类项：9x + 4 = 20。两边减 4：9x = 16，所以 x = 16/9。仔细展开并在合并前检查每一项。

4. Equations with Fractions | 含分数的方程

To solve equations with fractions, multiply every term by the lowest common denominator (LCD) to eliminate fractions. For example, solve x/3 + 1 = 5. Multiply all terms by 3: x + 3 = 15, then x = 12. If the equation is (x + 1)/2 = (x – 2)/3, cross-multiply or multiply by 6: 3(x +

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