Linear Equations and Inequalities | 线性方程与不等式

📚 Linear Equations and Inequalities | 线性方程与不等式

Linear equations and inequalities form the backbone of IGCSE Mathematics. They appear in almost every paper, both as standalone questions and inside larger problems on sequences, geometry, or word problems.

线性方程和不等式是 IGCSE 数学的主干内容。它们几乎出现在每份试卷中，既作为独立题目，也出现在数列、几何或文字题等更大的问题里。

In this revision guide, you will learn how to solve linear equations step by step, how to handle brackets and fractions, how to form equations from written problems, and how to solve and represent linear inequalities accurately.

在本复习指南中，你将学习如何逐步解线性方程、如何处理括号和分数、如何根据文字题建立方程，以及如何准确地求解并表示线性不等式。

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1. Solving One-Step Linear Equations | 解一步线性方程

An equation states that two expressions are equal. A linear equation has the unknown raised only to the power of 1, such as x, 2x + 3, or 5 − 4x.

方程表示两个表达式相等。线性方程的未知数只有 1 次幂，例如 x、2x + 3 或 5 − 4x。

The key idea is to isolate the unknown by performing the same inverse operation on both sides of the equation.

核心思路是对方程两边同时进行相同的逆运算，从而把未知数单独留在一边。

x + 5 = 12 → x = 12 − 5 → x = 7

by substituting your answer back into the original equation: 7 + 5 = 12, which is correct.

这里未知数加了 5，因此两边同时减去 5。一定要把答案代回原方程检验：7 + 5 = 12，结果正确。

er, divide both sides by that number. For example, 4x = 20 gives x = 5 because 20 ÷ 4 = 5.

如果未知数乘以一个数，那么两边同时除以这个数。例如 4x = 20，解得 x = 5，因为 20 ÷ 4 = 5。

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2. Solving Two-Step and Multi-Step Equations | 解两步与多步方程

Many linear equations require two operations: first undo addition or subtraction, then undo multiplication or division. The order follows the reverse of BODMAS.

许多线性方程需要两步运算：先消去加法或减法，再消去乘法或除法。顺序与 BODMAS 相反。

Example: solve 2x + 3 = 11. First subtract 3 from both sides, then divide by 2.

例题：解方程 2x + 3 = 11。先从两边减去 3，再除以 2。

2x + 3 = 11 → 2x = 8 → x = 4

by collecting like terms. For example, 5x − 2 + x = 16 becomes 6x − 2 = 16, giving x = 3.

对于多步方程，先通过合并同类项化简每一边。例如 5x − 2 + x = 16 化简为 6x − 2 = 16，解得 x = 3。

er. Solve 7x − 4 = 3x + 8 by subtracting 3x from both sides to get 4x − 4 = 8, then x = 3.

如果未知数出现在等号两边，就把所有含 x 的项移到一边，常数移到另一边。解 7x − 4 = 3x + 8，先两边减去 3x，得到 4x − 4 = 8，再解得 x = 3。

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3. Equations with Brackets | 含括号的方程

When an equation contains brackets, start by expanding them using the distributive property. Multiply each term inside the bracket by the term outside.

当方程含有括号时，先用分配律展开括号。用括号外的项乘以括号内的每一项。

Example: 3(x + 2) = 21. Expand to get 3x + 6 = 21.

例题：3(x + 2) = 21。展开得到 3x + 6 = 21。

3(x + 2) = 21 → 3x + 6 = 21 → 3x = 15 → x = 5

st expand −2(x − 3) to −2x + 6, then simplify the left side to 10 − 2x = 10, giving x = 0.

注意负号。对于 4 − 2(x − 3) = 10，先把 −2(x − 3) 展开为 −2x + 6，再把左边化简为 10 − 2x = 10，解得 x = 0。

Always check expansion before solving. A sign error in brackets will carry through the entire solution, so write the expanded form clearly in your working.

解题前一定要检查展开过程。括号中的符号错误会影响整个解题过程，因此在草稿中要清楚地写出展开后的形式。

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4. Equations with Fractions | 含分数的方程

To solve equations with fractions, multiply every term by the lowest common denominator to clear the fractions. This makes the equation easier to handle.

解含分数的方程时，把每一项都乘以最小公分母，以消去分数。这样方程会更容易处理。

ample: solve x/4 + 2 = 5. First subtract 2 from both sides, then multiply both sides by 4.

例题：解方程 x/4 + 2 = 5。先从两边减去 2，再在两边同时乘以 4。

x/4 + 2 = 5 → x/4 = 3 → x = 12

des by 3 first: 2x = 24, so x = 12. Always write the division bar clearly in your working.

对于更复杂的分数，例如 (2x)/3 = 8，先两边乘以 3：2x = 24，因此 x = 12。解题过程中要清楚地写出分数线。

mmon denominator. For x/2 + 3 = x/4 + 6, multiply by 4 to get 2x + 12 = x + 24, so x = 12.

如果方程两边都有分数，就把每一项都乘以公分母。对于 x/2 + 3 = x/4 + 6，两边乘以 4，得到 2x + 12 = x + 24，因此 x = 12。

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5. Forming Equations from Word Problems | 根据文字题建立方程

IGCSE exam questions often ask you to form an equation from a written situation. Define

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