📚 Solving Linear Equations and Inequalities | 解一次方程与不等式
Linear equations and inequalities are central to the IGCSE Mathematics algebra syllabus. A linear equation has the highest power of the variable equal to 1, and it can usually be solved using inverse operations. A linear inequality uses symbols such as <, >, ≤ and ≥, and follows most of the same solving rules with one key exception: reversing the sign when multiplying or dividing by a negative number.
一次方程与一次不等式是 IGCSE 数学代数部分的核心内容。一次方程中未知数的最高次数为 1，通常可以用逆运算求解。一次不等式使用 <、>、≤ 和 ≥ 等符号，其求解规则与方程基本相同，但有一个关键例外：当两边同乘或同除以负数时，必须改变不等号的方向。
1. What Is a Linear Equation? | 什么是一次方程
A linear equation is an algebraic statement in which the highest power of the variable is 1. In IGCSE Mathematics, typical examples include 3x + 5 = 20, 2(x – 4) = 10 and x/3 – 2 = 7. The word ‘linear’ refers to the fact that the graph of y = ax + b is a straight line.
一次方程是指未知数的最高次数为 1 的代数等式。在 IGCSE 数学中，常见例子包括 3x + 5 = 20、2(x – 4) = 10 和 x/3 – 2 = 7。”一次” 一词来源于 y = ax + b 的图像是一条直线。
3x + 5 = 20
The general form of a linear equation in one variable is ax + b = c, where a, b and c are constants and a ≠ 0. Finding the value of x is called solving the equation.
一元一次方程的一般形式是 ax + b = c，其中 a、b、c 为常数且 a ≠ 0。求出 x 的值的过程称为解方程。
2. Balancing Method: Inverse Operations | 平衡法：逆运算
To solve an equation, keep the equation balanced by applying the same operation to both sides. Each step should ‘undo’ what has been done to x. For 5x + 3 = 18, subtract 3 from both sides, then divide both sides by 5.
解方程时，要对等号两边同时进行相同运算以保持平衡。每一步都要”抵消”对 x 所做的运算。例如 5x + 3 = 18，先两边减 3，再两边除以 5。
5x + 3 = 18 → 5x = 15 → x = 3
Inverse operations are essential: addition is undone by subtraction, multiplication by division, and vice versa. Applying the same inverse operation to both sides keeps the equation true.
逆运算至关重要：加法由减法抵消，乘法由除法抵消，反之亦然。对等号两边同时进行相同的逆运算可以保证等式仍然成立。
| Operation | Inverse |
| Add 5 | Subtract 5 |
| Subtract 4 | Add 4 |
| Multiply by 3 | Divide by 3 |
| Divide by 7 | Multiply by 7 |
3. Equations with Brackets | 带括号的方程
When an equation contains brackets, expand them first using the distributive law, unless you prefer to divide first. Solve 3(x + 2) = 21 by expanding: 3x + 6 = 21, so 3x = 15 and x = 5.
当方程含括号时，通常先用分配律展开；有时也可以先除以括号前的系数。解 3(x + 2) = 21 时，展开得 3x + 6 = 21，因此 3x = 15，x = 5。
3(x + 2) = 21 → 3x + 6 = 21 → 3x = 15 → x = 5
Be careful with negative signs outside brackets. For 4 – 2(x – 3) = 10, expand -2(x – 3) as -2x + 6. The equation becomes 4 – 2x + 6 = 10, then 10 – 2x = 10, giving x = 0.
括号外带负号时要特别小心。例如 4 – 2(x – 3) = 10，应把 -2(x – 3) 展开为 -2x + 6。方程变为 4 – 2x + 6 = 10，即 10 – 2x = 10，解得 x = 0。
4. Equations with Fractions | 带分数的方程
For equations with fractions, multiply every term by the lowest common denominator (LCD) to clear fractions. For x/3 + 2 = 7, subtract 2 first, then multiply by 3, giving x = 15.
对于带分数的方程，通常将每一项乘以最小公分母以去分母。例如 x/3 + 2 = 7，先减 2，再乘以 3，得 x = 15。
x/3 + 2 = 7 → x/3 = 5 → x = 15
If the fraction has a binomial numerator, multiply both sides by the denominator. For (x + 1)/4 = 3, multiply both sides by 4 to get x + 1 = 12, so x = 11.
如果分数的分子为二项式，应把两边同时乘以分母。例如 (x + 1)/4 = 3，两边乘 4 得 x + 1 = 12，因此 x = 11。
(x + 1)/4 = 3 → x + 1 = 12 → x = 11
5. Unknown on Both Sides | 未知数在两边
If x appears on both sides of the equation, collect like terms so that all x terms are on one side. For 5x – 3 = 2x + 9, subtract 2x from both sides to get 3x – 3 = 9, then add 3 and divide by 3.
如果 x 同时出现在等号两边，先移项，使所有含 x 的项集中到一边。例如 5x – 3 = 2x + 9，两边减 2x 得 3x – 3 = 9，再加 3 并除以 3。
5x – 3 = 2x + 9 → 3x – 3 = 9 → 3x = 12 → x = 4
Always aim to make the x coefficient positive. If you have -x = 5, multiply both sides by -1 to get x = -5. This avoids sign errors in the final answer.
解题时尽量使 x 的系数为正。如果得到 -x = 5，两边乘 -1 得 x = -5。这样可以避免最终答案出现符号错误。
6. Word Problems Leading to Linear Equations | 由文字题建立一次方程
IGCSE questions often ask you to translate a written statement into an equation. Define the unknown clearly, write an equation, solve it, and answer in context. Example: three more than twice a number is 19. Let x be the number, so 2x + 3 = 19, giving x = 8.
IGCSE 题目经常要求把文字描述转化为方程。先清楚地设未知数，再列方程、求解并回到题目中作答。例如：一个数的两倍再加 3 等于 19。设这个数为 x，则 2x + 3 = 19，解得 x = 8。
2x + 3 = 19 → 2x = 16 → x = 8
For consecutive integer problems, represent the integers as n, n + 1 and n + 2. If their sum is 42, then n + (n + 1) + (n + 2) = 42, so 3n + 3 = 42 and n = 13. The integers are 13, 14 and 15.
对于连续整数问题，可把整数设为 n、n + 1 和 n + 2。如果它们的和为 42，则 n + (n + 1) + (n + 2) = 42，即 3n + 3 = 42，解得 n = 13。这三个整数为 13、14 和 15。
7. Introduction to Linear Inequalities | 一次不等式入门
An inequality compares two expressions using <, >, ≤ or ≥. Solving a linear inequality is similar to solving an equation, but the direction of the inequality must be reversed when multiplying or dividing both sides by a negative number.
不等式用 <、>、≤ 或 ≥ 比较两个表达式。解一次不等式与解方程类似，但当两边同乘或同除以一个负数时，必须改变不等号的方向。
| Symbol | Meaning | Example |
| < | Less than | x < 5 |
| > | Greater than | x > -2 |
| ≤ | Less than or equal to | x ≤ 7 |
| ≥ | Greater than or equal to | x ≥ 0 |
The solution to an inequality is usually a range of values rather than a single number. For example, x < 4 includes every real number less than 4.
不等式的解通常是一个取值范围，而不是一个单一的数。例如 x < 4 包括所有小于 4 的实数。
8. Solving Linear Inequalities | 解一次不等式
Solve 4x – 3 < 13 by adding 3 to both sides and dividing by 4: x < 4. The inequality sign does not change because we divided by a positive number.
解 4x – 3 < 13 时，两边加 3 再除以 4，得 x < 4。因为除以的是正数，不等号方向不变。
4x – 3 < 13 → 4x < 16 → x < 4
However, for -2x + 6 ≥ 14, subtract 6 from both sides to get -2x ≥ 8. Then divide by -2 and reverse the inequality, giving x ≤ -4.
但解 -2x + 6 ≥ 14 时，先两边减 6，得 -2x ≥ 8。然后两边除以 -2 并改变不等号方向，得 x ≤ -4。
-2x + 6 ≥ 14 → -2x ≥ 8 → x ≤ -4
The reversal rule is often tested in IGCSE papers. A useful check is to test a number in the original inequality.
不等号改变规则是 IGCSE 试卷中的常见考点。一个有效的检查方法是代入一个数到原不等式中进行验证。
9. Inequalities on a Number Line | 数轴上的不等式
Inequalities can be shown on a number line using open circles for strict inequalities (< or >) and closed circles for inclusive inequalities (≤ or ≥). For x > 2, draw an open circle at 2 and shade to the right.
不等式可在数轴上表示：严格不等式（< 或 >）用空心圆，包含不等式（≤ 或 ≥）用实心圆。例如 x > 2，在 2 处画空心圆并向右画阴影。
For a compound inequality such as -1 ≤ x < 3, use a closed circle at -1 and an open circle at 3, then shade the region between them. This represents all numbers from -1 up to but not including 3.
对于 -1 ≤ x < 3 这样的复合不等式，在 -1 处用实心圆，在 3 处用空心圆，然后涂满两者之间的区域。这表示从 -1 开始、到 3 为止但不包括 3 的所有数。
When listing integer solutions, use the shaded region and the circle types. For -1 ≤ x < 3, the integer solutions are -1, 0, 1 and 2.
列出整数解时，要根据阴影区域和圆圈类型判断。对于 -1 ≤ x < 3，整数解为 -1、0、1 和 2。
10. Checking Solutions and Common Errors | 验根与常见错误
Always substitute your answer back into the original equation to verify it. This is especially important when fractions, brackets or negative signs are involved.
始终把答案代回原方程进行验算。当涉及分数、括号或负号时，这一点尤其重要。
Common errors include forgetting to reverse the inequality sign when dividing by a negative, mishandling fractions, and losing the balance by applying an operation to only one side. Another frequent mistake is expanding brackets incorrectly when a negative sign is outside.
常见错误包括：除以负数时忘记改变不等号、分数处理错误、以及只对等号一边进行运算而失去平衡。另一个常见错误是括号外为负号时展开符号出错。
| Common error | Correct approach |
| Dividing by -2 without reversing the sign | Reverse ≤ to ≥ |
| Multiplying only one term by the LCD | Multiply every term |
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