📚 Simultaneous Equations Demystified | 联立方程组完全解析
Simultaneous equations are one of the most important topics in IGCSE Mathematics because they ask you to find values that satisfy two or more equations at the same time. Mastering them develops algebraic fluency, logical reasoning and the ability to model real-world problems in science, business and engineering.
联立方程组是 IGCSE 数学中最重要的主题之一，因为它要求你找出同时满足两个或更多方程的变量值。掌握它可以培养代数运算能力、逻辑推理能力，以及为科学、商业和工程中的现实问题建模的能力。
1. What Are Simultaneous Equations? | 什么是联立方程组？
es that make every equation in the system true at the same time. For example, the pair 2x + y = 7 and x − y = 2 is a linear system in x and y. The solution is x = 3, y = 1 because 2(3) + 1 = 7 and 3 − 1 = 2.
联立方程组是一组含有相同未知变量的两个或更多方程。解就是使方程组中每一个方程同时成立的这些变量的值。例如，2x + y = 7 和 x − y = 2 是一组关于 x 和 y 的线性方程组。解为 x = 3, y = 1，因为 2(3) + 1 = 7 且 3 − 1 = 2。
2x + y = 7
x − y = 2
⟹ x = 3, y = 1
In IGCSE Edexcel Mathematics, you will mainly meet linear simultaneous equations, whose graphs are straight lines, but you also need to handle some non-linear cases, such as one linear equation and one quadratic equation. The syllabus methods are elimination, substitution and graphical interpretation.
在 IGCSE Edexcel 数学中，你主要会遇到线性联立方程组，它们的图像是直线；但你也需要处理一些非线性情况，例如一个线性方程和一个二次方程。大纲要求的方法包括消元法、代入法和图像法。
2. Why Do We Need Them? | 为什么要学习联立方程组？
he meeting point of two roads represented by linear equations, identifying the break-even point in business where cost = revenue, or solving circuit problems using Kirchhoff's laws.
联立方程组不只是抽象的代数练习。它们可以模拟必须同时满足两个条件的现实情形，例如寻找由线性方程表示的两条道路的交汇点，确定商业中成本 = 收入的盈亏平衡点，或利用基尔霍夫定律解决电路问题。
Developing fluency with these techniques strengthens your algebraic manipulation and teaches systematic thinking, which is why they form a core part of the IGCSE algebra strand.
熟练掌握这些技巧可以加强你的代数运算能力，并教你系统性思考，这就是为什么它们构成 IGCSE 代数部分的核心内容。
- Geometry: intersection of two lines | 几何：两条直线的交点
- Business: break-even analysis, cost = revenue | 商业：盈亏平衡分析，成本 = 收入
- Physics: Kirchhoff's laws in circuits | 物理：电路中的基尔霍夫定律
- Economics: supply and demand equilibrium | 经济：供求均衡
3. The Three Core Methods at a Glance | 三大核心方法概览
Before working through full solutions, it helps to compare the three main methods. Each method has situations where it is faster or safer.
在完整解题之前，先比较三种主要方法会很有帮助。每种方法都有更快或更稳妥的适用场景。
| Method | 方法 | Best for | 最适合 | Key idea | 核心思想 | Watch out | 注意 |
|---|---|---|---|
| Elimination | 消元法 | Two linear equations with matching or easily matched coefficients | 两个线性方程，系数相同或容易配成相同 | Add or subtract to remove one variable | 通过相加或相减消去一个变量 | Align like terms and manage negative signs | 对齐同类项，处理好负号 |
| Substitution | 代入法 | One variable is already isolated, or one equation is quadratic | 一个变量已单独表示，或其中一个方程是二次方程 | Replace one variable with an expression | 用一个表达式替换另一个变量 | Solve the resulting equation fully, then back-substitute | 完整解出所得方程，再回代求另一个变量 |
| Graphical | 图像法 | Estimating solutions or when the question asks for a graph | 估算解，或题目要求画图 | Find the intersection point of the graphs | 找两条图像的交点 | Readings may be approximate; verify algebraically | 读图可能有误差，需用代数验证 |
4. Solving by Elimination | 消元法求解
Elimination works by adding or subtracting the two equations so that one variable cancels. The goal is to make the coefficients of one variable equal in magnitude but opposite in sign, then add.
消元法通过将两个方程相加或相减，使其中一个变量被消去。目标是让某个变量的系数大小相等、符号相反，然后相加。
The y-terms already have opposite signs and the same coefficient 2, so add the equations.
考虑方程组 3x + 2y = 14 和 2x − 2y = 6。y 项符号相反且系数都是 2，所以把两个方程相加。
3x + 2y = 14
2x − 2y = 6
────────────
5x = 20
x = 4
2y = 14: 3(4) + 2y = 14, so 12 + 2y = 14, 2y = 2 and y = 1. The solution is x = 4, y = 1.
将 x = 4 代回 3x + 2y = 14：3(4) + 2y = 14，所以 12 + 2y = 14，2y = 2，y = 1。解为 x = 4, y = 1。
- Step 1: Write both equations in the form ax + by = c. | 第一步：把两个方程写成 ax + by = c 的形式。
- Step 2: Multiply one or both equations to match coefficients. | 第二步：把一个或两个方程乘以适当的数，使系数匹配。
- Step 3: Add or subtract to eliminate one variable. | 第三步：相加或相减消去一个变量。
- Step 4: Solve for the remaining variable. | 第四步：解出剩下的变量。
- Step 5: Substitute back to find the other variable and check. | 第五步：回代求另一个变量并检验。
5. Solving by Substitution | 代入法求解
Substitution is especially useful when one equation already gives one variable in terms of the other, or when the system contains a quadratic equation. You replace one variable with an expression involving the other variable.
当一个方程已经用一个变量表示另一个变量时，或者方程组包含二次方程时，代入法特别有用。你用一个含另一个变量的表达式替换该变量。
Solve y = 2x + 1 and 3x + y = 11. Substitute y = 2x + 1 into the second equation.
解 y = 2x + 1 和 3x + y = 11。将 y = 2x + 1 代入第二个方程。
3x + (2x + 1) = 11
5x + 1 = 11
5x = 10
x = 2
= 5. Check in 3x + y = 11: 3(2) + 5 = 11, which is correct. The solution is x = 2, y = 5.
然后 y = 2(2) + 1 = 5。代入 3x + y = 11 检验：3(2) + 5 = 11，正确。解为 x = 2, y = 5。
, you usually need to rearrange it to = 0 and then factorise or use the quadratic formula.
当代入后得到二次方程时，通常需要整理成 = 0 的形式，然后因式分解或使用二次公式。
6. Graphical Interpretation | 图像法
Graphically, each linear equation represents a straight line. The solution of a simultaneous system is the point where the two lines intersect. If the lines are parallel, there is no solution. If they are the same line, there are infinitely many solutions.
从图像上看，每个线性方程表示一条直线。联立方程组的解就是两条直线的交点。如果两条直线平行，则无解。如果它们实际上是同一条直线，则有无穷多解。
ines cross at (3, 1), which matches the algebraic solution. A graph can give a quick estimate, but examiners usually require an exact algebraic answer.
对于方程组 2x + y = 7 和 x − y = 2，两条直线交于 (3, 1)，与代数解一致。图像可以给出快速估算，但考官通常要求精确的代数答案。
Intersection point = solution
交点 = 方程组的解
In IGCSE questions, a graph may be used to solve equations approximately, or to show why a system has no solution or infinitely many solutions.
在 IGCSE 题目中，图像可能用于近似求解，或用于说明为什么方程组无解或有无穷多解。
7. Linear vs Non-Linear Systems | 线性与非线性方程组
raight lines. A non-linear system may contain a quadratic, a circle or another curve. One common IGCSE case is one linear equation and one quadratic equation.
线性方程组包含 ax + by = c 形式的方程，图像是两条直线。非线性方程组可能包含二次方程、圆或其他曲线。IGCSE 中常见的一种情况是一个线性方程和一个二次方程。
and x² + y² = 13. From the first equation, y = 5 − x. Substitute into the second equation.
解 x + y = 5 和 x² + y² = 13。由第一个方程得 y = 5 − x，代入第二个方程。
x² + (5 − x)² = 13
x² + 25 − 10x + x² = 13
2x² − 10x + 12 = 0
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2 or x = 3
he solutions are (2, 3) and (3, 2). Always give all solution pairs for non-linear systems.
若 x = 2，则 y = 3。若 x = 3，则 y = 2。解为 (2, 3) 和 (3, 2)。对于非线性方程组，一定要给出所有解对。
8. Special Cases: No Solution and Infinite Solutions | 特殊情况：无解与无穷多解
Not every simultaneous system has exactly one solution. Two straight lines may be parallel, giving no solution. They may also be coincident, giving infinitely many solutions.
并非每个联立方程组都只有一个解。两条直线可能平行，从而无解；也可能重合，从而有无穷多解。
| Case | 情况 | Example | 例子 | Graph | 图像 | Solution | 解 |
|---|---|---|---|
| Unique solution | 唯一解 | 2x + y = 7, x − y = 2 | 2x + y = 7, x − y = 2 | Lines intersect once | 两条直线相交于一点 | x = 3, y = 1 | x = 3, y = 1 |
| No solution | 无解 | 2x + y = 5, 2x + y = 8 | 2x + y = 5, 2x + y = 8 | Parallel lines | 两条平行线 | No solution | 无解 |
| Infinite solutions | 无穷多解 | 2x + y = 5, 4x + 2y = 10 | 2x + y = 5, 4x + 2y = 10 | Same line | 同一条直线 | Infinitely many | 无穷多解 |
For no solution, the coefficients of x and y are proportional but the constants are not. For infinite solutions, both the coefficients and the constants are proportional.
无解时，x 和 y 的系数成比例，但常数项不成比例。无穷多解时，系数和常数项都成比例。
9. Word Problems and Modelling | 应用题与建模
Word problems require you to define variables, form two equations, solve them and interpret the answer in context. A common exam style involves tickets, ages, numbers or money.
应用题要求你定义变量、列出两个方程、求解，并在情境中解释答案。常见的考试风格涉及票务、年龄、数字或金钱问题。
A concert sells adult tickets for £8 and child tickets for £5.50. A total of 200 tickets are sold and the total income is £1350. Let x be adult tickets and y be child tickets.
一场音乐会成人票每张 £8，儿童票每张 £5.50。共售出 200 张票，总收入为 £1350。设 x 为成人票数，y 为儿童票数。
x + y = 200
8x + 5.5y = 1350
⟹ 16x + 11y = 2700
⟹ 5x = 500
⟹ x = 100, y = 100
So 100 adult tickets and
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