← 返回资讯屏轩教育 · 行业资讯

Solving Simultaneous Equations | 联立方程组求解

2026-08-30 00:20 · 本文部分内容由AI辅助创作

📚 Solving Simultaneous Equations | 联立方程组求解

Simultaneous equations are a core topic in the Edexcel IGCSE Mathematics syllabus (4MA1). They appear in both Paper 1 (non-calculator) and Paper 2 (calculator), at Foundation and Higher Tier. Mastering the elimination method, the substitution method, and the graphical method will help you secure marks quickly and accurately. This article walks through each method step by step, with worked examples, common pitfalls, and exam-style practice questions.

联立方程组是爱德思 IGCSE 数学大纲（4MA1）的核心考点，在 Paper 1（非计算器）和 Paper 2（计算器）中都会出现，涵盖基础级和高级。掌握消元法、代入法和图像法，可以帮助你快速而准确地拿分。本文将一步步讲解每种方法，并配有例题、常见错误和模拟练习题。

1. What Are Simultaneous Equations? | 什么是联立方程组

A pair of simultaneous equations contains two unknowns, usually x and y, and both equations must be true at the same time. The solution is a single ordered pair (x, y) that satisfies both equations. For example:

联立方程组包含两个未知数，通常为 x 和 y，并且两个方程必须同时成立。解是一个有序数对 (x, y)，它必须同时满足两个方程。例如：

x + y = 10
x − y = 4

ir satisfies both lines. The key idea is that we are looking for the unique point where the two equations agree.

这里数对 (7, 3) 对两个方程都成立：7 + 3 = 10 且 7 − 3 = 4。没有其他数对能同时满足这两个方程。核心思想是：我们要找到两个方程共同成立的唯一交点。

2. Solving by Elimination | 消元法

The elimination method works by adding or subtracting the equations so that one variable disappears. Follow these steps:

消元法的原理是通过两个方程相加或相减，使其中一个未知数被消去。步骤如下：

Example 1 — no scaling needed:

例 1——无需扩大倍数：

2x + y = 8
x − y = 1

hence x = 3. Substitute x = 3 into x − y = 1: 3 − y = 1, so y = 2. The solution is (3, 2).

将两个方程相加以消去 y：(2x + x) + (y − y) = 8 + 1，即 3x = 9，所以 x = 3。把 x = 3 代入 x − y = 1：3 − y = 1，所以 y = 2。解为 (3, 2)。

Example 2 — scaling required:

例 2——需要扩大倍数：

3x + 2y = 14
2x − y = 7

28, hence x = 4. Substitute x = 4 into 2x − y = 7: 8 − y = 7, so y = 1. Solution: (4, 1).

这里 y 的系数分别是 2 和 −1。将第二个方程整体乘以 2：4x − 2y = 14。现在相加：(3x + 4x) + (2y − 2y) = 14 + 14，即 7x = 28，所以 x = 4。把 x = 4 代入 2x − y = 7：8 − y = 7，所以 y = 1。解为 (4, 1)。

3. Solving by Substitution | 代入法

The substitution method is especially useful when one equation already has y or x as the subject. It also prepares you for solving one linear and one quadratic equation together.

代入法特别适合其中一个方程已经写成 y = 或 x = 的形式。它也是解“一个线性方程 + 一个二次方程”的基础。

Example:

例题：

y = 2x − 1
3x + 2y = 19

ce x = 3. Then y = 2(3) − 1 = 5. Solution: (3, 5). Always check: 3(3) + 2(5) = 9 + 10 = 19 ✓.

由于 y 已经是主语形式，将第二个方程中的 y 替换为 2x − 1：3x + 2(2x − 1) = 19。展开：3x + 4x − 2 = 19，即 7x = 21，所以 x = 3。于是 y = 2(3) − 1 = 5。解为 (3, 5)。务必检验：3(3) + 2(5) = 9 + 10 = 19 ✓。

equation first. For example, from 2x + y = 9 you can write y = 9 − 2x and then substitute.

如果两个方程都没有写成 y = 或 x = 的形式，可以先改写其中一个方程。例如由 2x + y = 9 可得 y = 9 − 2x，然后代入即可。

4. One Linear and One Quadratic | 一个线性方程与一个二次方程

This is a Higher Tier requirement in Edexcel IGCSE Mathematics. The method is always substitution: use the linear equation to express one variable, then substitute into the quadratic equation. This produces a quadratic in one variable, which you solve by factorising or using the quadratic formula.

这是爱德思 IGCSE 数学高级别的要求。解法永远是代入法：先用线性方程表示一个未知数，再代入二次方程。这样会得到一个一元二次方程，通过因式分解或求根公式来解。

Example:

例题：

y = x + 3
y = x² + 2x − 3

side zero: 0 = x² + x − 6, which factorises as (x + 3)(x − 2) = 0. Hence x = −3 or x = 2.

把 y = x + 3 代入二次方程：x + 3 = x² + 2x − 3。移项使一边为零：0 = x² + x − 6，因式分解得 (x + 3)(x − 2) = 0。因此 x = −3 或 x = 2。

. When x = 2, y = 5. The solutions are (−3, 0) and (2, 5). Both pairs must be written down — missing one solution is a common mark-loser.

再用线性方程 y = x + 3 求出对应的 y 值。当 x = −3 时，y = 0；当 x = 2 时，y = 5。所以解为 (−3, 0) 和 (2, 5)。两组解都必须写出来——漏写一组是最常见的失分点。

广告 / Sponsored

🛒
TI-Nspire CX II CAS 图形计算器 A-Level数学
¥1298
📱 去京东购买
京东精选

5. Graphical Method | 图像法

For two linear equations, plot both straight lines on the same axes. The coordinates of the point where the lines cross give the solution. For a line and a curve, the intersection points are the solutions.

对于两个线性方程，在同一坐标系中画出两条直线。两条直线的交点坐标就是方程组的解。对于直线与曲线，交点就是解。

Example: Solve x + y = 6 and 2x − y = 3 graphically.

例题：用图像法解 x + y = 6 和 2x − y = 3。

First rearrange: y = 6 − x and y = 2x − 3. Then plot a table of values.

先变形：y = 6 − x 和 y = 2x − 3，然后列表取值。

x 0 1 2 3 4 5
y = 6 − x 6 5 4 3 2 1
y = 2x − 3 −3 −1 1 3 5 7

, 3). In an exam, always draw the lines carefully and mark the intersection point clearly.

从表格可以看出，两条直线在 x = 3，y = 3 处相交。所以解为 (3, 3)。在考试中，一定要画准直线，并清楚标出交点。

6. Forming Equations from Word Problems | 由应用题列方程

Edexcel IGCSE often tests your ability to convert a word problem into a pair of simultaneous equations. Define your variables first, then translate each sentence into an equation.

爱德思 IGCSE 经常考查将文字题转化为联立方程组的能力。先定义未知数，再把每句话翻译成方程。

Example: "3 pens and 2 notebooks cost £13. 5 pens and 2 notebooks cost £19. Find the cost of one pen and one notebook."

例题：“3 支笔和 2 个笔记本共 13 英镑。5 支笔和 2 个笔记本共 19 英镑。求一支笔和一个笔记本各自的价格。”

Let p = price of one pen and n = price of one notebook. Then:

设 p 为一支笔的价格，n 为一个笔记本的价格。于是：

3p + 2n = 13
5p + 2n = 19

= 6, hence p = 3. Substitute p = 3 into 3p + 2n = 13: 9 + 2n = 13, so n = 2. One pen costs £3 and one notebook costs £2.

用第二个方程减去第一个方程：(5p − 3p) + (2n − 2n) = 19 − 13，即 2p = 6，所以 p = 3。将 p = 3 代入 3p + 2n = 13：9 + 2n = 13，所以 n = 2。一支笔 3 英镑，一个笔记本 2 英镑。

7. Common Mistakes | 常见错误

Many students lose easy marks on simultaneous equations through avoidable slips. Watch out for these:

很多学生在联立方程题目上因为可以避免的失误而丢分。请注意以下几点：

8. Exam Tips | 考试技巧

To maximise your marks in the Edexcel IGCSE exam, use a systematic approach:

在爱德思 IGCSE 考试中，要想拿满分数，请使用系统化的解题流程：