Solving Simultaneous Equations | 解联立方程

📚 Solving Simultaneous Equations | 解联立方程

Simultaneous equations are one of the most frequently tested topics in the IGCSE Mathematics syllabus. They appear in both Paper 2 and Paper 4, and they form the foundation for many higher-level problems. In this revision article, we will explore the elimination method, the substitution method, the graphical method, word problems, and non-linear cases, together with common pitfalls and exam-style practice questions.

联立方程是 IGCSE 数学大纲中考查频率最高的知识点之一。它在 Paper 2 和 Paper 4 中都会出现，并且是许多高难度题目的基础。在本篇复习文章中，我们将系统讲解消元法、代入法、图像法、应用题以及非线性联立方程，并总结常见易错点和考试风格练习题。

---

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

A simultaneous equation system consists of two or more equations that share the same variables, usually x and y. To solve the system means to find the values of x and y that satisfy both equations at the same time.

联立方程组由两个或两个以上包含相同变量的方程组成，通常变量为 x 和 y。求解该方程组意味着找到同时满足所有方程的 x 和 y 的数值。

A linear equation in two variables has the general form:

ax + by = c

where a, b and c are constants. The graph of such an equation is a straight line. When we solve two linear equations simultaneously, we are effectively finding the point where the two lines intersect.

二元一次方程的一般形式为：

ax + by = c

其中 a、b、c 为常数。该方程的图像是一条直线。当我们联立求解两个一次方程时，实际上就是在寻找这两条直线的交点坐标。

There are three main methods of solution: elimination, substitution and graphical. Each has its own advantages, and you should be comfortable with all three.

求解联立方程主要有三种方法：消元法、代入法和图像法。每种方法各有优势，建议同学们熟练掌握这三种方法。

---

2. The Elimination Method | 消元法

The elimination method works by adding or subtracting the two equations to cancel out one of the variables. Here is a step-by-step example.

消元法的核心思想是通过两式相加或相减，消去其中一个未知数。我们通过以下例题逐步说明。

Example 1: Solve the system

2x + 3y = 12 (1)
x − y = 1 (2)

Step 1: Look at the coefficients. The coefficient of y in equation (1) is 3, and in equation (2) it is −1. Multiply equation (2) by 3 so that the y terms become opposites: 3x − 3y = 3.

第一步：观察各项系数。方程 (1) 中 y 的系数为 3，方程 (2) 中 y 的系数为 −1。将方程 (2) 乘以 3，使 y 项的系数互为相反数：3x − 3y = 3。

Step 2: Add the two equations to eliminate y:

2x + 3y + 3x − 3y = 12 + 3
5x = 15
x = 3

Step 3: Substitute x = 3 into equation (2): 3 − y = 1, so y = 2.

第三步：将 x = 3 代入方程 (2)：3 − y = 1，解得 y = 2。

Step 4: Check the solution in equation (1): 2(3) + 3(2) = 6 + 6 = 12 ✓. The solution is x = 3, y = 2.

第四步：将解代入方程 (1) 检验：2(3) + 3(2) = 6 + 6 = 12 ✓。因此解为 x = 3，y = 2。

Always multiply every term in the equation when scaling it, and pay close attention to signs when subtracting equations. A common error is forgetting to distribute the negative sign.

在对方程进行倍数放大时，务必逐项乘以该倍数；在相减时特别注意符号变化。常见错误是忘记分配负号。

---

3. The Substitution Method | 代入法

The substitution method involves rearranging one equation to make either x or y the subject, and then substituting this expression into the other equation.

代入法是将其中一个方程变形，使其成为用另一个变量表示某个变量的形式，然后代入另一个方程求解。

Example 2: Solve the same system using substitution.

2x + 3y = 12 (1)
x − y = 1 (2)

Step 1: Rearrange equation (2) to make x the subject: x = y + 1.

第一步：将方程 (2) 变形，用 y 表示 x：x = y + 1。

Step 2: Substitute x = y + 1 into equation (1):

2(y + 1) + 3y = 12
2y + 2 + 3y = 12
5y = 10
y = 2

Step 3: Substitute y = 2 back into x = y + 1, giving x = 3.

第三步：将 y = 2 代回 x = y + 1，得 x = 3。

The substitution method is especially useful when one equation already has a single variable on one side, or when dealing with non-linear equations, as we will see later.

代入法特别适用于某个方程中已经有一个变量单独位于一侧的情形，也适用于我们稍后要讲的非线性联立方程。

---

4. Comparing the Two Methods | 两种方法的比较

Feature | 特点	Elimination | 消元法	Substitution | 代入法
Best used when | 最适合场景	Coefficients match neatly or with simple multiplication | 系数能够方便地配平	One variable is already isolated | 某个变量已经被单独表示
Speed | 速度	Fast when coefficients are friendly | 系数友好时速度最快	Slightly slower for messy decimals | 遇到复杂小数时略慢
Error risk | 出错风险	Sign errors when subtracting | 相减时容易弄错符号	Bracket expansion errors | 去括号时容易出错
Non-linear systems | 非线性系统	Not usually suitable | 通常不适用	The preferred method | 首选方法

In the IGCSE examination, you are free to choose either method. However, you must show your working clearly: a correct answer without working may only earn partial marks. Develop the habit of writing every line of your manipulation.

在 IGCSE 考试中，你可以自由选择任一种方法。但是必须清晰展示计算过程：只有正确答案而没有过程，往往只能拿到部分分数。请大家养成逐步书写计算过程的好习惯。

---

5. The Graphical Method | 图像法

To solve simultaneous equations graphically, plot both lines on the same set of axes. The coordinates of the point where the two lines cross give the solution.

图像解法是在同一坐标系中画出两条直线，两直线交点的坐标就是联立方程的解。

For the system x + y = 5 and x − y = 1, we can rearrange each equation into the form y = mx + c:

对于方程组 x + y = 5 和 x − y = 1，我们可以将每个方程转化为 y = mx + c 的形式：

y = −x + 5 and y = x − 1

Drawing these two lines on the same graph, they intersect at the point (3, 2). Hence the solution is x = 3, y = 2.

在同一直角坐标系中画出这两条直线，它们的交点为 (3, 2)。因此方程组的解为 x = 3，y = 2。

Graphical methods are rarely required for exact answers in the non-calculator paper, but you must know how to choose a suitable scale, plot accurately, and read the intersection point correctly. In Paper 4, a graphical question may ask you to estimate the solution to one decimal place.

在不使用计算器的试卷中，图像法很少要求精确答案，但你必须知道如何选择合适刻度、准确画出图像并正确读取交点。在 Paper 4 中，图像题可能要求你估算精确到一位小数的解。

---

6. Setting Up Equations from Word Problems | 从应用题建立方程

Many students find word problems difficult because they struggle to translate English statements into mathematical equations. The key is to define variables clearly at the start and look for relationships such as “sum”, “difference”, “twice”, and “total”.

许多同学觉得应用题很难，因为难以将文字表述转化为数学方程。关键在于先明确设定变量，并寻找诸如”和”、”差”、”两倍”、”总共”等数量关系。

Example 3: Two numbers have a sum of 10 and a difference of 4. Find the numbers.

例 3：两个数之和为 10，之差为 4，求这两个数。

Let the larger number be x and the smaller number be y. Then:

x + y = 10 (1)
x − y = 4 (2)

Add the equations: 2x = 14, so x = 7. Substitute into (1): 7 + y = 10, so y = 3. The numbers are 7 and 3.

两式相加：2x = 14，故 x = 7。代入 (1)：7 + y = 10，得 y = 3。所以这两个数为 7 和 3。

Example 4: A shop sells pencils and pens. Two pencils and three pens cost $5.50, while four pencils and two pens cost $6.00. Find the price of one pencil and one pen.

例 4：某商店售卖铅笔和钢笔。2 支铅笔和 3 支钢笔共 5.50 美元，4 支铅笔和 2 支钢笔共 6.00 美元。求一支铅笔和一支钢笔的单价。

Let a pencil cost p dollars and a pen cost q dollars:

2p + 3q = 5.50 (1)
4p + 2q = 6.00 (2)

Multiply equation (1) by 2: 4p + 6q = 11.00. Subtract equation (2): 4q = 5.00, so q = 1.25. Substitute into (1): 2p + 3.75 = 5.50, so 2p = 1.75 and p = 0.875. A pencil costs $0.88 (to the nearest cent) and a pen costs $1.25.

将方程 (1) 乘以 2：4p + 6q = 11.00。减去方程 (2)：4q = 5.00，得 q = 1.25。代回 (1)：2p + 3.75 = 5.50，解得 2p = 1.75，p = 0.875。因此一支铅笔约 0.88 美元（精确到分），一支钢笔 1.25 美元。

When dealing with money, always check whether your final answer makes practical sense. A negative price signals an error somewhere.

在涉及金钱问题时，务必检查最终答案是否符合实际意义。如果出现负价格，说明计算过程中某个步骤有误。

---

7. Non-Linear Simultaneous Equations | 非线性联立方程

In the Extended IGCSE syllabus, you must also be able to solve one linear equation combined with one quadratic equation. The substitution method is always used here.

在 IGCSE 扩展课程中，你还需要掌握一个一次方程与一个二次方程联立的解法。此时应使用代入法。

Example 5: Solve the system

y = x² (1)
y = x + 2 (2)

Step 1: Since both expressions equal y, equate them: x² = x + 2.

第一步：因为两个表达式的值都等于 y，所以令它们相等：x² = x + 2。

Step 2: Rearrange to standard form: x² − x − 2 = 0.

第二步：整理为标准形式：x² − x − 2 = 0。

Step 3: Factorise: (x − 2)(x + 1) = 0, so x = 2 or x = −1.

第三步：因式分解：(x − 2)(x + 1) = 0，因此 x = 2 或 x = −1。

Step 4: Substitute each value back into y = x + 2:

• When x = 2, y = 4.
• When x = −1, y = 1.

第四步：将每个 x 值代回 y = x + 2：

• 当 x = 2 时，y = 4。
• 当 x = −1 时，y = 1。

The solutions are (2, 4) and (−1, 1). Graphically, these are the two intersection points between the parabola y = x² and the straight line y = x + 2.

解为 (2, 4) 和 (−1, 1)。在图像上，这两个坐标就是抛物线 y = x² 与直线 y = x + 2 的两个交点。

A quadratic equation can have two, one or zero real roots, meaning the line may cross the parabola twice, touch it once, or not meet it at all. In the “touch once” case, we say the line is tangent to the curve.

二次方程可能有两个、一个或零个实数根，这意味着直线与抛物线可能有两个交点、仅相切一次，或者完全没有交点。当只有一个交点时，我们称该直线为曲线的切线。

---

8. Common Pitfalls and How to Avoid Them | 常见易错点及应对策略

Pitfall 1: Sign errors in elimination. When subtracting one equation from another, every term changes sign. Write the subtracted equation in brackets, then expand carefully.

易错点 1：消元中的符号错误。当一个方程减去另一个方程时，被减方程的每一项符号都会改变。建议将所减方程写在括号内，再去括号展开。

Pitfall 2: Forgetting to check the answer. Always substitute your solution into both original equations. If it does not satisfy both, you have made a mistake.

易错点 2：忘记验算。务必将解分别代入两个原方程进行检验。如果不能满足两个方程，说明计算有误。

Pitfall 3: Mixing up x and y. When you find a value for x, do not rush to write it as y. Clearly label each variable in your final answer as an ordered pair (x, y).

易错点 3：混淆 x 和 y。当你求出 x 的值后，不要急于将其写成 y。最终答案应以有序数对 (x, y) 的形式明确写出。

Pitfall 4: Incorrect factorisation. In non-linear problems, an incorrect factorisation will produce wrong roots. If factorisation is difficult, you may use the quadratic formula:

x = [−b ± √(b² − 4ac)] / 2a

易错点 4：因式分解错误。在非线性问题中，因式分解错误会导致根求错。如果难以因式分解，可以使用求根公式：

x = [−b ± √(b² − 4ac)] / 2a

Pitfall 5: Rounding too early. In word problems involving money or measurement, do not round intermediate values. Round only at the very end.

易错点 5：过早四舍五入。在涉及金钱或测量的应用题中，不要对中间过程进行四舍五入，只在最后一步取近似值。

---

9. Exam-Style Practice Questions | 考试风格练习题

Try these questions in 15 minutes without a calculator. Solutions are given after each question, but attempt them yourself first.

请在 15 分钟内不使用计算器完成以下练习。每道题后附有答案，但请务必先独立尝试。

Q1. Solve the simultaneous equations:

3x + 2y = 8
2x + 5y = 9

Solution: Multiply the first equation by 2 and the second by 3:

6x + 4y = 16
6x + 15y = 27

Subtract: 11y = 11, so y = 1. Substitute into 3x + 2(1) = 8: 3x = 6, so x = 2. Answer: x = 2, y = 1.

解答：将第一个方程乘以 2，第二个方程乘以 3：

6x + 4y = 16
6x + 15y = 27

两式相减：11y = 11，故 y = 1。代入 3x + 2(1) = 8：3x = 6，得 x = 2。答案：x = 2，y = 1。

Q2. Solve the simultaneous equations:

2x + y = 7
x − 3y = −7

Solution: From the first equation, y = 7 − 2x. Substitute into the second:

x − 3(7 − 2x) = −7
x − 21 + 6x = −7
7x = 14
x = 2

Then y = 7 − 2(2) = 3. Answer: x = 2, y = 3.

解答：由第一个方程得 y = 7 − 2x。代入第二个方程：

x − 3(7 − 2x) = −7
x − 21 + 6x = −7
7x = 14
x = 2

于是 y = 7 − 2(2) = 3。答案：x = 2，y = 3。

Q3. The line y = x intersects the curve y = x² − 2 at two points. Find the coordinates of both intersection points.

Q3. 直线 y = x 与曲线 y = x² − 2 有两个交点，求这两个交点的坐标。

Solution: Equate the two expressions: x = x² − 2, so x² − x − 2 = 0. Factorise: (x − 2)(x + 1) = 0, giving x = 2 or x = −1. Substitute into y = x:

• When x = 2, y = 2.
• When x = −1, y = −1.

The points are (2, 2) and (−1, −1).

解答：令两式相等：x = x² − 2，整理得 x² − x − 2 = 0。因式分解：(x − 2)(x + 1) = 0，解得 x = 2 或 x = −1。代入 y = x：

• 当 x = 2 时，y = 2。
• 当 x = −1 时，y = −1。

交点坐标为 (2, 2) 和 (−1, −1)。

---

10. Final Revision Checklist | 考前最终检查清单

Before the exam, make sure you can confidently answer the following questions:

考试前，请确保你能自信地回答以下问题：

• Can I solve two linear equations using both elimination and substitution? | 我能否分别用消元法和代入法求解二元一次方程组？
• Can I rearrange equations into the form y = mx + c for graphing? | 我能否将方程变形为 y = mx + c 的形式以便作图？
• Can I interpret the point of intersection of two graphs as the solution? | 我能否将两条图像的交点理解为方程组的解？
• Can I set up equations from a word problem with clear variable definitions? | 我能否从应用题中通过清晰定义变量来建立方程？
• Can I solve a linear-quadratic system using substitution and factorisation? | 我能否用代入法和因式分解求解一次与二次方程组成的系统？
• Do I always check my answers in both original equations? | 我是否总是将答案代回两个原方程进行验算？

Mastering simultaneous equations is not only an exam requirement; it is also a key skill for future study of matrices, calculus and optimisation problems. The more you practise, the faster and more accurate you will become. Aim to complete a full set of practice questions at least three times before your exam day.

掌握联立方程不仅是考试要求，也是未来学习矩阵、微积分和优化问题的重要基础。练习越多，解题速度和准确率就越高。建议在考试前至少完整练习三套相关题目。

Published by TutorHao | 数学 Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）