Simultaneous Equations | 联立方程

📚 Simultaneous Equations | 联立方程

Simultaneous equations are two or more equations that share the same unknown variables. In IGCSE Mathematics, you will most often work with two linear equations in two unknowns, x and y, and occasionally with one linear and one quadratic equation. The key idea is that a single solution must satisfy every equation in the system at the same time.

联立方程是指共享相同未知数的两个或多个方程。在 IGCSE 数学中，最常见的是含两个未知数 x 和 y 的两个线性方程，偶尔也会遇到一个线性方程与一个二次方程组成的方程组。关键思想是：一个解必须同时满足方程组中的每一个方程。

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

A linear equation in two variables, such as 2x + y = 7, has infinitely many solutions. When you pair it with another line, such as x − y = 2, the pair usually has exactly one common solution.

形如 2x + y = 7 的二元一次方程有无数个解。当它与另一个方程如 x − y = 2 配对时，这一对通常只有一个公共解。

An ordered pair (x, y) is a solution if it makes both equations true when substituted in.

有序数对 (x, y) 如果在代入后能使两个方程都成立，就是方程组的解。
Graphically, the solution is the point where the two straight lines intersect.

从图象上看，解就是两条直线交点的坐标。

2x + y = 7 and x − y = 2 → solution: x = 3, y = 1

Check: 2(3) + 1 = 7 ✓ and 3 − 1 = 2 ✓.

检验：2(3) + 1 = 7 ✓，且 3 − 1 = 2 ✓。

2. The Elimination Method | 消元法

The elimination method removes one variable by adding or subtracting the two equations. It works best when the coefficients of one variable are the same or opposite.

消元法通过将两个方程相加或相减来消去一个变量。当一个变量的系数相同或互为相反数时，此法最为高效。

Example: Solve 3x + y = 11 and 2x − y = 4.

例：解方程组 3x + y = 11 和 2x − y = 4。

Step 1: The y-coefficients are +1 and −1, so add the equations to eliminate y.

第一步：y 的系数是 +1 和 −1，将两个方程相加以消去 y。

3x + y = 11
+ 2x − y = 4
5x = 15

Step 2: Solve for x: x = 15 ÷ 5 = 3.

第二步：解得 x = 15 ÷ 5 = 3。

Step 3: Substitute x = 3 into either original equation. Using 3x + y = 11 gives 9 + y = 11, so y = 2.

第三步：将 x = 3 代入任意一个原方程。代入 3x + y = 11 得 9 + y = 11，所以 y = 2。

Solution: x = 3, y = 2

If the coefficients are not equal, multiply one or both equations by a constant first, then eliminate.

如果系数不相等，应先将一个或两个方程乘以适当的常数，再进行消元。

3. The Substitution Method | 代入法

The substitution method is used when one equation has a variable with coefficient 1, or when the system is not easy to align for elimination.

代入法适用于某个变量的系数为 1 的情况，或者当两个方程不方便对齐进行消元时使用。

Example: Solve y = 2x + 1 and 3x + 2y = 23.

例：解方程组 y = 2x + 1 和 3x + 2y = 23。

Step 1: Substitute y = 2x + 1 into the second equation.

第一步：将 y = 2x + 1 代入第二个方程。

3x + 2(2x + 1) = 23

Step 2: Expand and solve: 3x + 4x + 2 = 23 → 7x = 21 → x = 3.

第二步：展开并求解：3x + 4x + 2 = 23 → 7x = 21 → x = 3。

Step 3: Back-substitute: y = 2(3) + 1 = 7.

第三步：回代：y = 2(3) + 1 = 7。

Solution: x = 3, y = 7

4. The Graphical Method | 图象法

You can solve a pair of linear equations by drawing both lines on the same set of axes. The coordinates of the intersection point give the solution.

你可以把两条直线画在同一坐标平面上来解方程组。交点坐标就是方程组的解。

Rearrange each equation into the form y = mx + c before plotting.

先将每个方程改写为 y = mx + c 的形式，再作图。
Draw both lines accurately using a ruler; read the intersection coordinates carefully.

用直尺精确画出两条直线，仔细读出交点坐标。
The graphical method is useful for checking answers, but it relies on accurate drawing.

图象法适合用来检验答案，但它依赖于绘图的准确性。

y = x + 1 and y = −2x + 7 intersect at (2, 3)

5. Solving Linear–Quadratic Systems | 解线性–二次方程组

IGCSE Extended candidates must solve a linear equation and a quadratic equation simultaneously. The quadratic may be given as x² + y² = r² or as y = ax² + bx + c.

IGCSE 扩展课程要求考生解一个线性方程与一个二次方程组成的方程组。二次方程可能以 x² + y² = r² 或 y = ax² + bx + c 的形式给出。

Example: Solve y = x + 1 and x² + y² = 25.

例：解方程组 y = x + 1 和 x² + y² = 25。

Step 1: Substitute y = x + 1 into x² + y² = 25.

第一步：将 y = x + 1 代入 x² + y² = 25。

x² + (x + 1)² = 25

Step 2: Expand and rearrange: x² + x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0 → divide by 2: x² + x − 12 = 0.

第二步：展开并整理：x² + x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0 → 两边除以 2：x² + x − 12 = 0。

Step 3: Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3.

第三步：因式分解：(x + 4)(x − 3) = 0，所以 x = −4 或 x = 3。

Step 4: Substitute each x-value into y = x + 1: when x = −4, y = −3; when x = 3, y = 4.

第四步：将每个 x 值代入 y = x + 1：当 x = −4 时，y = −3；当 x = 3 时，y = 4。

Solutions: (−4, −3) and (3, 4)

6. Interpreting Solutions | 解的几何意义

A linear–quadratic system can have 0, 1, or 2 solutions. This corresponds to how the line meets the curve.

线性–二次方程组可能有 0 个、1 个或 2 个解，这对应着直线与曲线的位置关系。

Number of solutions 解的个数	Geometric meaning 几何意义
2	The line cuts the curve at two distinct points. 直线与曲线相交于两个不同点。
1	The line is tangent to the curve. 直线与曲线相切于一点。
0	The line misses the curve completely. 直线与曲线没有交点。

You can check which case you have by looking at the discriminant b² − 4ac of the resulting quadratic equation. If it is positive, there are two solutions; if zero, one solution; if negative, no real solutions.

你可以通过所得二次方程的判别式 b² − 4ac 来判断属于哪种情况。若判别式为正，有两个解；为零，有一个解；为负，则无实数解。

7. Word Problems | 应用题

Real-world problems can be translated into a pair of equations. Define two variables, form two equations from the given conditions, then solve.

实际问题可以转化为一对方程。先设定两个变量，再根据题目条件列出两个方程，最后求解。

Example: A shop sells pens and notebooks. Two pens and one notebook cost $11. One pen and three notebooks cost $18. Find the price of one pen and one notebook.

例：一家商店出售笔和笔记本。两支笔和一本笔记本共 11 美元；一支笔和三本笔记本共 18 美元。求一支笔和一本笔记本的价格。

Let the price of a pen be p dollars and a notebook be n dollars.

设一支笔的价格为 p 美元，一本笔记本的价格为 n 美元。

2p + n = 11
p + 3n = 18

From the first equation, n = 11 − 2p. Substitute into the second: p + 3(11 − 2p) = 18 → p + 33 − 6p = 18 → −5p = −15 → p = 3. Then n = 11 − 2(3) = 5.

由第一个方程得 n = 11 − 2p。代入第二个方程：p + 3(11 − 2p) = 18 → p + 33 − 6p = 18 → −5p = −15 → p = 3。于是 n = 11 − 2(3) = 5。

A pen costs $3 and a notebook costs $5. 一支笔 3 美元，一本笔记本 5 美元。

8. Common Mistakes to Avoid | 常见错误

Many marks are lost through small but avoidable errors. Here are the most frequent ones.

许多分数都因为小而可以避免的错误而丢失。以下是最高频的错误。

Sign errors when subtracting equations: always write down each line carefully.

相减方程时符号出错：务必一步一步仔细写下每一步。
Forgetting to multiply every term when scaling an equation.

乘以常数时遗漏某些项。
Only solving for one variable and not substituting back to find the other.

只求出一个变量，没有回代求另一个变量。
When solving a linear–quadratic system, losing one of the two solutions by cancelling a factor.

解线性–二次方程组时，因约去公因式而丢掉一个解。
Not checking the final answer in both original equations.

没有把最终答案代回两个原方程检验。

9. Exam-Style Worked Example | 考试例题精解

The following question is typical of IGCSE Paper 2 or Paper 4.

下面的题目是 IGCSE 卷二或卷四的典型题型。

Question: Solve the simultaneous equations:

题目：解下列联立方程组：

4x + 3y = 10
2x − y = 0

From 2x − y = 0, we have y = 2x. Substitute into 4x + 3y = 10:

由 2x − y = 0 可得 y = 2x。代入 4x + 3y = 10：

4x + 3(2x) = 10 → 4x + 6x = 10 → 10x = 10 → x = 1

Then y = 2(1) = 2. Check in the first equation: 4(1) + 3(2) = 4 + 6 = 10 ✓.

于是 y = 2(1) = 2。代入第一个方程检验：4(1) + 3(2) = 4 + 6 = 10 ✓。

Final answer: x = 1, y = 2

Always present your substitution steps clearly; IGCSE mark schemes reward method marks even if a small arithmetic slip occurs.

一定要清晰地写出代入步骤；IGCSE 评分标准即使在出现小的计算错误时也会给方法分。

10. Quick Revision Checklist | 快速复习清单

Use this checklist before your exam to confirm you are ready.

考试前请用这张清单确认自己已经准备好。

I can solve two linear equations by elimination. 我能用消元法解两个线性方程。
I can solve two linear equations by substitution. 我能用代入法解两个线性方程。
I can interpret the solution as the intersection point of two lines. 我能把解理解为两条直线的交点。
I can solve one linear and one quadratic equation together. 我能联立求解一个线性方程和一个二次方程。
I know how to use the discriminant to predict the number of solutions. 我知道如何用判别式判断解的个数。
I can translate word problems into simultaneous equations. 我能把应用题转化为联立方程。
I always check my answers in the original equations. 我总会在原方程中检验答案。

Mastering simultaneous equations unlocks many other topics, from coordinate geometry to calculus. Keep practising until the methods become automatic.

掌握联立方程会为许多其他主题铺平道路，从坐标几何到微积分。请持续练习，直到这些方法变得得心应手。

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Simultaneous Equations | 联立方程