Solving Simultaneous Equations | 解联立方程

📚 Solving Simultaneous Equations | 解联立方程

Simultaneous equations are a fundamental topic in IGCSE Mathematics. A system of simultaneous equations consists of two or more equations that share variables, and we seek values of these variables that satisfy all equations at the same time.

联立方程是 IGCSE 数学中的一个基础考点。联立方程组由两个或多个共享变量的方程组成，我们需要找到同时满足所有方程的变量值。

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

In IGCSE Maths, you will most often encounter two linear equations with two unknown variables, typically x and y. For example:

在 IGCSE 数学中，最常见的题型是两个含有两个未知变量（通常为 x 和 y）的线性方程。例如：

3x + 2y = 12
x – y = 1

The solution is the pair of values (x, y) that makes both equations true simultaneously. There are three main methods: elimination, substitution, and graphical.

解就是同时使两个方程成立的 (x, y) 数值对。主要有三种解法：消元法、代入法和图像法。

2. The Elimination Method | 消元法

The elimination method involves adding or subtracting the equations to eliminate one variable. This is often the fastest method when the coefficients are simple.

消元法通过将两个方程相加或相减来消去一个变量。当系数比较简单时，这通常是最快的方法。

Step 1: Make the coefficients of one variable the same in both equations by multiplying if necessary.

步骤 1：如有必要，通过乘法使两个方程中某个变量的系数相同。

Step 2: Add or subtract the equations to eliminate that variable.

步骤 2：将两式相加或相减以消去该变量。

Step 3: Solve the resulting linear equation for the remaining variable.

步骤 3：解所得的一元线性方程，求出一个变量的值。

Step 4: Substitute back to find the other variable.

步骤 4：代回原方程求出另一个变量的值。

Consider the example above:

考虑上面的例子：

3x + 2y = 12 … (1)
x – y = 1 … (2)

Multiply equation (2) by 2 to match the y coefficient:

将方程 (2) 乘以 2，使 y 的系数匹配：

2x – 2y = 2 … (3)

Now add equations (1) and (3):

现在将 (1) 和 (3) 相加：

3x + 2y + 2x – 2y = 12 + 2
5x = 14
x = 2.8

Substitute x = 2.8 into equation (2):

将 x = 2.8 代入方程 (2)：

2.8 – y = 1
y = 1.8

So the solution is x = 2.8, y = 1.8.

因此解为 x = 2.8，y = 1.8。

3. The Substitution Method | 代入法

The substitution method is particularly useful when one variable has a coefficient of 1. You solve one equation for one variable, then substitute that expression into the other equation.

当某个变量的系数为 1 时，代入法尤其好用。先从一个方程中解出一个变量，然后将表达式代入另一个方程。

Using the same example:

使用相同的例子：

x – y = 1 → x = y + 1

Substitute x = y + 1 into 3x + 2y = 12:

将 x = y + 1 代入 3x + 2y = 12：

3(y + 1) + 2y = 12
3y + 3 + 2y = 12
5y = 9
y = 1.8

Then x = y + 1 = 1.8 + 1 = 2.8. The same solution is obtained.

然后 x = y + 1 = 1.8 + 1 = 2.8。得到相同的解。

4. The Graphical Method | 图像法

Graphically, each linear equation represents a straight line on the Cartesian plane. The solution to the simultaneous equations is the point where the two lines intersect.

在图像上，每个线性方程都代表笛卡尔平面上的一条直线。联立方程的解就是两条直线的交点。

To solve graphically:

用图像法求解的步骤：

Rearrange each equation into the form y = mx + c.
将每个方程改写为 y = mx + c 的形式。
Plot both lines on the same axes.
在同一坐标轴中画出两条直线。
Read off the coordinates of the intersection point.
读出交点的坐标。

For our example, the lines y = 6 – 1.5x and y = x – 1 intersect at the point (2.8, 1.8).

在我们的例子中，直线 y = 6 – 1.5x 和 y = x – 1 相交于点 (2.8, 1.8)。

Note: The graphical method may give approximate answers if the intersection point does not fall on grid lines. Examiners expect algebraic methods for exact values.

注意：如果交点不在网格线上，图像法可能会给出近似答案。考试中期望使用代数方法获得精确值。

5. Linear and Quadratic Systems | 线性与二次方程组

At IGCSE level, you may also need to solve a system where one equation is linear and the other is quadratic. The substitution method is the recommended approach.

在 IGCSE 阶段，你可能还需要解一个线性方程与一个二次方程组成的方程组。推荐使用代入法。

Consider the following example:

考虑以下例子：

y = x² – 4x + 5 … (1)
y = 2x – 1 … (2)

Since both expressions are equal to y, set them equal to each other:

由于两个式子都等于 y，令它们相等：

x² – 4x + 5 = 2x – 1
x² – 6x + 6 = 0

Solve using the quadratic formula:

使用二次公式求解：

x = [6 ± √(36 – 24)] ÷ 2 = [6 ± √12] ÷ 2 = 3 ± √3

Then y = 2x – 1 gives y = 5 ± 2√3. So the two intersection points are (3 + √3, 5 + 2√3) and (3 – √3, 5 – 2√3).

然后代入 y = 2x – 1 得到 y = 5 ± 2√3。因此两个交点为 (3 + √3, 5 + 2√3) 和 (3 – √3, 5 – 2√3)。

6. Word Problems | 应用题

Simultaneous equations are often presented in problem context. Let us look at a classic example.

联立方程常常以实际问题的形式出现。让我们看一个经典例子。

Problem: The sum of the ages of a father and his son is 50. The father is 4 times as old as the son. Find their ages.

题目：父子年龄之和为 50 岁。父亲的年龄是儿子的 4 倍。求他们的年龄。

Let f be the father’s age and s be the son’s age.

设 f 为父亲的年龄，s 为儿子的年龄。

f + s = 50
f = 4s

Substitute f = 4s into the first equation:

将 f = 4s 代入第一个方程：

4s + s = 50
5s = 50
s = 10

Then f = 4 × 10 = 40. The father is 40 years old and the son is 10 years old.

然后 f = 4 × 10 = 40。父亲 40 岁，儿子 10 岁。

7. Common Mistakes to Avoid | 常见错误与避坑指南

Here are the most frequent errors students make:

以下是学生最常犯的错误：

Forgetting to multiply every term when scaling an equation.
在对方程进行倍数缩放时，忘记乘以每一项。
Sign errors when subtracting equations, especially when dealing with negative numbers.
在方程相减时出现符号错误，尤其是处理负数时。
Substituting into the same equation instead of the other one, which gives an identity like 0 = 0.
代入了同一个方程而不是另一个方程，导致出现 0 = 0 这样的恒等式。
Not checking the solution in both original equations.
没有把解代回两个原方程进行验证。
When solving linear-quadratic systems, forgetting to find both x and y values for both solutions.
在解线性-二次方程组时，忘记为两个解分别求出对应的 x 和 y。

8. Practice Questions | 练习巩固

Try these problems on your own before checking with a teacher or calculator:

请独立完成以下练习，再向老师或计算器核对答案：

1. Solve: 2x + 3y = 7 and x – y = 1.

练习 1：解方程组 2x + 3y = 7 和 x – y = 1。

2. Solve: y = x² – 3 and y = x + 1.

练习 2：解方程组 y = x² – 3 和 y = x + 1。

3. The perimeter of a rectangle is 44 cm. The length is 2 cm more than twice the width. Find the dimensions.

练习 3：一个长方形的周长为 44 cm。长比宽的 2 倍多 2 cm。求长方形的长与宽。

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Solving Simultaneous Equations | 解联立方程