Solving Simultaneous Equations | 解二元一次方程组

📚 Solving Simultaneous Equations | 解二元一次方程组

Simultaneous equations are a fundamental topic in IGCSE Mathematics. When we have two unknown variables, a single equation is not enough to find a unique solution. We need at least two independent equations, which we solve together – hence the name “simultaneous equations”.

联立方程组是 IGCSE 数学中的一个基础考点。当我们面对两个未知数时，仅凭一个方程无法求出唯一解。我们需要至少两个独立的方程，并将它们联合求解，因此得名”联立方程组”。

1. Core Concepts | 核心概念

In IGCSE Mathematics, simultaneous equations involve two or more equations that share the same unknown variables. The most common case at Core level is a pair of linear equations with two unknowns, x and y.

在 IGCSE 数学中，联立方程组涉及两个或更多共享相同未知数的方程。在 Core 级别最常见的情况是含有两个未知数 x 和 y 的一对线性方程。

The word “simultaneous” means “at the same time”. When we solve such a system, we are looking for a single pair (x, y) that makes both equations true at the same time.

“simultaneous” 一词意为”同时”。当我们求解这样的方程组时，我们是在寻找一对能同时使两个方程都成立的 (x, y)。

A linear equation in two variables has the general form ax + by = c, where a, b and c are constants, and a and b are not both zero.

含两个变量的线性方程一般形式为 ax + by = c，其中 a、b、c 为常数，且 a 和 b 不能同时为零。

Each linear equation represents a straight line on the x-y plane.
每个线性方程在 xy 平面上表示一条直线。
The solution is the point where the two lines intersect.
方程组的解就是两条直线交点的坐标。

To find a unique solution, we need as many independent equations as there are unknowns – usually two equations for two unknowns.

要求出唯一解，方程的数目必须等于未知数的数目，通常是两个未知数需要两个方程。

2. The Substitution Method | 代入消元法

The substitution method is most useful when one equation already has a variable isolated, for example y = 2x + 1, or when rearrangement is easy.

代入消元法最适用于某个方程中已经有一个变量被单独表示的情况，例如 y = 2x + 1，或者方程很容易通过变形做到这一点时。

Solve: x + y = 10 and y = 2x + 1

Step 1: Substitute y = 2x + 1 into the first equation. Replace every y in x + y = 10 with 2x + 1.

第 1 步：将 y = 2x + 1 代入第一个方程。用 2x + 1 替换 x + y = 10 中所有的 y。

x + (2x + 1) = 10

Step 2: Simplify and solve for x.

第 2 步：化简并求出 x。

3x + 1 = 10 → 3x = 9 → x = 3

Step 3: Substitute x = 3 back into y = 2x + 1 to find y.

第 3 步：将 x = 3 代回 y = 2x + 1 求出 y。

y = 2 × 3 + 1 = 7

Step 4: Check your answer in both original equations: 3 + 7 = 10 ✓ and 7 = 2 × 3 + 1 ✓.

第 4 步：将答案代入两个原方程检验：3 + 7 = 10 ✓，7 = 2 × 3 + 1 ✓。

The solution is x = 3, y = 7. Always write the final answer clearly as a coordinate pair (3, 7).

解为 x = 3，y = 7。务必以坐标对 (3, 7) 的形式清晰写出最终答案。

3. The Elimination Method | 加减消元法

The elimination method, also called the addition–subtraction method, removes one variable by adding or subtracting the two equations.

加减消元法，也称加减法，通过将两个方程相加或相减来消去一个变量。

Solve: 3x + 2y = 13 and 2x − y = 4

Step 1: Make the coefficients of y the same in magnitude. Multiply the second equation by 2.

第 1 步：使 y 的系数绝对值相同。将第二个方程乘以 2。

4x − 2y = 8

Step 2: Add the two equations together so that the y terms cancel out.

第 2 步：将两个方程相加，使 y 项相互抵消。

(3x + 2y) + (4x − 2y) = 13 + 8 → 7x = 21 → x = 3

Step 3: Substitute x = 3 into either original equation. Using 2x − y = 4 gives 6 − y = 4.

第 3 步：将 x = 3 代入任意一个原方程。代入 2x − y = 4 得 6 − y = 4。

y = 2

Step 4: Check in the first equation: 3(3) + 2(2) = 9 + 4 = 13 ✓.

第 4 步：代入第一个方程检验：3(3) + 2(2) = 9 + 4 = 13 ✓。

The solution is x = 3, y = 2. Notice that when the signs of the terms are opposite, we add; when they are the same, we subtract.

解为 x = 3，y = 2。注意：当对应项的符号相反时用加法，符号相同时用减法。

4. Comparing the Methods | 两种方法的比较

Both the substitution and elimination methods are valid for any pair of linear simultaneous equations. Your choice depends on the form of the equations.

代入法和加减消元法对任何一对线性联立方程都有效。具体选择哪种方法取决于方程的形式。

Method	Best used when	Advantage
Substitution	One variable is already isolated	Direct and easy to follow
Elimination	Coefficients are simple multiples	Avoids fractions early on

方法选择：如果某个变量已经单独在等号一侧，优先考虑代入法；如果两个方程形式规整，优先考虑加减消元法。

In the exam, you may use either method. Examiners award method marks for correct substitution or manipulation, so always show your working line by line.

考试中两种方法均可使用。阅卷官会按正确的代入或变形步骤给方法分，因此务必逐行写出你的运算过程。

5. The Graphical Method | 图像法

Both equations can be plotted as straight lines on the same axes. The x- and y-coordinates of the intersection point give the solution of the simultaneous equations.

我们可以将两个方程在同一坐标系中画出直线，交点的 x、y 坐标就是联立方程组的解。

To draw a line from an equation such as x + y = 10, rearrange it into the form y = mx + c, then plot it using a table of values.

要画出 x + y = 10 这类方程的图像，先将其变形为 y = mx + c 的形式，再利用数值表取点作图。

x + y = 10 → y = −x + 10

When the line y = −x + 10 and the line y = 2x + 1 are drawn on the same grid, they intersect at the point (3, 7). Therefore x = 3 and y = 7.

当直线 y = −x + 10 与直线 y = 2x + 1 画在同一坐标网格中时，它们相交于点 (3, 7)，因此 x = 3，y = 7。

The graphical method is useful for estimating solutions and for checking a result, but it is less precise unless the solution has integer coordinates. In the exam, use an algebraic method for exact answers.

图像法适合估算解或检验结果，但除非解恰好是整数坐标，否则精度较低。考试中求精确答案时应使用代数方法。

6. Word Problems | 应用题

Simultaneous equations are often used to model real-life situations. The key skill is translating words into algebraic equations.

联立方程组常用于建立实际问题的数学模型。关键在于将文字信息转化为代数方程。

Three apples and two bananas cost $11. One apple and one banana cost $4. Find the price of each fruit.

Step 1: Define the variables: let a be the price of one apple and b be the price of one banana.

第 1 步：定义变量：设 a 为一个苹果的价格，b 为一个香蕉的价格。

3a + 2b = 11 and a + b = 4

Step 2: Rearrange the second equation: a = 4 − b. Substitute this into the first equation.

第 2 步：变形第二个方程：a = 4 − b，并将其代入第一个方程。

3(4 − b) + 2b = 11 → 12 − 3b + 2b = 11 → b = 1

Step 3: Substitute b = 1 back into a = 4 − b to get a = 3.

第 3 步：将 b = 1 代回 a = 4 − b，得 a = 3。

So one apple costs $3 and one banana costs $1. Always check: 3(3) + 2(1) = 11 ✓ and 3 + 1 = 4 ✓.

因此一个苹果 3 美元，一个香蕉 1 美元。务必检验：3(3) + 2(1) = 11 ✓，3 + 1 = 4 ✓。

In every word problem, clearly state what each variable represents before writing any equations.

在每一道应用题中，写出方程前务必先说明每个变量代表什么。

7. Special Cases | 特殊情况

Not every pair of linear equations has a unique solution. There are three possible situations you may encounter.

并非每一对线性方程都有唯一解。你可能会遇到三种情况。

Case 1: A unique solution. The two lines intersect at exactly one point, which happens when the equations are independent.

情况一：唯一解。两条直线相交于一点，此时方程为相互独立的方程。

Case 2: No solution. The lines are parallel and never meet, for example x + y = 5 and x + y = 8.

情况二：无解。两条直线平行且永不相交，例如 x + y = 5 和 x + y = 8。

x + y = 5 and x + y = 8 → no solution

When you try to solve this by elimination, you obtain a false statement such as 0 = 3, which indicates no solution.

当你尝试用加减消元法求解时，会得到一个错误的等式如 0 = 3，这说明方程组无解。

Case 3: Infinite solutions. The two equations represent the same line, for example y = 2x + 1 and 2y = 4x + 2. Every point on the line satisfies both equations.

情况三：无穷多解。两个方程表示同一条直线，例如 y = 2x + 1 和 2y = 4x + 2。直线上的每一个点都同时满足两个方程。

In the IGCSE exam, you are normally expected to identify these cases by inspection or by noticing the contradiction during elimination.

在 IGCSE 考试中，通常要求你通过观察或在消元过程中发现矛盾来识别这些情况。

8. Quadratic Simultaneous Equations | 联立二次方程组

In the Extended (Higher) IGCSE paper, you may be asked to solve a linear equation together with a quadratic equation. The substitution method is usually the most efficient approach.

在 IGCSE Extended（Higher）试卷中，你可能需要联立一个线性方程和一个二次方程求解。代入法通常是最有效的方法。

Solve: y = x + 2 and y = x²

Step 1: Substitute y = x² into y = x + 2, giving x² = x + 2.

第 1 步：将 y = x² 代入 y = x + 2，得到 x² = x + 2。

x² − x − 2 = 0 → (x − 2)(x + 1) = 0 → x = 2 or x = −1

Step 2: Find the corresponding y-values by substituting each x into the linear equation y = x + 2.

第 2 步：将每个 x 代入线性方程 y = x + 2，求出对应的 y 值。

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Solving Simultaneous Equations | 解二元一次方程组