Simultaneous Equations | 联立方程

📚 Simultaneous Equations | 联立方程

Simultaneous equations are a cornerstone of IGCSE Mathematics, appearing in both Paper 1 (non-calculator) and Paper 2 (calculator). This guide covers the substitution method, the elimination method, graphical interpretation, and practical problem-solving — everything you need to secure full marks.

联立方程是 IGCSE 数学的核心内容，在 Paper 1（不能使用计算器）和 Paper 2（可使用计算器）中都会出现。本指南涵盖代入消元法、加减消元法、图像解释以及实际应用题的解法，助你拿到满分。

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

A linear equation in two variables, such as 2x + y = 7, is not enough on its own to find unique values of x and y. In fact, it has infinitely many solutions: if x = 0, then y = 7; if x = 1, then y = 5, and so on. Every solution (x, y) is a point on a straight line.

二元一次方程，如 2x + y = 7，仅凭它本身无法求出 x 和 y 的唯一值。事实上，它有无数多组解：若 x = 0，则 y = 7；若 x = 1，则 y = 5，以此类推。每一组解 (x, y) 都是直线上的一点。

When we combine two linear equations that share the same variables, we call them simultaneous equations (or a system of equations). The solution is the ordered pair (x, y) that makes both equations true at the same time.

当我们把两个含有相同变量的线性方程放在一起时，就称为联立方程组。方程组的解是能同时满足两个方程的坐标对 (x, y)。

2. Why Two Equations Are Needed | 为什么需要两个方程

A single equation with two unknowns gives infinitely many solutions. Adding a second equation restricts the solution to a single point — provided the two lines are not parallel and not identical. This is why a system of “two equations, two unknowns” is generally solvable.

一个含有两个未知数的方程给出无数多组解。增加第二个方程后，解就被限制为唯一一点——前提是两条直线不平行且不重合。这就是”两个方程、两个未知数”通常可以求解的原因。

Geometrically, the solution is the intersection point of the two lines. For the system x + y = 10 and x − y = 4, the lines cross at (7, 3).

从几何角度看，解就是两条直线的交点。对于方程组 x + y = 10 和 x − y = 4，两条直线相交于点 (7, 3)。

3. The Substitution Method | 代入消元法

The substitution method has three steps. Step 1: rearrange one equation so that one variable is isolated, for example x = 10 − y. Step 2: substitute this expression into the other equation. Step 3: solve the resulting one-variable equation, then back-substitute to find the other variable.

代入消元法分三步：第一步，变换其中一个方程，使一个变量单独出现在等号一边，例如 x = 10 − y；第二步，把这个表达式代入另一个方程；第三步，解出得到的一元一次方程，再回代求出另一个变量。

Worked example: solve x + y = 10 and 2x − y = 5.

实例：求解 x + y = 10 和 2x − y = 5。

From the first equation, y = 10 − x. Substituting into the second:

由第一个方程得 y = 10 − x。代入第二个方程：

2x − (10 − x) = 5

Simplify: 2x − 10 + x = 5, so 3x = 15, hence x = 5. Then y = 10 − 5 = 5. The solution is (5, 5). Verify: 2(5) − 5 = 5. ✓

化简得 2x − 10 + x = 5，即 3x = 15，所以 x = 5。进而 y = 10 − 5 = 5。解为 (5, 5)。验证：2(5) − 5 =

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