Solving Two Simultaneous Equations − One Linear and One Non-Linear | 解二元联立方程：线性方程与非线性方程

📚 Solving Two Simultaneous Equations − One Linear and One Non-Linear | 解二元联立方程：线性方程与非线性方程

In this revision guide, we will explore how to solve a pair of simultaneous equations where one equation is linear (of degree 1) and the other is non-linear (typically quadratic, of degree 2). This is a core skill in the Edexcel IGCSE Mathematics syllabus and appears frequently in both Paper 1 and Paper 2.

在本复习指南中，我们将探讨如何求解一对联立方程，其中一个是线性方程（一次方程），另一个是非线性方程（通常是二次方程）。这是 Edexcel IGCSE 数学大纲中的核心技能，在试卷1和试卷2中经常出现。

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

Simultaneous equations are a set of two or more equations that share the same variables. The solution is the set of values that satisfies all equations at the same time. When one equation is linear and the other is non-linear, we cannot simply use elimination by adding or subtracting — we must use substitution.

联立方程是共享相同变量的两个或多个方程所组成的集合。解就是同时满足所有方程的变量值。当一个方程是线性方程而另一个是非线性方程时，我们不能简单地通过加减来消元——我们必须使用代入法。

For example:

例如：

y = x² + 3x − 2 and y = 2x + 1

The first equation is quadratic (non-linear), and the second is linear. They can intersect at up to two points, meaning there may be two pairs of solutions.

第一个方程是二次的（非线性），第二个是线性的。它们最多可以相交于两个点，这意味着可能有两组解。

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2. Why Substitution Works | 为什么代入法有效

The substitution method works by replacing one variable in the non-linear equation with an expression from the linear equation. Since both equations equal y (or both equal the same variable), we can set them equal to each other and solve for x.

代入法的原理是用线性方程中的表达式替换非线性方程中的一个变量。由于两个方程都等于 y（或都等于同一个变量），我们可以令它们彼此相等，然后解出 x。

Steps to follow:

操作步骤如下：

• Step 1: Rearrange the linear equation to make one variable the subject (if not already done).
• 步骤1：将线性方程变形，使一个变量成为主项（如果尚未如此）。
• Step 2: Substitute this expression into the non-linear equation.
• 步骤2：将该表达式代入非线性方程。
• Step 3: Solve the resulting equation (usually a quadratic) to find the first variable.
• 步骤3：解得到的方程（通常是二次方程）以求出第一个变量。
• Step 4: Substitute each value back into the linear equation to find the corresponding second variable.
• 步骤4：将每个值代回线性方程，求出对应的第二个变量。

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3. Worked Example 1: Quadratic | 例题1：二次方程

Solve the simultaneous equations:

解下列联立方程：

y = x² − 4x + 5 and y = x + 1

Since both equations are already expressed with y as the subject, we set the right-hand sides equal:

由于两个方程都已将 y 作为主项，我们令等号右边的表达式相等：

x² − 4x + 5 = x + 1

Rearrange into standard form:

整理为标准形式：

x² − 5x + 4 = 0

Factorise:

因式分解：

(x − 1)(x − 4) = 0

So x = 1 or x = 4. Now substitute each value into the linear equation y = x + 1:

因此 x = 1 或 x = 4。将每个值代入线性方程 y = x + 1：

• When x = 1, y = 1 + 1 = 2
• 当 x = 1 时，y = 1 + 1 = 2
• When x = 4, y = 4 + 1 = 5
• 当 x = 4 时，y = 4 + 1 = 5

Solutions: (1, 2) and (4, 5)

Always check your answers by substituting into the original non-linear equation as well.

务必通过将答案代回原非线性方程来检验结果。

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4. The Graphical Meaning | 图像意义

Graphically, solving these equations means finding the intersection points between a straight line and a curve (often a parabola). A line and a parabola can:

从图像上看，求解这些方程意味着找出直线与曲线（通常是抛物线）的交点。直线与抛物线可以有：

Case 情况	Number of Solutions 解的个数
Line intersects curve at two points 直线与曲线相交于两点	2 solutions 两组解
Line touches curve at one point (tangent) 直线与曲线相切于一点	1 solution 一组解
Line does not touch curve 直线与曲线无交点	0 solutions 无解

The discriminant b² − 4ac of the resulting quadratic tells you which case you have:

所得二次方程的判别式 b² − 4ac 可以告诉你属于哪种情况：

• b² − 4ac > 0 → two distinct solutions
• b² − 4ac > 0 → 两个不同解
• b² − 4ac = 0 → one (repeated) solution
• b² − 4ac = 0 → 一个（重）解
• b² − 4ac < 0 → no real solutions
• b² − 4ac < 0 → 无实数解

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5. Worked Example 2: Rearranging the Linear Equation | 例题2：重组线性方程

Sometimes the linear equation is not given with y as the subject. Consider:

有时线性方程并未以 y 为主项给出。考虑：

y = 2x² − 3x and x + y = 4

Step 1: Rearrange the linear equation:

步骤1：重组

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