Solving Simultaneous Equations: Linear-Quadratic Systems | 解方程组：线性与二次方程联立

📚 Solving Simultaneous Equations: Linear-Quadratic Systems | 解方程组：线性与二次方程联立

Simultaneous equations are a core topic in IGCSE Mathematics, and while most students feel comfortable solving two linear equations, the exam often raises the difficulty by pairing a linear equation with a quadratic one. This article explains the two standard methods — elimination and substitution — and then focuses on linear-quadratic systems, including how to predict the number of solutions using the discriminant.

联立方程是 IGCSE 数学的核心考点。大多数学生对两个线性方程组成的方程组比较熟练，但考试中常常会提高难度，把一个线性方程与一个二次方程结合起来。本文将讲解两种标准方法——消元法和代入法——并重点分析线性-二次方程组，包括如何利用判别式预判解的个数。

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

A simultaneous equation is a set of equations with the same unknown variables. To solve them, we look for values of the variables that satisfy all equations at the same time. At IGCSE the unknowns are usually called x and y, and the solution is either an ordered pair (x, y) or, for some practical problems, a single value for each variable.

联立方程就是包含相同未知数的一组方程。解联立方程，就是找到能同时满足所有方程的未知数值。在 IGCSE 中，未知数通常用 x 和 y 表示，解可以写成有序数对 (x, y)，在某些应用题中也可能是一个具体数值。

Two linear equations → one intersection point (one solution).
One linear + one quadratic → up to two intersection points.
No solution sometimes, when the graphs do not meet.

两个线性方程 → 一个交点（唯一解）。
一个线性 + 一个二次 → 最多两个交点。
当图像不相交时，也可能无解。

2. Solving Linear Systems by Elimination | 用消元法解线性方程组

The elimination method works by adding or subtracting the equations so that one variable cancels out. The key is to make the coefficients of x (or y) the same, or exact opposites.

消元法的核心思想是通过两式相加或相减，使其中一个未知数抵消。关键在于让 x（或 y）的系数相等或互为相反数。

Worked example | 例题：

2x + y = 7
x − y = 2

Here the y-coefficients are +1 and −1, so adding the two equations eliminates y:

这里 y 的系数分别为 +1 和 −1，两式相加即可消去 y：

2x + y + x − y = 7 + 2
3x = 9
x = 3

Substitute x = 3 into x − y = 2: 3 − y = 2, so y = 1. The solution is (3, 1).

把 x = 3 代入 x − y = 2：3 − y = 2，所以 y = 1。解为 (3, 1)。

3. The Substitution Method | 代入法

The substitution method involves making one variable the subject of an equation and replacing it in the other equation. This method is essential for linear-quadratic systems because elimination is rarely useful when one equation contains x².

代入法是指把一个方程中的某个变量表示出来，再代入另一个方程。在处理线性-二次方程组时，这种方法尤其重要，因为当一个方程中含有 x² 时，消元法通常并不好用。

Worked example | 例题：

y = 2x − 1
x + y = 5

Since y is already the subject in the first equation, substitute 2x − 1 for y in the second equation:

第一个方程中 y 已经是主项，把 2x − 1 代入第二个方程中的 y：

x + (2x − 1) = 5
3x = 6
x = 2

Then y = 2(2) − 1 = 3. The solution is (2, 3).

于是 y = 2(2) − 1 = 3。解为 (2, 3)。

4. Linear–Quadratic Systems: When Parabolas Meet Lines | 线性-二次联立：抛物线与直线相交

A linear-quadratic system typically looks like this: one equation is linear, y = mx + c, and the other is quadratic, y = ax² + bx + c. Geometrically, the line may cut the parabola twice, touch it once as a tangent, or miss it completely.

线性-二次方程组通常形如：一个方程是线性的 y = mx + c，另一个是二次的 y = ax² + bx + c。从几何上看，直线可能与抛物线有两个交点、相切于一个点，也可能完全不相交。

The general solving strategy:

一般解题策略如下：

Substitute the linear expression for y into the quadratic equation.
Rearrange to obtain a quadratic equation in x: ax² + bx + c = 0.
Solve for x (factorisation, completing the square, or the quadratic formula).
For each x, substitute back to find the corresponding y.

把线性方程中 y 的表达式代入二次方程。
整理得到关于 x 的一元二次方程：ax² + bx + c = 0。
解出 x（因式分解、配方法或求根公式）。
对每个 x，代回原式求出对应的 y。

5. Worked Example: Step-by-Step | 例题精解

Solve the system | 解方程组：

y = x² − 2x + 3
y = x + 1

Step 1: Substitution. Since both expressions equal y, set them equal to each other:

第一步：代入。因为两个式子都等于 y，令它们相等：

x² − 2x + 3 = x + 1

Step 2: Rearrange into a quadratic equation:

第二步：整理成标准一元二次方程：

x² − 3x + 2 = 0

Step 3: Factorise:

第三步：因式分解：

(x − 1)(x − 2) = 0

So x = 1 or x = 2.

所以 x = 1 或 x = 2。

Step 4: Substitute back into y = x + 1:

第四步：代回 y = x + 1：

x = 1 → y = 2
x = 2 → y = 3

The solutions are (1, 2) and (2, 3). Always present both pairs, since a linear-quadratic system can produce two points of intersection.

解为 (1, 2) 和 (2, 3)。两组解都要完整写出，因为线性-二次方程组可能产生两个交点。

6. The Discriminant: Predicting Solutions | 判别式：预判解的个数

After substituting and rearranging, you get a quadratic equation ax² + bx + c = 0. Its discriminant, Δ = b² − 4ac, tells you how many solutions the system has without fully solving it.

代入并整理后，得到 ax² + bx + c = 0。它的判别式 Δ = b² − 4ac 能在不完全求解的情况下，判断方程组解的个数。

Discriminant Δ	Number of Solutions	Geometric Meaning
Δ > 0	Two distinct solutions	Line cuts the parabola at two points
Δ = 0	Exactly one solution	Line is tangent to the parabola
Δ < 0	No real solutions	Line and parabola do not meet

判别式 Δ	解的个数	几何意义
Δ > 0	两个不同解	直线与抛物线相交于两点
Δ = 0	恰好一个解	直线与抛物线相切
Δ < 0	没有实数解	直线与抛物线不相交

Example: y = x² − 4x + 4 and y = 2x − 5 give x² − 6x + 9 = 0, with Δ = (−6)² − 4(1)(9) = 0. The line is a tangent, so there is one solution. Example: y = x² − 4x + 5 and y = x − 2 give x² − 5x + 7 = 0, with Δ = 25 − 28 = −3 < 0, so there is no solution.

例如：y = x² − 4x + 4 与 y = 2x − 5 得到 x² − 6x + 9 = 0，Δ = (−6)² − 4(1)(9) = 0，直线是切线，因此只有一个解。再如：y = x² − 4x + 5 与 y = x − 2 得到 x² − 5x + 7 = 0，Δ = 25 − 28 = −3 < 0，因此无解。

7. Graphical Interpretation | 图像意义

Sketching the graphs can confirm the algebraic result. A positive quadratic curve opens upward; a line with slope m approaches it from one side. Two solutions appear as two crossing points; one solution appears at the touching point; zero solutions mean the graphs stay apart.

画图可以帮助验证代数结果。开口向上的抛物线与一条斜率为 m 的直线，可能相交于两点、相切于一点，或完全不相交。两点对应两个解，切点对应一个解，分离则无解。

When working graphically, always label:

画图时务必标出：

The coordinates of any intersection points.
The y-intercept of each curve.
The vertex of the parabola, usually found at x = −b/2a.

所有交点的坐标。
每条曲线的 y 截距。
抛物线的顶点，通常在 x = −b/2a 处。

Remember that the algebraic method gives exact answers, while a sketch only gives an estimate. Use the sketch to check, but rely on algebra for the final answer.

请注意，代数方法给出精确答案，而草图只能给出近似值。用草图来检查，但最终答案要以代数计算为准。

8. Common Mistakes to Avoid | 常见错误

Many students lose marks not because they cannot solve, but because of small avoidable errors. Here are the most frequent mistakes in linear-quadratic systems:

很多学生丢分不是因为不会解，而是因为一些可以避免的小错误。以下是线性-二次方程组中最常见的错误：

Forgetting to find y: after solving for x, always substitute back to find y for every x value.
Sign errors when expanding: be careful with (x − 2)² = x² − 4x + 4.
Losing the quadratic coefficient: keep the equation in the form ax² + bx + c = 0 before applying the quadratic formula.
Incorrect substitution: when substituting a bracketed expression, keep the brackets until you expand.
Writing only one solution: if the discriminant is positive, two solution pairs must be given.

忘记求 y：求出 x 后，必须代回原式求每个 x 对应的 y。
展开时符号错误：注意 (x − 2)² = x² − 4x + 4。
丢失二次项系数：使用求根公式前，必须先化成 ax² + bx + c = 0。
代入错误：代入带括号的表达式时，先保留括号再展开。
只写一个解：当判别式为正时，必须写出两个解对。

9. Exam-Style Practice Questions | 考试风格练习

Try these questions on your own, then check the answers below.

请先独立完成以下练习，再对照下方答案。

Question 1 | 第 1 题：Solve the simultaneous equations | 解方程组：

y = x² − 2x − 3
y = x + 1

Question 2 | 第 2 题：Solve the simultaneous equations | 解方程组：

y = 3x − 2
y = x² − 2x + 2

Question 3 | 第 3 题：Find how many solutions this system has, without solving it fully: | 不解完全程，判断方程组有几个解：

y = x² + 4x + 5
y = 2x + 1

Answers | 参考答案：

Q1: x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0 → x = 4, y = 5 and x = −1, y = 0. Solutions: (4, 5) and (−1, 0).

第 1 题：x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0 → x = 4, y = 5；x = −1, y = 0。解为 (4, 5) 和 (−1, 0)。

Q2: x² − 5x + 4 = 0 → (x − 1)(x − 4) = 0 → x = 1, y = 1 and x = 4, y = 10. Solutions: (1, 1) and (4, 10).

第 2 题：x² − 5x + 4 = 0 → (x − 1)(x − 4) = 0 → x = 1, y = 1；x = 4, y = 10。解为 (1, 1) 和 (4, 10)。

Q3: x² + 4x + 5 = 2x + 1 → x² + 2x + 4 = 0; Δ = 4 − 16 = −12 < 0, so there are no real solutions.

第 3 题：x² + 4x + 5 = 2x + 1 → x² + 2x + 4 = 0；Δ = 4 − 16 = −12 < 0，因此没有实数解。

10. Summary | 总结

For any simultaneous equation problem, decide between elimination and substitution. For linear-quadratic systems, substitution is the reliable path. After substituting, rearrange into ax² + bx + c = 0 and solve. Use the discriminant to check how many solutions to expect, and always find the matching y-values. With careful algebraic habits, these questions become predictable and easy marks in the exam.

无论面对哪种联立方程题，都要先决定用消元法还是代入法。对于线性-二次方程组，代入法是最稳妥的路径。代入后整理成 ax² + bx + c = 0 并求解。利用判别式来预判解的个数，并始终求出对应的 y 值。只要养成严谨的代数习惯，这类题目就会成为考试中可预测的送分题。

Keep practicing with a variety of coefficient values until the process becomes automatic.

请坚持用不同系数的题目练习，直到整个解题流程完全熟练为止。

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Solving Simultaneous Equations: Linear-Quadratic Systems | 解方程组：线性与二次方程联立