T-K-1026: Mastering Quadratic-Linear Simultaneous Equations | T-K-1026：掌握二次-线性联立方程

📚 T-K-1026: Mastering Quadratic-Linear Simultaneous Equations | T-K-1026：掌握二次-线性联立方程

Welcome to this TutorHao revision session on T-K-1026. This question type is a favourite in IGCSE Mathematics Extended papers because it combines algebraic manipulation with quadratic solving. By the end of this article, you will be able to solve this class of problem confidently and avoid common pitfalls.

欢迎来到 TutorHao 的 T-K-1026 复习专题。这一题型是 IGCSE 数学（Extended 课程）中的常见考点，因为它结合了代数化简与二次方程求解。学完这篇文章后，你将能够自信地解决这类问题，并避开常见误区。

1. Understanding T-K-1026 | 理解 T-K-1026

T-K-1026 is our internal code for a classic simultaneous equation problem: one quadratic equation and one linear equation. In your exam, you might see a pair of equations such as y = x² – 4x + 5 and y = 2x – 3. The word ‘simultaneous’ simply means the two equations are true at the same time.

T-K-1026 是我们内部对这一经典联立方程题的编号：一个二次方程和一个一次方程。在实际考试中，你可能会看到类似 y = x² – 4x + 5 和 y = 2x – 3 这样的一对方程。”联立”意味着这两个方程在同一个时刻同时成立。

You are asked to find the values of x and y that satisfy both equations. Because one equation is quadratic, there are usually two pairs of solutions, but sometimes only one or none.

题目要求你找到同时满足两个方程的 x 和 y 的值。因为其中一个是二次方程，通常会有两组解，但有时可能只有一组解，甚至无解。

2. The Problem Statement | 题目陈述

Here is the exact style of T-K-1026 we will solve today:

下面就是我们今天要解决的 T-K-1026 题型示例：

y = x² – 4x + 5

y = 2x – 3

Notice that in this example, the first equation is quadratic and the second is linear. A linear equation has x to the power of 1, while a quadratic has x².

注意：在这个例子中，第一个方程是二次的，第二个方程是一次的。一次方程中 x 的指数是 1，而二次方程中含有 x²。

3. Why Do We Equate Them? | 为什么令两方程相等？

Both equations are written in the form y = something. Since the left-hand side of each is y, the right-hand sides must also be equal to each other. This is the key step: we equate the two expressions.

两个方程都以 y = 某个表达式的形式给出。既然每个方程的左边都是 y，那么右边的表达式彼此也必须相等。这就是关键步骤：令两个表达式相等。

Equating gives us a single equation in x, which we can solve using factorisation, the quadratic formula, or completing the square.

令它们相等后，我们就得到了一个只含 x 的方程，可以用因式分解、求根公式或配方法来解。

4. Step-by-Step Solution | 逐步解答

Step 1: Equate the two expressions.

第一步：令两个表达式相等。

x² – 4x + 5 = 2x – 3

Step 2: Rearrange the equation to make one side zero.

第二步：移项，使等式一边为零。

x² – 4x + 5 – 2x + 3 = 0

x² – 6x + 8 = 0

Step 3: Factorise the quadratic.

第三步：对二次式进行因式分解。

(x – 2)(x – 4) = 0

This tells us that either x – 2 = 0 or x – 4 = 0.

这说明要么 x – 2 = 0，要么 x – 4 = 0。

Step 4: Solve for x.

第四步：解出 x。

x = 2 or x = 4

Step 5: Substitute each x value back into the linear equation to find y.

第五步：把每个 x 值代回一次方程，求出 y。

Using y = 2x – 3:

使用 y = 2x – 3：

If x = 2, then y = 2(2) – 3 = 1.
如果 x = 2，则 y = 2(2) – 3 = 1。
If x = 4, then y = 2(4) – 3 = 5.
如果 x = 4，则 y = 2(4) – 3 = 5。

Therefore the solutions are (2, 1) and (4, 5).

因此方程组的解为 (2, 1) 和 (4, 5)。

5. Checking Your Answers | 检验答案

Always check your answers in both original equations. This catches arithmetic errors.

一定要把答案代入原方程检验，这样可以发现计算错误。

For (2, 1):

对于 (2, 1)：

In y = x² – 4x + 5: 1 = (2)² – 4(2) + 5 = 4 – 8 + 5 = 1 ✅
在 y = x² – 4x + 5 中：1 = (2)² – 4(2) + 5 = 4 – 8 + 5 = 1 ✅
In y = 2x – 3: 1 = 2(2) – 3 = 1 ✅
在 y = 2x – 3 中：1 = 2(2) – 3 = 1 ✅

For (4, 5):

对于 (4, 5)：

In y = x² – 4x + 5: 5 = (4)² – 4(4) + 5 = 16 – 16 + 5 = 5 ✅
在 y = x² – 4x + 5 中：5 = (4)² – 4(4) + 5 = 16 – 16 + 5 = 5 ✅
In y = 2x – 3: 5 = 2(4) – 3 = 5 ✅
在 y = 2x – 3 中：5 = 2(4) – 3 = 5 ✅

6. Graphical Interpretation | 图像解释

A quadratic equation graphs as a curve called a parabola. A linear equation graphs as a straight line. The solutions of the simultaneous equations are the coordinates of the points where the parabola and the line intersect.

二次方程对应的图像是一条抛物线；一次方程对应的图像是一条直线。联立方程的解就是抛物线与直线的交点坐标。

In our example, the parabola y = x² – 4x + 5 and the line y = 2x – 3 meet at exactly two points: (2, 1) and (4, 5). This is why we find two solutions.

在我们的例子中，抛物线 y = x² – 4x + 5 和直线 y = 2x – 3 恰好交于两点：(2, 1) 和 (4, 5)。这就是为什么我们得到两组解。

If the line is tangent to the parabola, there is only one solution. If the line does not touch the parabola, there are no real solutions.

如果直线与抛物线相切，那么只有一组解；如果直线与抛物线没有交点，则没有实数解。

7. Common Mistakes | 常见错误

Here are the most frequent errors students make in T-K-1026 questions:

以下是学生在 T-K-1026 题中最常犯的错误：

Wrong rearrangement: forgetting to change signs when moving terms across the equals sign.
移项错误：把项移到等号另一边时忘记变号。
Incomplete factorisation: writing (x – 2)(x – 4) = 0 and immediately jumping to x = 2 only.
因式分解不完整：写出 (x – 2)(x – 4) = 0 后只得到 x = 2，漏掉了 x = 4。
Using the wrong equation for substitution: substituting x back into the quadratic instead of the linear equation. This often gives extra incorrect solutions.
代错方程：把 x 代回二次方程而不是一次方程，这样容易产生多余的错误解。
No check: not verifying the final pairs in the original equations.
不检验：没有把最终结果代回原方程验证。

8. Exam Tips | 考试技巧

Here is a simple strategy to maximise your marks on T-K-1026 in the real exam:

这里有一个简单的策略，帮助你在真实考试中应对 T-K-1026 拿到高分：

Write down the two equations clearly.
清晰写出两个方程。
Equate the two expressions and show your rearrangement step by step.
令两个表达式相等，并逐步展示移项过程。
Try to factorise first; if it does not factorise, use the quadratic formula.
优先尝试因式分解；若无法分解，则使用求根公式。
Always substitute x back into the linear equation to find y.
求 y 时一定要代回一次方程。
Give your final answers as coordinate pairs, e.g. (2, 1) and (4, 5).
用坐标对写出最终答案，例如 (2, 1) 和 (4, 5)。
If the question is worth 5 marks, your working should show equating, rearranging, factorising, solving, and substituting.
如果题目分值是 5 分，你的解题过程应包含：令相等、移项、因式分解、求解、代回。

9. Practice Question | 练习题目

Try the following question on your own before looking at the solution.

请先独立完成下面这道题，再看答案。

y = x² – 2x – 3

y = x – 5

Find both pairs of solutions. Show all your working.

求两组解，并展示完整的解题过程。

10. Solution to Practice | 练习题解答

Step 1: Equate the two expressions.

第一步：令两个表达式相等。

x² – 2x – 3 = x – 5

Step 2: Rearrange.

第二步：移项。

x² – 2x – 3 – x + 5 = 0

x² – 3x + 2 = 0

Step 3: Factorise.

第三步：因式分解。

(x – 1)(x – 2) = 0

Step 4: Solve for x.

第四步：解出 x。

x = 1 or x = 2

Step 5: Find y using y = x – 5.

第五步：用 y = x – 5 求 y。

If x = 1, then y = 1 – 5 = -4. Solution: (1, -4).
如果 x = 1，则 y = 1 – 5 = -4。解为：(1, -4)。
If x = 2, then y = 2 – 5 = -3. Solution: (2, -3).
如果 x = 2，则 y = 2 – 5 = -3。解为：(2, -3)。

Quick check: For (1, -4), the quadratic gives 1 – 2 – 3 = -4 ✅. For (2, -3), the quadratic gives 4 – 4 – 3 = -3 ✅.

快速检验：对于 (1, -4)，二次式给出 1 – 2 – 3 = -4 ✅。对于 (2, -3)，二次式给出 4 – 4 – 3 = -3 ✅。

Therefore the final answers are (1, -4) and (2, -3).

因此最终答案为 (1, -4) 和 (2, -3)。

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T-K-1026: Mastering Quadratic-Linear Simultaneous Equations | T-K-1026：掌握二次-线性联立方程