📚 Example 3.3.2: Solving Linear and Quadratic Simultaneous Equations | 例题 3.3.2：线性与二次联立方程的求解
In this article, we work through Example 3.3.2 from the AQA A-Level Mathematics specification. The problem asks you to solve a pair of simultaneous equations in which one equation is linear and the other is quadratic. We will explain every step in both English and Chinese so that you can confidently apply this essential exam technique.
在本文中，我们将详细解析 AQA A-Level 数学大纲中的例题 3.3.2。该问题要求我们求解一个由线性方程和二次方程组成的联立方程组。我们会用中英文逐一解释每一个步骤，帮助你自信地运用这一重要的考试技巧。
1. The Problem | 问题陈述
Solve the following simultaneous equations.
求解下列联立方程。
y = 2x + 1
x² + y² = 10
The first equation is linear, and the second is quadratic. We need to find all ordered pairs (x, y) that satisfy both equations at the same time.
第一个方程是线性方程，第二个方程是二次方程。我们需要找到所有同时满足两个方程的有序数对 (x, y)。
2. Understanding the Equations | 理解方程结构
n y = 2x + 1 has degree 1, so its graph is a straight line with slope 2 and y-intercept 1.
线性方程 y = 2x + 1 的次数为 1，因此其图像是一条斜率为 2、y 截距为 1 的直线。
uation x² + y² = 10 has degree 2 and represents a circle centred at the origin with radius √10.
二次方程 x² + y² = 10 的次数为 2，表示一个以原点为圆心、半径为 √10 的圆。
A solution to this system is any point where the line and the circle intersect.
该方程组的一个解就是直线与圆的任意一个交点。
3. Why Use Substitution? | 为什么使用代入法？
Because the first equation already expresses y in terms of x, we can substitute 2x + 1 for y in the second equation.
因为第一个方程已经用 x 表示了 y，所以我们可以将第二个方程中的 y 替换为 2x + 1。
This substitution converts the system into a single quadratic equation in one variable, which we already know how to solve.
这种代入将方程组转化为只含一个变量的二次方程，而我们已经知道如何求解这类方程。
Substitution is usually the most reliable method when one equation is linear and the other is quadratic.
当一个方程是线性、另一个是二次时，代入法通常是最可靠的方法。
4. Step 1: Substitute | 第一步：代入
Replace y in the quadratic equation with 2x + 1.
将二次方程中的 y 替换为 2x + 1。
x² + (2x + 1)² = 10
Now we have a single equation in x only.
现在我们得到了一个只含 x 的方程。
5. Step 2: Expand and Rearrange | 第二步：展开并整理
Expand the squared bracket.
展开平方项。
(2x + 1)² = 4x² + 4x + 1
Substitute this back into the equation.
将其代回原方程。
x² + 4x² + 4x + 1 = 10
Combine like terms and subtract 10 from both sides.
合并同类项，并在方程两边同时减去 10。
5x² + 4x − 9 = 0
This is the quadratic equation we need to solve.
这就是我们需要求解的二次方程。
6. Step 3: Solve the Quadratic | 第三步：解二次方程
Factorise the quadratic expression.
对二次表达式进行因式分解。
5x² + 4x − 9 = (5x + 9)(x − 1)
Apply the zero product property: if the product is zero, at least one factor must be zero.
应用零乘积性质：如果乘积为零，那么至少有一个因式为零。
5x + 9 = 0 or x − 1 = 0
So the two x solutions are:
因此两个 x 解为：
x = −9/5 or x = 1
7. Step 4: Find the Corresponding y Values | 第四步：求对应的 y 值
Use the linear equation y = 2x + 1 to find the y value for each x solution.
利用线性方程 y = 2x + 1，求出每个 x 解对应的 y 值。
For x = 1:
当 x = 1 时：
y = 2(1) + 1 = 3
For x = −9/5:
当 x = −9/5 时：
y = 2(−9/5) + 1 = −18/5 + 5/5 = −13/5
Thus the two solution pairs are (1, 3) and (−9/5, −13/5).
因此两组解分别为 (1, 3) 和 (−9/5, −13/5)。
8. Verification | 验证答案
Check the pair (1, 3).
验证解 (1, 3)。
y = 2(1) + 1 = 3 ✓
x² + y² = 1² + 3² = 1 + 9 = 10 ✓
Check the pair (−9/5, −13/5).
验证解 (−9/5, −13/5)。
y = 2(−9/5) + 1 = −13/5 ✓
x² + y² = (−9/5)² + (−13/5)² = 81/25 + 169/25 = 250/25 = 10 ✓
Both pairs satisfy the original equations, so the solutions are correct.
两组解都满足原方程，因此解得正确。
9. Graphical Interpretation | 图形解释
The line y = 2x + 1 and the circle x² + y² = 10 intersect at exactly two points.
直线 y = 2x + 1 与圆 x² + y² = 10 恰好相交于两个点。
The discriminant of the resulting quadratic 5x² + 4x − 9 = 0 is:
所得二次方程 5x² + 4x − 9 = 0 的判别式为：
Δ = 4² − 4 × 5 × (−9) = 16 + 180 = 196
Since Δ > 0, there are two distinct real roots, meaning two distinct intersection points.
因为 Δ > 0，所以存在两个不同的实数根，即两个不同的交点。
line would be tangent to the circle. If Δ < 0, the line would not meet the circle at all.
如果 Δ = 0，则直线与圆相切；如果 Δ < 0，则直线与圆没有交点。
10. Common Mistakes and Exam Tips | 常见错误与考试建议
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Mistake: Forgetting to expand (2x + 1)² correctly.
错误：忘记正确展开 (2x + 1)²。
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Mistake: Forgetting to subtract 10 from both sides before solving.
错误：在求解之前忘记在方程两边同时减去 10。
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Mistake: Finding x values but not substituting back to find the corresponding y values.
错误：求出 x 值后，没有代回原方程求对应的 y 值。
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Tip: Always write both solution pairs clearly as coordinates.
建议：务必以坐标形式清晰地写出两组解。
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Tip: Check your answers by substituting both pairs into the original quadratic equation.
建议：将两组解分别代入原二次方程进行验算。
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Tip: If the quadratic does not factorise easily, use the quadratic formula.
建议：如果二次式不容易因式分解，可使用二次求根公式。
11. Practice Question | 练习题目
Try this AQA-style question on your own.
请独立尝试下面这道 AQA 风格题目。
y = x − 2
x² + y² = 10
− 2 into the second equation. You should obtain a quadratic in x which factorises simply.
提示：将 y = x − 2 代入第二个方程，你会得到一个可以简单因式分解的二次方程。
The solutions are (3, 1) and (−1, −3).
答案为 (3, 1) 和 (−1, −3)。
12. Conclusion | 结语
Example 3.3.2 shows a standard method for solving simultaneous equations where one equation is linear and the other is quadratic. The key steps are substitution, expanding, rearranging, solving the quadratic, and then finding the matching y values.
例题 3.3.2 展示了求解一个线性方程与一个二次方程组成的联立方程组的标准方法。关键步骤包括代入、展开、整理、求解二次方程，然后求出对应的 y 值。
Always verify your final answers, and remember that intercepts can be read directly from the linear equation because it gives a unique y value for each x.
始终记得验算最终答案，并注意线性方程每个 x 只对应唯一的 y 值，因此可以直接由线性方程求出 y。
With regular practice, these questions become quick marks in the AQA examination.
通过定期练习，这类题目在 AQA 考试中会成为容易得分的题型。
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