📚 Topic 249: Simultaneous Equations | Edexcel IGCSE Maths | 主题 249：联立方程组 | 爱德思 IGCSE 数学
Simultaneous equations are a fundamental topic in the Edexcel IGCSE Mathematics syllabus, appearing in both Foundation and Higher tier papers. The ability to find values that satisfy more than one equation at the same time is essential for solving real-world problems, from calculating costs to analysing motion. This article, Topic 249, will guide you through solving linear simultaneous equations using elimination, substitution and graphical methods, as well as tackling one linear and one quadratic equation - a key skill for Higher tier students. We will also look at typical exam questions, mark-scheme habits and common mistakes to help you achieve top marks.
联立方程组是爱德思 IGCSE 数学课程的基础主题，在基础卷和进阶卷中均有考查。能够找到同时满足多个方程的值，对于解决从成本计算到运动分析等实际问题至关重要。本文（主题 249）将带您学习使用消元法、代入法和图像法求解线性联立方程组，以及处理一个线性方程与一个二次方程的问题——这是进阶卷学生必须掌握的关键技能。我们还将分析典型考题、评分习惯和常见错误，帮助您取得高分。
1. What Are Simultaneous Equations? | 什么是联立方程组？
Simultaneous equations are two (or more) equations that share the same variables and must all be true at the same time. In IGCSE, you will usually work with two equations involving two unknowns, almost always called x and y. The word "simultaneous" simply means "at the same moment": a single value of x and a single value of y must satisfy every equation in the system at once.
联立方程组是指两个（或更多）共享同一组未知数的方程，它们必须同时成立。在 IGCSE 中，您通常处理的是含两个未知数的两个方程，未知数几乎总是记作 x 和 y。"联立"一词的意思就是"同时"：同一个 x 值和同一个 y 值必须一次性满足系统中每一个方程。
m below. The pair x = 6, y = 4 works because both statements become true at the same time.
例如，看下面这个方程组。数对 x = 6、y = 4 成立，因为两个等式同时变为真命题。
+ y = 10 and x − y = 2 → x = 6, y = 4
ecomes 4 − 6 = −2, which is false. The order matters, and both coordinates must be stated.
因此，方程组的解不是一个单独的数，而是一个有序数对 (x, y)。如果把两个数交换，写成 x = 4、y = 6，第二个方程就变成 4 − 6 = −2，不成立。顺序很重要，两个解值都必须写出。
- Linear equation: highest power of any variable is 1, e.g. 3x + 2y = 12. | 线性方程：任何未知数的最高次数都是 1，例如 3x + 2y = 12。
- Quadratic equation: contains a squared term such as x² or xy. | 二次方程：含有 x² 或 xy 之类的平方项。
- Solution set: the value or values that satisfy all equations simultaneously. | 解集：同时满足所有方程的那个（些）值。
2. Why This Topic Matters in the Exam | 为什么这个主题在考试中很重要
Simultaneous equations sit inside the Algebra strand of the Edexcel IGCSE Mathematics specification. At Foundation tier, you are expected to solve two linear equations in two unknowns using elimination, substitution or a graph. At Higher tier, the same skills are examined plus the harder case of one linear and one quadratic equation, where you must produce two pairs of solutions.
联立方程组属于爱德思 IGCSE 数学大纲中的代数模块。在基础卷中，您需要用消元法、代入法或图像法求解两个未知数的两个线性方程。在进阶卷中，除同样的技能外，还会考查更难的情形：一个线性方程与一个二次方程，此时必须给出两组解。
The topic also appears disguised inside other questions. A question about the meeting point of two taxis, the break-even point of a small business, or the point where a thrown ball crosses a wall is really a simultaneous-equations question in words. Recognising the structure is half the battle; the other half is choosing an efficient method and showing clean working.
这个主题还会伪装成其他题型出现。关于两辆出租车相遇点、小店盈亏平衡点，或抛出的球飞越墙面的位置等问题，本质上都是用文字包装的联立方程组题。识别出结构是成功的一半，另一半是选择高效方法并写出清晰的步骤。
Typical mark weighting is 3 to 5 marks: usually 1 mark for a correct method, 1 mark for one correct variable, and further marks for the second variable or for correct interpretation in context. Method marks are available even if your arithmetic slips, so always write down your working line by line.
典型分值通常为 3 至 5 分：一般 1 分给正确方法，1 分给求出的第一个未知数，其余分数给第二个未知数或结合情境的正确解释。即使算术出错，方法分依然可以拿到，所以务必逐行写出过程。
3. The Elimination Method | 消元法
Elimination removes one variable by adding or subtracting the two equations. The golden rule is: if the coefficients of one variable are the same, subtract the equations; if they are the same size but opposite in sign, add them.
消元法通过把两个方程相加或相减来消去一个未知数。黄金法则是：如果某个未知数的系数相同，就把两式相减；如果系数绝对值相同而符号相反，就把两式相加。
Worked example: solve 3x + 2y = 16 and 2x − 2y = 4.
例题：求解 3x + 2y = 16 与 2x − 2y = 4。
- Step 1: the y coefficients are +2 and −2, so add the equations: 5x = 20. | 第一步：y 的系数为 +2 和 −2，故两式相加：5x = 20。
- Step 2: divide by 5 to get x = 4. | 第二步：两边除以 5，得 x = 4。
- Step 3: substitute into the simpler equation: 2(4) − 2y = 4, so 8 − 2y = 4. | 第三步：代入较简单的方程：2(4) − 2y = 4，即 8 − 2y = 4。
- Step 4: rearrange to −2y = −4, so y = 2. | 第四步：整理得 −2y = −4，故 y = 2。
- Step 5: check in the other equation: 3(4) + 2(2) = 12 + 4 = 16. Correct. | 第五步：代入另一个方程检验：3(4) + 2(2) = 12 + 4 = 16，正确。
solve 3x + 4y = 26 and 2x + 3y = 19, multiply the first equation by 2 and the second by 3:
当系数不匹配时，先给一个或两个方程乘以适当的数，使某个未知数的系数对齐。例如解 3x + 4y = 26 与 2x + 3y = 19，可把第一式乘 2、第二式乘 3：
6x + 8y = 52 and 6x + 9y = 57
gives y = 5; substituting back gives 3x + 20 = 26, so x = 2. The solution is x = 2, y = 5.
两式相减得 y = 5；再代回得 3x + 20 = 26，故 x = 2。所以解为 x = 2、y = 5。
4. The Substitution Method | 代入法
the other equation with the whole expression, producing a single equation in one unknown.
当其中一个方程已经写成 y = ... 或 x = ... 的形式时，代入法最快。您用整个表达式替换另一个方程中的该未知数，从而得到只含一个未知数的一元方程。
Worked example: solve y = 3x − 1 and 2x + y = 14.
例题：求解 y = 3x − 1 与 2x + y = 14。
- Substitute the expression for y into the second equation: 2x + (3x − 1) = 14. | 把 y 的表达式代入第二个方程：2x + (3x − 1) = 14。
- Collect like terms: 5x − 1 = 14. | 合并同类项：5x − 1 = 14。
- Solve: 5x = 15, so x = 3. | 求解：5x = 15，故 x = 3。
- Substitute back into y = 3x − 1: y = 3(3) − 1 = 8. | 回代到 y = 3x − 1：y = 3(3) − 1 = 8。
- Check in 2x + y = 14: 2(3) + 8 = 14. Correct. | 在 2x + y = 14 中检验：2(3) + 8 = 14，正确。
case such as y = 5 − 2x, forgetting the bracket in 3x − (5 − 2x) is one of the most common sources of lost marks.
代入时一定要加括号，特别是当表达式为负或含有减号时。此处写成 2x + 3x − 1 没问题，但在 y = 5 − 2x 这类情形中，若在 3x − (5 − 2x) 里漏掉括号，就是最常见的失分原因之一。
5. The Graphical Method | 图像法
Each linear equation represents a straight line. Because a solution must satisfy both equations, it must lie on both lines - that is, at their point of intersection. To solve graphically, plot both lines accurately on the same axes and read off the coordinates where they cross.
每个线性方程都代表一条直线。由于解必须同时满足两个方程，它必须同时位于两条直线上——也就是在它们的交点处。用图像法解题时，请在同一坐标系中准确画出两条直线，然后读出它们相交处的坐标。
- Make a table of values for each equation, using three points to guard against slips. | 为每个方程列出取值表，取三个点以防出错。
- Use a ruler and a sharp pencil; a wobbly line changes the answer. | 用直尺和削尖的铅笔；线条画歪会改变答案。
- State the answer as a coordinate pair, e.g. (3, 8), not just "3 and 8". | 答案要写成坐标对，例如 (3, 8)，而不是只写"3 和 8"。
- If the question supplies the graph, do not redraw it - just read and check. | 如果题目已给出图像，不要重画——只需读取并检验。
The graphical method is also the cleanest way to see why some systems have no solution: parallel lines never meet. It is, however, only as accurate as your drawing, so examiners usually expect an algebraic check or an exact solution when the answer is not a whole number.
图像法也是看清某些方程组为何无解的最直观方式：平行线永不相交。不过它的精度取决于作图，因此当答案不是整数时，考官通常期望您用代数方法检验或给出精确解。
6. Forming Your Own Equations from Word Problems | 从应用题中自行建立方程
Context questions rarely hand you the equations. Your first job is to define the variables in words, then translate each sentence into an equation.
情境题很少直接把方程给您。您的第一步是用文字定义未知数，然后把每个句子翻译成一个方程。
Worked example: three adult tickets and two child tickets cost £31. Two adult tickets and five child tickets cost £39. Find the cost of each ticket.
例题：3 张成人票与 2 张儿童票共 31 英镑；2 张成人票与 5 张儿童票共 39 英镑。求每种票的价格。
Let A be the cost of an adult ticket and C the cost of a child ticket, both in pounds.
设 A 为成人票价格，C 为儿童票价格，单位均为英镑。
3A + 2C = 31 and 2A + 5C = 39
Subtracting gives 11C = 55, so C = 5. Then 3A + 10 = 31 gives A = 7. An adult ticket costs £7 and a child ticket costs £5.
把第一式乘 2、第二式乘 3，使 A 的系数一致：6A + 4C = 62 与 6A + 15C = 117。相减得 11C = 55，故 C = 5。再由 3A + 10 = 31 得 A = 7。成人票 7 英镑，儿童票 5 英镑。
Finish by writing the answer in the context of the question, with units. A bare "A = 7, C = 5" may lose the final interpretation mark.
最后要用题目的情境和单位写出答案。只写 "A = 7, C = 5" 可能会丢掉最后的分结论分。
7. One Linear and One Quadratic Equation | 一个线性方程与一个二次方程
Higher tier students must also solve a system in which one equation is linear and the other is quadratic. Substitution is the standard method: rearrange the linear equation, substitute it into the quadratic, and solve the resulting quadratic equation.
进阶卷学生还必须求解一个方程是线性、另一个是二次的方程组。标准方法是代入法：先把线性方程整理好，代入二次方程，再解所得的二次方程。
Worked example: solve y = x + 1 and y = x² − 1.
例题：求解 y = x + 1 与 y = x² − 1。
Since both expressions equal y, set them equal to each other.
由于两个表达式都等于 y，可令它们相等。
x² − 1 = x + 1 → x² − x − 2 = 0
on y = x + 1 gives y = 3 and y = 0 respectively. The two solutions are (2, 3) and (−1, 0).
因式分解得 (x − 2)(x + 1) = 0，故 x = 2 或 x = −1。把每个值代入线性方程 y = x + 1，分别得 y = 3 与 y = 0。两组解为 (2, 3) 与 (−1, 0)。
- Geometrically, the line cuts the curve at two points, which is why two pairs of solutions appear. | 几何上，直线与曲线相交于两点，这正是出现两组解的原因。
- If the quadratic does not factorise, use the formula x = (−b ± √b² − 4ac) ÷ 2a and give answers to 3 significant figures unless told otherwise. | 若二次式无法因式分解，就用公式 x = (−b ± √b² − 4ac) ÷ 2a，除另有要求外，答案保留 3 位有效数字。
- Never discard the negative root. Both values of x give valid points of intersection. | 绝不要舍去负根。两个 x 值都对应有效的交点。
- Always give both coordinates for both solutions; the question almost always asks for "the coordinates". | 两组解都必须写出两个坐标；题目几乎总是要求给出"坐标"。
8. Special Cases: No Solution and Infinite Solutions | 特殊情况：无解与无穷多解
Not every pair of linear equations has exactly one solution. Comparing the coefficients tells you immediately what to expect.
并非每一对线性方程都恰好有一组解。比较系数就能立刻判断会出现哪种情况。
| Situation | 情形 | Condition | 条件 | Graph | 图像 | Solutions | 解的个数 |
| Unique solution | 唯一解 | Different gradients | 斜率不同 | Lines cross once | 两线相交一次 | One pair (x, y) | 一组 (x, y) |
| No solution | 无解 | Same gradient, different intercept | 斜率相同、截距不同 | Parallel lines | 平行线 | None | 无 |
| Infinite solutions | 无穷多解 | One equation is a multiple of the other | 一个方程是另一个的倍数 | Same line | 同一条直线 | Infinitely many | 无穷多 |
so every point on that line is a solution. You can detect either case during elimination: the variables vanish and you are left with a false statement such as 0 = 4 (no solution) or a true statement such as 0 = 0 (infinite solutions).
例如，y = 2x + 1 与 y = 2x + 5 平行，所以没有任何一组值能同时满足两者。相反，2x + 3y = 6 与 4x + 6y = 12 是同一条直线的两种写法，因此该直线上每一点都是解。用消元法也能识别这两种情况：未知数被消掉后，若剩下 0 = 4 这样的假命题，则无解；若剩下 0 = 0 这样的真命题，则有无穷多解。
9. Typical Exam Questions and How They Are Marked | 典型考题与评分方式
Edexcel IGCSE questions on this topic follow recognisable patterns. Knowing the pattern tells you which method to reach for and how many answers to expect.
爱德思 IGCSE 关于本主题的题目有可辨认的模式。了解模式就能知道该用哪种方法、应给出几个答案。
| Question type | 题型 | Typical wording | 常见问法 | Method | 方法 |
| Two linear equations | 两个线性方程 | "Solve the simultaneous equations ..." | "求解联立方程组……" | Elimination or substitution | 消元法或代入法 |
| Word context | 文字情境 | "Form two equations and solve them ..." | "建立两个方程并求解……" | Define variables, then eliminate | 定义未知数后消元 |
| Graph reading | 读图 | "Use the graph to solve ..." | "利用图像求解……" | Read the intersection | 读取交点 |
| Linear and quadratic | 线性与二次 | "Solve ... give your answers to 3 s.f." | "求解……答案保留 3 位有效数字" | Substitute, then factorise or use the formula | 代入后因式分解或用公式 |
Marks are awarded for method as well as accuracy. A clear statement such as "adding the equations eliminates y" or "substituting y = 3x − 1 into equation 2" nothing if it is incorrect, and long unsupported workings are hard for a marker to follow.
分数既给方法也给准确度。像"两式相加可消去 y"或"把 y = 3x − 1 代入方程 2"这样的明确表述，即使后面数字算错也能拿方法分。相反，若最终答案错误又没有解释，就一分不得；而冗长却缺乏说明的过程也会让阅卷人难以跟上。
10. Common Mistakes to Avoid | 需要避免的常见错误
Most lost marks in this topic come from a small number of repeated slips
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