Solving Simultaneous Equations | 解联立方程组

📚 Solving Simultaneous Equations | 解联立方程组

Simultaneous equations are one of the most important topics in IGCSE Mathematics. They appear in both Paper 2 and Paper 4, either as direct algebraic questions or inside real-world problem solving. Mastering this topic means understanding why a ‘solution’ is the point where two conditions meet, and being confident with both the substitution and elimination methods.

联立方程组是IGCSE数学中最重要的考点之一，在Paper 2和Paper 4中都会出现，既会直接考查代数运算，也会融入实际应用题。掌握这一考点，关键在于理解“解”就是两个条件同时成立的那组数值，并熟练运用代入消元法和加减消元法。

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

A linear equation in two variables, such as 2x + 3y = 7, has infinitely many solutions. Each solution is a pair of values (x, y) that makes the equation true. When we have two such equations, we want the single pair (x, y) that satisfies both equations at the same time.

一个含有两个变量的线性方程，例如 2x + 3y = 7，有无数多组解。每一组解都是一对数值 (x, y)，使该方程成立。当我们有两个这样的方程时，我们要求的是能同时满足这两个方程的唯一一对 (x, y)。

For example, consider the system 2x + 3y = 7 and x – y = 1. Substituting x = 2 and y = 1 checks out: 2(2) + 3(1) = 7 and 2 – 1 = 1. So the solution is x = 2, y = 1. There are three main ways to find such solutions: substitution, elimination, and graphical methods.

例如，考虑方程组 2x + 3y = 7 和 x – y = 1。代入 x = 2、y = 1 验证：2(2) + 3(1) = 7，且 2 – 1 = 1，所以解是 x = 2，y = 1。求解这类方程组主要有三种方法：代入消元法、加减消元法和图像法。

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2. The Substitution Method | 代入消元法

The substitution method is ideal when one equation is already written in the form y = … or x = …. You replace that variable in the other equation with the expression given, leaving a single equation in one variable.

代入消元法适用于其中一个方程已经写成 y = … 或 x = … 的形式。此时只需将该表达式代入另一个方程，就可以得到一个只含一个变量的一元方程。

Example: Solve

y = 2x – 1 (1)

3x + 2y = 12 (2)

Step 1: Substitute y = 2x – 1 into equation (2):

3x + 2(2x – 1) = 12 → 3x + 4x – 2 = 12 → 7x = 14 → x = 2

Step 2: Substitute x = 2 back into equation (1):

y = 2(2) – 1 = 3

Therefore the solution is x = 2, y = 3. Always finish by checking that x = 2, y = 3 satisfies equation (2): 3(2) + 2(3) = 6 + 6 = 12 ✓

因此解为 x = 2，y = 3。最后一定要验证：把 x = 2，y = 3 代入方程 (2)：3(2) + 2(3) = 6 + 6 = 12 ✓

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3. The Elimination Method | 加减消元法

The elimination method works by adding or subtracting the two equations so that one variable cancels out. This works best when the coefficient of x or y is the same (or opposite) in both equations.

加减消元法通过将两个方程相加或相减，使其中一个变量的系数相互抵消，从而消去一个未知数。当两个方程中 x 或 y 的系数相同（或互为相反数）时，此法最为高效。

Example: Solve

2x + 3y = 7 (1)

2x – 3y = 1 (2)

Notice that the coefficients of y are +3 and -3. Adding the two equations eliminates y:

(2x + 3y) + (2x – 3y) = 7 + 1 → 4x = 8 → x = 2

Substitute x = 2 into equation (1):

2(2) + 3y = 7 → 4 + 3y = 7 → 3y = 3 → y = 1

The solution is x = 2, y = 1. When coefficients do not match, multiply one or both equations by a suitable number first. For example, to solve 3x + 2y = 12 and 2x – y = 1, multiply the second equation by 2 to get 4x – 2y = 2, then add to eliminate y.

解为 x = 2，y = 1。当系数不一致时，先给一个或两个方程乘以适当的数。例如解 3x + 2y = 12 和 2x – y = 1 时，先将第二个方程乘以 2，得到 4x – 2y = 2，再两式相加消去 y。

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4. Choosing the Right Method | 如何选择合适的方法

In an exam, you should quickly decide which method will save time and reduce sign errors. The table below summarises the best strategy.

在考试中，你应快速判断哪种方法更省时、更不容易出错。下表总结了常见情况的最佳策略。

Situation | 情形	Recommended method | 推荐方法
One equation is already y = … or x = …	Substitution | 代入消元法
Coefficients of one variable are equal or opposite	Elimination | 加减消元法
Coefficients are large and unrelated	Substitution (after rearranging) | 代入消元法
The system includes a quadratic equation	Substitution | 代入消元法

In many cases either method will work; what matters is that you show clear working and check your answer. A correct answer with no working may lose method marks.

很多情况下两种方法都能解，关键在于书写清楚过程并验证答案。只有答案而没有过程，可能会被扣掉方法分。

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5. Graphical Solution | 图像法求解

Each linear equation can be plotted as a straight line on an xy-plane. The solution of the simultaneous equations is the coordinates of the point where the two lines intersect.

每个线性方程都可以在 xy 坐标平面上画成一条直线。联立方程组的解，就是两条直线交点的坐标。

To draw the line of 2x + 3y = 7, find two or three points. When x = 0, y = 7/3 ≈ 2.33; when y = 0, x = 3.5; when x = 2, y = 1. Plot these points and join them with a straight line. Do the same for the second equation; the intersection gives the solution.

要画出 2x + 3y = 7 的直线，先找两三个点：当 x = 0 时，y = 7/3 ≈ 2.33；当 y = 0 时，x = 3.5；当 x = 2 时，y = 1。描点并用直尺连线，再用同样方法画出第二条直线，交点坐标就是方程组的解。

The graphical method is rarely required for exact answers in Paper 2, but it is useful for checking your algebra and for understanding the concept. On graph paper, read the intersection coordinates as accurately as possible.

图像法在 Paper 2 中很少要求精确答案，但它非常适合验证代数计算是否合理，同时能帮助你理解解的几何意义。在坐标纸上作图时，要尽可能准确地读出交点坐标。

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6. Word Problems: Forming Equations | 应用题建模

Many IGCSE questions present a real-world situation and ask you to define two variables, form two equations, and solve them. This tests both algebraic skill and reading comprehension.

许多IGCSE题目会提供一个现实情境，要求你先设两个未知数、列出两个方程并求解。这类题同时考查代数能力和阅读理解能力。

Example: A shop sells pens for $2 each and pencils for $1 each. A customer buys 8 items for a total of $13. How many pens and how many pencils did the customer buy?

例：某商店每支钢笔售 2 美元，每支铅笔售 1 美元。一位顾客买了 8 件文具，共花费 13 美元。请问他买了几支钢笔、几支铅笔？

Let p = number of pens and c = number of pencils. From the item count, p + c = 8. From the cost, 2p + c = 13. Subtract the first equation from the second:

(2p + c) – (p + c) = 13 – 8 → p = 5

Then c = 8 – 5 = 3. So the customer bought 5 pens and 3 pencils. Check: 5 + 3 = 8 items, 2(5) + 1(3) = 13 dollars ✓

设 p 为钢笔数量、c 为铅笔数量。由总件数得 p + c = 8；由总花费得 2p + c = 13。用第二个方程减去第一个方程：

(2p + c) – (p + c) = 13 – 8 → p = 5

于是 c = 8 – 5 = 3，所以顾客买了 5 支钢笔和 3 支铅笔。验证：5 + 3 = 8 件，2(5) + 1(3) = 13 美元 ✓

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7. Non-Linear Simultaneous Equations | 非线性联立方程组

At IGCSE level, you may also meet a system where one equation is linear and the other is quadratic, such as y = x + 2 and y = x². The substitution method is always the way forward here.

在IGCSE阶段，你还会遇到一个线性方程与一个二次方程组成的方程组，例如 y = x + 2 和 y = x²。此时应使用代入消元法。

Since both expressions equal y, we can set them equal:

x + 2 = x² → x² – x – 2 = 0 → (x – 2)(x + 1) = 0

So x = 2 or x = -1. Substitute into y = x + 2 to find the corresponding y values:

当 x = 2 时，y = 4；当 x = -1 时，y = 1。因此解为 (2, 4) 和 (-1, 1)。

When x = 2, y = 4; when x = -1, y = 1

The solutions are (2, 4) and (-1, 1). Notice that we obtain two solution pairs, because a straight line can intersect a parabola twice. If the quadratic has a repeated root, there is only one solution, meaning the line is tangent to the curve.

注意这里得到了两组解，因为一条直线与抛物线可能有两个交点。如果二次方程出现重根，则只有一组解，表示直线与曲线相切。

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8. Special Cases | 特殊情况讨论

Not every pair of simultaneous equations has a unique solution. There are two special cases you should recognise.

并非所有联立方程组都有唯一解。有两种特殊情况你需要会识别。

No solution (inconsistent system): The two lines are parallel. For example, 2x + 3y = 6 and 2x + 3y = 12 have the same left-hand side but different constants. Subtracting gives 0 = 6, a contradiction. In algebra, if your elimination step produces a false statement like 0 = 6, the system has no solution.

无解（矛盾方程组）：两条直线平行。例如 2x + 3y = 6 和 2x + 3y = 12，左边相同而常数不同。相减得 0 = 6，矛盾。在代数中，如果消元后出现 0 = 6 这样的假命题，说明方程组无解。

Infinite solutions (dependent system): The two equations represent the same line. For example, 2x + 3y = 6 and 4x + 6y = 12 are multiples of each other. Elimination would give 0 = 0, which is always true; every point on the line is a solution.

无穷多解（相关方程组）：两个方程表示同一条直线。例如 2x + 3y = 6 与 4x + 6y = 12 互为倍数关系。消元后得到 0 = 0，恒成立；直线上的每一个点都是解。

In IGCSE examinations, inconsistent or dependent systems are rarely asked as full questions, but recognising them can save you from getting confused, especially in multiple-choice items.

在IGCSE考试中，无解或无穷多解很少作为完整大题出现，但识别这两种情况能帮助你在选择题中快速排除干扰项。

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9. Common Mistakes and Exam Tips | 常见错误与考试技巧

Here are the most frequent errors students make when solving simultaneous equations, together with advice on how to avoid them.

以下是最常见的失分点，以及相应的避错建议。

• Sign errors when subtracting: When you subtract one equation from another, distribute the minus sign across every term. Write the subtracted equation in brackets first.

  相减时符号出错：用一个方程减去另一个方程时，负号要分配给每一项。先把被减的方程整体写在括号里，再去括号。

• Not substituting back: After finding one variable, you must substitute back into one of the original equations to find the other variable. Do not skip this step.

  忘记回代：求出一个变量后，必须代回原方程之一求出另一个变量。这一步不能省略。

• Making fractions prematurely: If a coefficient does not divide evenly, keep the value as a fraction instead of a rounded decimal. Rounding can lead to errors in the final answer.

  过早使用小数：当系数不能整除时，保留分数形式，而不要四舍五入成小数，否则会导致最终答案不准确。

• Not checking the answer: Always substitute both values into both original equations. This takes 20 seconds and prevents careless mistakes.

  不检查答案：务必把两组值分别代回两个原方程验证。这只需20秒，却能避免粗心失分。

In the exam, write each equation with a clear label (1) and (2), and show every substitution step. This makes it easy for the examiner to award method marks even if you make a small arithmetic slip.

考试中，给每个方程标上编号 (1)、(2)，并写出每一步代入过程。这样即使最后有小的计算失误，考官也容易根据过程给方法分。

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10. Practice Questions | 巩固练习

Try these questions before checking the answers. They cover the full range of methods discussed above.

请先独立完成以下练习，再核对答案。题目覆盖了上文讨论的所有方法。

Question 1: Solve by elimination: 3x + y = 11 and 2x – y = 4.

题目1：用加减消元法解：3x + y = 11 和 2x – y = 4。

Question 2: Solve by substitution: y = 3x – 2 and 2x + y = 8.

题目2：用代入消元法解：y = 3x – 2 和 2x + y = 8。

Question 3: The sum of two numbers is 15 and their difference is 3. Find the numbers.

题目3：两数之和为 15，之差为 3，求这两个数。

Question 4: Solve the non-linear system: y = x + 1 and y = x² – 1.

题目4：解非线性方程组：y = x + 1 和 y = x² – 1。

Question | 题目	Answer | 答案
1	x = 3, y = 2
2	x = 2, y = 4
3	9 and 6 | 9 和 6
4	x = 2, y = 3 或 x = -1, y = 0

For question 3, let the two numbers be a and b. Then a + b = 15 and a – b = 3. Adding gives 2a = 18, so a = 9; then b = 6. For question 4, set x + 1 = x² – 1, rearrange to x² – x – 2 = 0, factor as (x – 2)(x + 1) = 0, giving x = 2 or x = -1.

第3题：设两数为 a 和 b，则 a + b = 15，a – b = 3，相加得 2a = 18，所以 a = 9，进而 b = 6。第4题：由 x + 1 = x² – 1，整理得 x² – x – 2 = 0，因式分解为 (x – 2)(x + 1) = 0，所以 x = 2 或 x = -1。

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Simultaneous equations reward methodical, neat working. Practise until the two core methods, substitution and elimination, feel automatic, and always verify your solution point. With consistent practice, this topic becomes a guaranteed source of marks

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