Solving Simultaneous Equations | 解联立方程

📚 Solving Simultaneous Equations | 解联立方程

Simultaneous equations are one of the most frequently tested topics in IGCSE Mathematics (0580 and 0606). Whether you are taking the Core or Extended paper, mastering this skill is essential for achieving top marks in algebra and problem-solving questions. This article will guide you through every method you need, step by step, and highlight the traps that examiners love to set.

联立方程是IGCSE数学（0580和0606）中最常考的知识点之一。无论你参加Core（核心）还是Extended（扩展）试卷，掌握这一技巧都是取得代数与应用题高分的关键。本文将一步一步带你掌握所有必需的方法，并指出考官最爱设置的陷阱。

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

A system of simultaneous equations is a set of two or more equations that involve two or more unknown variables. In IGCSE Mathematics, the most common form is two linear equations in two unknowns, usually written as x and y. The word “simultaneous” means that the equations must be solved together: the same pair of values must satisfy both equations at the same time.

联立方程是包含两个或多个未知数的一组方程。在IGCSE数学中，最常见的形式是含有两个未知数（通常记作 x 和 y）的两个线性方程。”联立”意味着这些方程必须一起求解：同一组数值必须同时满足两个方程。

For example, consider the system below. We need to find one value of x and one value of y that make both equations true.

例如，考虑下面的方程组。我们需要找出一组 x 和 y 的值，使两个方程同时成立。

2x + 3y = 8
x − y = −1

By inspection or by solving, we find x = 1 and y = 2. Check: 2(1) + 3(2) = 2 + 6 = 8 ✓, and 1 − 2 = −1 ✓. The solution is the ordered pair (1, 2).

通过观察或求解，我们得到 x = 1，y = 2。验证：2(1) + 3(2) = 2 + 6 = 8 ✓，且 1 − 2 = −1 ✓。解为有序数对 (1, 2)。

2. The Elimination Method | 消元法

The elimination method is usually the fastest way to solve linear simultaneous equations. The idea is simple: we add or subtract the two equations so that one variable cancels out, leaving a single equation in one unknown.

消元法通常是解线性联立方程最快的方法。思路很简单：我们将两个方程相加或相减，使一个未知数被消去，从而得到一个只含一个未知数的方程。

Step 1: “Align” the equations. Write both equations one above the other, with x-terms, y-terms and constants aligned in columns.

第一步：对齐方程。将两个方程上下排列，使x项、y项和常数项分别对齐。

Step 2: Make the coefficients of one variable the same. If the coefficients of x or y are already equal, you can proceed directly. If not, multiply one or both equations by a suitable number.

第二步：使某一个未知数的系数相同。如果x或y的系数已经相等，可直接进行下一步。若不相等，则将其中一个或两个方程乘以适当的数。

Step 3: Add or subtract. If the signs are the same, subtract; if the signs are opposite, add. This eliminates one variable.

第三步：相加或相减。若符号相同，则相减；若符号相反，则相加。这样即可消去一个未知数。

Step 4: Solve for the remaining variable, then substitute back.

第四步：解出剩下的未知数，再代回求另一个未知数。

Let us apply this to our earlier example.

下面我们用前面的例子来说明。

2x + 3y = 8 …… ①
x − y = −1 …… ②

Multiply equation ② by 3 so that the y-coefficients become the same size with opposite signs: 3x − 3y = −3. Now add ① and the new ②.

将方程②乘以3，使y的系数大小相等且符号相反：3x − 3y = −3。然后将①与新的②相加。

2x + 3y = 8
+ (3x − 3y = −3)
5x = 5 → x = 1

Substitute x = 1 into ②: 1 − y = −1 → y = 2. The solution is x = 1, y = 2.

将 x = 1 代入②：1 − y = −1，得 y = 2。解为 x = 1，y = 2。

Example where you must multiply both equations: Solve 3x + 2y = 12 and 2x + 5y = 19.

需要同时乘以两个方程的例子：解方程组 3x + 2y = 12 和 2x + 5y = 19。

To eliminate x, multiply the first equation by 2 and the second by 3:

为了消去x，将第一个方程乘以2，第二个方程乘以3：

6x + 4y = 24
6x + 15y = 57
Subtract: 11y = 33 → y = 3

Substitute y = 3 into the first equation: 3x + 2(3) = 12 → 3x = 6 → x = 2. Solution: (2, 3).

将 y = 3 代入第一个方程：3x + 2(3) = 12，得 3x = 6，x = 2。解为 (2, 3)。

3. The Substitution Method | 代入法

The substitution method is especially useful when one equation has a single variable with coefficient 1, such as x = 2y + 3 or y = 5 − 2x. In this method, we isolate one variable and “substitute” the expression into the other equation.

代入法在一个方程中某个变量的系数为1时特别有用，例如 x = 2y + 3 或 y = 5 − 2x。此方法先解出一个变量，然后将表达式”代入”另一个方程。

Worked example: Solve y = 2x − 1 and 3x + 2y = 12.

例题：解方程组 y = 2x − 1 和 3x + 2y = 12。

Since y is already the subject of the first equation, replace y in the second equation with 2x − 1:

由于 y 已经是第一个方程的主项，将第二个方程中的 y 替换为 2x − 1：

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

Now substitute x = 2 back into y = 2x − 1: y = 2(2) − 1 = 3. Solution: x = 2, y = 3.

再将 x = 2 代回 y = 2x − 1：y = 2(2) − 1 = 3。解为 x = 2，y = 3。

The substitution method is also the preferred approach when you need to solve a system involving a non-linear equation, as we will see in Section 6.

代入法也是解含非线性方程组的首选方法，我们将在第6节中看到。

4. The Graphical Method | 图像法

Another way to solve simultaneous equations is to draw both lines on the same set of axes. The point where the two lines intersect represents the solution, because it lies on both lines and therefore satisfies both equations.

解联立方程的另一种方法是把两条直线画在同一坐标系中。两条直线的交点就是方程组的解，因为这个交点同时位于两条直线上，即同时满足两个方程。

Worked example: Solve graphically y = 2x + 1 and y = −x + 4.

例题：用图像法解 y = 2x + 1 和 y = −x + 4。

For the first line, plot three points: when x = 0, y = 1; when x = 1, y = 3; when x = 2, y = 5. For the second line: when x = 0, y = 4; when x = 1, y = 3; when x = 2, y = 2. Draw both lines carefully.

对第一条直线，取三个点：x = 0 时 y = 1；x = 1 时 y = 3；x = 2 时 y = 5。对第二条直线：x = 0 时 y = 4；x = 1 时 y = 3；x = 2 时 y = 2。仔细画出两条直线。

Both lines pass through the point (1, 3), so the solution is x = 1, y = 3.

两条直线都经过点 (1, 3)，所以解为 x = 1，y = 3。

The graphical method has clear limitations:

图像法有明显的局限：

It relies on accurate drawing, which is difficult under exam pressure.
它依赖绘图的准确性，在考场压力下很难画得精确。
If the intersection point has non-integer coordinates, such as (1.3, 2.7), reading the exact answer from a graph is almost impossible.
如果交点坐标为非整数，例如 (1.3, 2.7)，从图上读出精确答案几乎不可能。
Graph paper is not always provided; freehand sketches are often too rough.
考卷不总是提供坐标纸；手绘草图往往过于粗糙。

Therefore, in an exam you are strongly advised to use the elimination or substitution method for exact answers. Use the graphical method only if the question explicitly asks you to do so.

因此，在考试中强烈建议使用消元法或代入法来获得精确答案。只有当题目明确要求作图时，才使用图像法。

5. Solving Word Problems | 解应用题

Word problems are a favourite question type in IGCSE exams. The challenge is not the algebra — it is converting English sentences into equations. A good strategy is to define your variables clearly at the start, then translate each statement one by one.

应用题是IGCSE考试中的常客。难点不在于代数本身，而在于把英文句子转换成方程。一个好的策略是：先清楚地定义未知数，然后逐句翻译成方程。

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

例题：两个数之和为15，差为3。求这两个数。

Let the larger number be x and the smaller number be y. Then:

设较大的数为 x，较小的数为 y。则：

x + y = 15
x − y = 3

Add the two equations: 2x = 18 → x = 9. Substitute back: 9 + y = 15 → y = 6. The two numbers are 9 and 6.

两式相加：2x = 18，得 x = 9。代回：9 + y = 15，得 y = 6。两个数分别是9和6。

Example: Tickets at a theatre cost $8 for adults and $5 for children. A family of 6 people pays a total of $39. How many adults and children are there?

例题：剧院成人票每张8美元，儿童票每张5美元。一个6人家庭共支付39美元。问有多少成人和儿童？

Let a be the number of adults and c the number of children.

设成人数为 a，儿童数为 c。

a + c = 6
8a + 5c = 39

From the first equation, a = 6 − c. Substitute into the second: 8(6 − c) + 5c = 39 → 48 − 8c + 5c = 39 → 48 − 3c = 39 → 3c = 9 → c = 3. Then a = 3. So there are 3 adults and 3 children.

由第一个方程得 a = 6 − c。代入第二个方程：8(6 − c) + 5c = 39，即 48 − 8c + 5c = 39，48 − 3c = 39，3c = 9，c = 3。于是 a = 3。所以有3个成人和3个儿童。

Always check that your answers make sense in the original context. A negative number of children, or a fractional number of adults, should immediately signal an error.

务必检查答案是否符合实际语境。如果出现负数的人数或分数的成人数，应立即警惕计算有误。

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

In the Extended paper, you will also meet systems where one equation is linear and the other is quadratic, for example x² + y² = 25 and x + y = 7. These are solved by substitution: rearrange the linear equation to make one variable the subject, then substitute into the quadratic equation.

在Extended试卷中，你还会遇到一个方程为线性、另一个为二次的方程组，例如 x² + y² = 25 和 x + y = 7。这类方程组用代入法求解：先从线性方程中解出一个变量，然后代入二次方程。

Worked example: Solve the system x² + y² = 25 and x + y = 7.

例题：解方程组 x² + y² = 25 和 x + y = 7。

From the linear equation, y = 7 − x. Substitute into the quadratic:

由线性方程得 y = 7 − x。代入二次方程：

x² + (7 − x)² = 25
x² + 49 − 14x + x² = 25
2x² − 14x + 24 = 0
x² − 7x + 12 = 0

Factorise: (x − 3)(x − 4) = 0, so x = 3 or x = 4. Substitute each value back into y = 7 − x:

因式分解：(x − 3)(x − 4) = 0，所以 x = 3 或 x = 4。将每个值代回 y = 7 − x：

When x = 3, y = 4. When x = 4, y = 3. The solutions are (3, 4) and (4, 3).

当 x = 3 时，y = 4。当 x = 4 时，y = 3。解为 (3, 4) 和 (4, 3)。

Important point: A linear-quadratic system can have up to two solutions. If the discriminant is negative, the system has no real solutions; if the discriminant is zero, there is exactly one solution (the line is tangent to the curve).

重要提示：线性-二次方程组最多有两个解。若判别式为负，则方程组无实数解；若判别式为零，则恰好有一个解（直线与曲线相切）。

7. Common Mistakes and How to Avoid Them | 常见错误及防范

Many students lose marks not because they do not understand the method, but because of careless algebra. Here are the most frequent pitfalls, together with practical advice.

许多学生失分并非因为不懂方法，而是因为粗心的代数运算。下面是最常见的失误及实用建议。

Sign errors when subtracting. When you subtract one equation from another, remember to subtract every term, not just the first. For example, (2x + 3y) − (2x − 5y) = 8y, not −2y.
减法中的符号错误。用一个方程减去另一个方程时，要减去每一项而不能只减第一项。例如 (2x + 3y) − (2x − 5y) = 8y，而不是 −2y。
Forgetting to multiply every term. When you multiply an equation by a number, you must multiply the constant term as well. A classic error is turning x + y = 5 into 2x + 2y = 5 after multiplying by 2.
忘记乘每一项。将方程乘以一个数时，必须连同常数项一起乘。一个典型错误是将 x + y = 5 乘以2后写成 2x + 2y = 5。
Substituting back into the wrong equation. Always substitute into an equation that you have not modified, to avoid carrying forward an error.
代回错误的方程。始终代入未经过修改的方程，以免将错误传递下去。
Not checking your final answer. A 10-second check of both original equations can catch half of all errors. Always verify your pair of values satisfies both equations.
没有检验最终答案。用10秒钟检查两个原方程就能发现一半的错误。务必验证你的这对数值满足两个方程。

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

In the exam, you should be able to pick the best method quickly. The table below provides a quick reference.

在考试中，你要能快速选出最佳方法。下表提供快速参考。

Situation / 情形	Recommended Method / 推荐方法
Coefficients already equal or easily made equal / 系数已相同或容易化为相同	Elimination / 消元法
One variable already the subject, e.g. y = 2x + 1 / 某个变量已是主项，如 y = 2x + 1	Substitution / 代入法
One equation is quadratic / 其中一个方程是二次方程	Substitution / 代入法
Question asks to estimate by drawing / 题目要求通过画图估计	Graphical / 图像法

Remember: the graphical method is the only one that gives approximate answers; the algebraic methods always give exact answers when the solution exists.

切记：只有图像法给出近似答案；当解存在时，代数方法总能给出精确答案。

9. Exam Tips from an Examiner’s Perspective | 从考官角度看的考试技巧

Having marked thousands of IGCSE scripts, I can tell you exactly what examiners look for. First, they award method marks: even if your final answer is wrong, a correct elimination step earns you credit. Therefore, always write down the new equation after multiplying or subtracting; do not do the algebra in your head.

批改过数千份IGCSE答卷后，我可以确切告诉你考官看重什么。首先，方法是给过程分的：即使最终答案错了，正确的消元步骤依然能得分。因此，务必写出乘除或相减后的新方程，不要在脑子里进行代数运算。

Second, use a sharp pencil and a ruler for the graphical method, and plot at least three points for each line. Third, when a word problem asks for “two numbers”, state the final answer clearly as a pair, such as “x = 9 and y = 6”. Do not leave the examiner to guess which variable is which.

其次，使用削尖的铅笔和直尺画图，每条直线至少取三个点。再次，当应用题要求”两个数”时，要清楚地以数对形式写出答案，例如”x = 9, y = 6″。不要让考官猜测哪个变量对应哪个数。

Finally, remember the checking step. A quick substitution of your answers into both original equations should take no more than fifteen seconds. It is the cheapest insurance policy you will ever buy in a mathematics exam.

最后，牢记检验这一步。把你的答案快速代入两个原方程只需不到十五秒。这是数学考试中最划算的”保险”。

10. Practice Questions | 练习

Try these questions under timed conditions. The answers are provided below so you can check your working.

请在限时条件下完成以下练习。文末附有答案，便于你核对过程。

Question 1. Solve the simultaneous equations:

练习1。解联立方程：

5x + 2y = 16
3x − y = 3

Question 2. A rectangle has perimeter 30 cm. Its length is 3 cm more than twice its width. Find the length and the width.

练习2。一个长方形的周长为30厘米。它的长比宽的2倍多3厘米。求长和宽。

Question 3. Solve the system:

练习3。解方程组：

y = x² + 2x − 3
y = 2x + 5

Answers:

答案：

Q1. From the second equation, y = 3x − 3. Substitute: 5x + 2(3x − 3) = 16 → 5x + 6x − 6 = 16 → 11x = 22 → x = 2, y = 3. Check: 5(2) + 2(3) = 16 ✓, 3(2) − 3 = 3 ✓.

练习1答案：由第二个方程得 y = 3x − 3。代入：5x + 2(3x − 3) = 16，得 5x + 6x − 6 = 16，11x = 22，所以 x = 2，y = 3。检验：5(2) + 2(3) = 16 ✓，3(2) − 3 = 3 ✓。

Q2. Let the width be w and the length be l. Then l = 2w + 3 and 2(l + w) = 30 → l + w = 15. Substitute: (2w + 3) + w = 15 → 3w = 12 → w = 4 cm, l = 11 cm.

练习2答案：设宽为 w，长为 l。则 l = 2w + 3，且 2(l + w) = 30，即 l + w = 15。代入：(2w + 3) + w = 15，得 3w = 12，所以 w = 4 厘米，l = 11 厘米。

Q3. Set x² + 2x − 3 = 2x + 5 → x² − 8 = 0 → x = ±2√2. Then y = 2x + 5. Solutions are (2√2, 5 + 4√2) and (−2√2, 5 − 4√2).

练习3答案：令 x² + 2x − 3 = 2x + 5，得 x² − 8 = 0，所以 x = ±2√2。于是 y = 2x + 5。解为 (2√2, 5 + 4√2) 和 (−2√2, 5 − 4√2)。

11. Conclusion | 结语

Simultaneous equations are a core skill in IGCSE Mathematics. Master the elimination and substitution methods until they become second nature, keep the graphical method ready for questions that demand it, and always check

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Solving Simultaneous Equations | 解联立方程