Solving Simultaneous Equations | 解联立方程组

📚 Solving Simultaneous Equations | 解联立方程组

Simultaneous equations are a set of equations that share two or more unknown variables and must be solved together. They appear frequently in IGCSE Mathematics examinations, testing both algebraic manipulation and graphical understanding.

联立方程组是一组共享两个或多个未知数、必须同时求解的方程。它们在 IGCSE 数学考试中频繁出现，既考查代数运算能力，也考查图形理解能力。

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

A linear equation in two variables, such as 2x + y = 7, has infinitely many solutions. To find a unique pair (x, y), we need two independent equations that are satisfied by the same values of x and y.

像 2x + y = 7 这样含两个变量的线性方程有无穷多组解。要找到唯一的 (x, y)，我们需要两个独立方程，并由同一组 x 和 y 值同时满足它们。

For example, the system below has a unique solution:

例如，下面这个方程组有唯一解：

2x + y = 7
x − y = 2

Solving both equations together gives x = 3 and y = 1. We will show how to obtain this result systematically in the following sections.

联立求解两个方程可得 x = 3，y = 1。我们将在后续小节中系统地展示如何得出这个结果。

2. The Substitution Method | 代入法

The substitution method involves rearranging one equation to express one variable in terms of the other, then substituting this expression into the second equation.

代入法的步骤是：先改写其中一个方程，把一个变量用另一个变量表示出来，再把该表达式代入第二个方程。

Step 1: Choose one equation and make x or y the subject.

步骤 1：选择其中一个方程，把 x 或 y 表示成另一个变量的函数。
Step 2: Substitute the expression into the other equation.

步骤 2：将该表达式代入另一个方程。
Step 3: Solve the resulting linear equation in one variable.

步骤 3：解这个只含一个变量的线性方程。
Step 4: Substitute the value back to find the second variable.

步骤 4：把求出的值代回去，求出第二个变量。

Worked Example 1: Solve the system 2x + y = 7 and x − y = 2.

例题 1：解方程组 2x + y = 7 和 x − y = 2。

From the second equation, y = x − 2. Substituting into the first equation gives:

由第二个方程得 y = x − 2。将它代入第一个方程得：

2x + (x − 2) = 7
3x − 2 = 7
3x = 9
x = 3

Now substitute x = 3 into y = x − 2, giving y = 1. The solution is x = 3, y = 1.

再将 x = 3 代入 y = x − 2，得 y = 1。所以解为 x = 3，y = 1。

3. The Elimination Method | 消元法

The elimination method removes one variable by adding or subtracting the equations. This works best when the coefficients of one variable are the same or opposites, or can be made so by multiplying one or both equations.

消元法通过相加或相减两个方程来消去一个变量。当一个变量的系数相同或互为相反数时最适用；若不相同，可先对其中一个或两个方程乘以适当的数。

Step 1: Multiply equations as needed so that the coefficients of x (or y) match in magnitude.

步骤 1：根据需要对方程整体乘以某个数，使 x（或 y）的系数绝对值相等。
Step 2: Add or subtract the equations to eliminate that variable.

步骤 2：将两个方程相加或相减，消去该变量。
Step 3: Solve the remaining one-variable equation.

步骤 3：解剩下的含一个变量的方程。
Step 4: Substitute back to find the other variable.

步骤 4：代回原方程求出另一个变量。

Worked Example 2: Solve 3x + 2y = 12 and 5x − 2y = 4.

例题 2：解方程组 3x + 2y = 12 和 5x − 2y = 4。

Notice that the y-coefficients are +2 and −2. Adding the two equations eliminates y:

注意 y 的系数分别为 +2 和 −2。将两式相加即可消去 y：

(3x + 2y) + (5x − 2y) = 12 + 4
8x = 16
x = 2

Substitute x = 2 into 3x + 2y = 12: 6 + 2y = 12, so 2y = 6, giving y = 3. The solution is x = 2, y = 3.

把 x = 2 代入 3x + 2y = 12：6 + 2y = 12，所以 2y = 6，得 y = 3。解为 x = 2，y = 3。

If coefficients do not match, multiply first. For example, to solve 2x + 3y = 8 and 3x + 2y = 7, multiply the first equation by 2 and the second by 3 to make the y-coefficients 6 and 6, then subtract.

如果系数不相同，先乘数。例如解 2x + 3y = 8 和 3x + 2y = 7 时，将第一个方程乘以 2，第二个方程乘以 3，使 y 的系数都为 6，再相减即可。

4. Graphical Interpretation | 图形解释

Each linear equation in two variables represents a straight line on the Cartesian plane. The solution to a system of two linear equations is the point where the two lines intersect.

每个含两个变量的线性方程在笛卡尔平面上都代表一条直线。两个线性方程组成的方程组的解，就是这两条直线的交点。

To solve graphically, draw both lines accurately on the same set of axes and read off the coordinates of the intersection point.

用图形法求解时，在同一个坐标系中精确地画出两条直线，然后读出交点的坐标。

For the system 2x + y = 7 and x − y = 2, the first line passes through (0, 7) and (3.5, 0), while the second passes through (2, 0) and (0, −2). Their intersection is (3, 1), which matches our algebraic answer.

对于方程组 2x + y = 7 和 x − y = 2，第一条直线经过 (0, 7) 和 (3.5, 0)，第二条经过 (2, 0) 和 (0, −2)。它们的交点是 (3, 1)，与代数求解的结果一致。

Graphical methods are helpful for estimating solutions, but in an exam you must show algebraic working to obtain exact answers unless the question asks for a graphical estimate.

图形法有助于估算解，但考试中除非题目明确要求用图形估算，否则你必须写出代数过程以获得精确答案。

5. 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 a unique solution. Two special cases arise depending on the relationship between the two lines.

并非每一对线性方程都有唯一解。根据两条直线的位置关系，会出现两种特殊情况。

Situation	Line Relationship	Number of Solutions
Parallel lines	Same gradient, different y-intercept	0 solutions
Coincident lines	Same gradient and same y-intercept	Infinite solutions

Parallel lines never meet, so the system has no solution. For example, y = 2x + 1 and y = 2x − 3 have the same gradient 2 but different intercepts.

平行线永不相交，因此方程组无解。例如 y = 2x + 1 和 y = 2x − 3 具有相同的斜率 2，但截距不同。

Coincident lines overlap completely, meaning every point on the line satisfies both equations. For example, x + y = 3 and 2x + 2y = 6 represent the same line when simplified.

重合直线完全重叠，意味着直线上的每一点都同时满足两个方程。例如 x + y = 3 和 2x + 2y = 6 化简后表示同一条直线。

In algebraic terms, you will recognise these cases when the elimination step produces a false statement like 0 = 4 (no solution) or an identity like 0 = 0 (infinite solutions).

在代数运算中，当消元步骤出现 0 = 4 这类矛盾等式时为无解；出现 0 = 0 这类恒等式时为无穷多解。

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

IGCSE examinations often include one linear equation and one quadratic equation, such as y = x² + 3x − 1 combined with 2x + y = 5. These systems are solved using substitution.

IGCSE 考试中经常出现一个线性方程与一个二次方程的组合，例如 y = x² + 3x − 1 与 2x + y = 5。这类方程组用代入法求解。

Step 1: Rearrange the linear equation to make one variable the subject.

步骤 1：改写线性方程，把一个变量表示出来。

Step 2: Substitute this expression into the quadratic equation. This produces a quadratic equation in one variable, which may have two solutions.

步骤 2：将该表达式代入二次方程。这会得到一个只含一个变量的二次方程，可能有两个解。

Step 3: Solve the quadratic equation using factorisation, completing the square, or the quadratic formula.

步骤 3：用因式分解、配方法或求根公式解这个二次方程。

Step 4: Substitute each value back into the linear equation to find the corresponding second coordinate.

步骤 4：把每个值代回线性方程，求出对应的第二个坐标。

7. Non-Linear Worked Example | 非线性例题详解

Solve the system y = x² + 1 and y = 2x + 2.

解方程组 y = x² + 1 和 y = 2x + 2。

Since both equations already have y as the subject, we set the right-hand sides equal:

由于两个方程都已经把 y 表示出来，我们令等号右侧相等：

x² + 1 = 2x + 2

Rearrange to standard form:

移项化为标准形式：

x² − 2x − 1 = 0

This does not factorise simply, so use the quadratic formula with a = 1, b = −2, c = −1:

该方程不易因式分解，所以用求根公式，其中 a = 1，b = −2，c = −1：

x = [−b ± √(b² − 4ac)] ÷ 2a
x = [2 ± √(4 + 4)] ÷ 2
x = [2 ± √8] ÷ 2 = 1 ± √2

So x = 1 + √2 ≈ 2.414 or x = 1 − √2 ≈ −0.414. Substitute into y = 2x + 2:

因此 x = 1 + √2 ≈ 2.414 或 x = 1 − √2 ≈ −0.414。代入 y = 2x + 2：

y = 2(1 + √2) + 2 = 4 + 2√2
y = 2(1 − √2) + 2 = 4 − 2√2

The two solutions are (1 + √2, 4 + 2√2) and (1 − √2, 4 − 2√2).

两组解分别为 (1 + √2, 4 + 2√2) 和 (1 − √2, 4 − 2√2)。

8. Setting Up Word Problems | 列方程解应用题

Exam questions often present a real-world context and ask you to form and solve a pair of simultaneous equations. The key skill is translating words into algebra.

考试题目常常给出实际情境，要求你列出一组联立方程并求解。关键技能是把文字转化为代数式。

Worked Example 3: Two pencils and three notebooks cost $7.50. Four pencils and one notebook cost $6.50. Find the price of one pencil and one notebook.

例题 3：两支铅笔和三本笔记本共 7.50 美元。四支铅笔和一本笔记本共 6.50 美元。求一支铅笔和一本笔记本的价格。

Let p be the price of a pencil and n the price of a notebook. The equations are:

设 p 为铅笔单价，n 为笔记本单价。可列出方程：

2p + 3n = 7.50
4p + n = 6.50

Multiply the first equation by 2: 4p + 6n = 15.00. Subtract the second equation:

将第一个方程乘以 2：4p + 6n = 15.00。减去第二个方程：

(4p + 6n) − (4p + n) = 15.00 − 6.50
5n = 8.50
n = 1.70

Substitute n = 1.70 into 4p + n = 6.50: 4p + 1.70 = 6.50, so 4p = 4.80, giving p = 1.20. A pencil costs $1.20 and a notebook costs $1.70.

将 n = 1.70 代入 4p + n = 6.50：4p + 1.70 = 6.50，所以 4p = 4.80，得 p = 1.20。铅笔每支 1.20 美元，笔记本每本 1.70 美元。

9. Common Exam Mistakes | 常见考试错误

Students lose valuable marks in simultaneous equation questions for a few recurring reasons. Being aware of these pitfalls will help you avoid them.

学生在联立方程题目中因几个反复出现的原因丢分。了解这些陷阱有助于避开它们。

Sign errors when subtracting equations: always distribute the minus sign across every term.

相减方程时出现符号错误：务必把负号分配到每一项上。
Forgetting to substitute back into the original equation to find the second variable.

忘记代回原方程求第二个变量。
When multiplying an equation, forgetting to multiply the constant term on the right-hand side.

方程乘以某个数时，忘记乘右侧的常数项。
In non-linear systems, losing one of the two solutions because the quadratic has two roots.

在非线性方程组中，因为二次方程有两个根而漏掉其中一个解。
Not stating the final answer clearly as an ordered pair (x, y).

没有以有序数对 (x, y) 的形式清晰写出最终答案。

Always check your solution by substituting both values into both original equations. For example, x = 3, y = 1 should satisfy both 2x + y = 7 and x − y = 2.

务必把两组值代入两个原方程进行检验。例如 x = 3，y = 1 应同时满足 2x + y = 7 和 x − y = 2。

10. Practice Questions | 练习巩固

Attempt these questions on your own before checking the answers below.

请先独立思考完成以下练习，再核对答案。

Q1: Solve 5x + 2y = 16 and 2x − y = 1 using substitution.

题 1：用代入法解 5x + 2y = 16 和 2x − y = 1。
Q2: Solve 4x + 3y = 10 and 6x − y = 7 using elimination.

题 2：用消元法解 4x + 3y = 10 和 6x − y = 7。
Q3: Solve y = x² − 2x and y = 3x − 4. Give both solution pairs.

题 3：解 y = x² − 2x 和 y = 3x − 4，写出两组解。

Answers: Q1: x = 2, y = 3. Q2: Multiply the second equation by 3 to get 18x − 3y = 21; adding gives 22x = 31, so x = 1.409…, then y ≈ 1.455. Q3: Setting x² − 2x = 3x − 4 gives x² − 5x + 4 = 0, so (x − 1)(x − 4) = 0, giving x = 1 or x = 4, with y = −1 or y = 8.

答案：题 1：x = 2，y = 3。题 2：第二个方程乘以 3 得 18x − 3y = 21；相加得 22x = 31，所以 x ≈ 1.409，y ≈ 1.455。题 3：令 x² − 2x = 3x − 4 得 x² − 5x + 4 = 0，即 (x − 1)(x − 4) = 0，所以 x = 1 或 x = 4，对应 y = −1 或 y = 8。

11. Summary | 总结

To solve any system of linear simultaneous equations, choose the most efficient method: substitution when one variable is already isolated or easy to isolate, and elimination when coefficients align neatly. For systems involving a quadratic, substitution is the only reliable algebraic approach.

解任何线性联立方程组时，选择最高效的方法：当一个变量已经单独表示或易于单独表示时用代入法；当系数对齐整齐时用消元法。对于含二次项的方程组，代入法是唯一可靠的代数方法。

Graphically, the solution is the intersection point of the lines or curves. Remember that parallel lines give no solution, coincident lines give infinitely many solutions, and a straight line may intersect a quadratic at zero, one, or two points.

从图形上看，解就是直线或曲线的交点。记住：平行直线无解，重合直线有无穷多解，而直线与二次曲线可能有零个、一个或两个交点。

With consistent practice and careful checking, simultaneous equations become one of the most predictable scoring topics in the IGCSE Mathematics paper.

通过持续练习和细致检验，联立方程组会成为 IGCSE 数学考试中最容易拿分的题型之一。

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Solving Simultaneous Equations | 解联立方程组