Simultaneous Equations | 联立方程

📚 Simultaneous Equations | 联立方程

Simultaneous equations are a group of two or more equations that share the same unknown variables. In IGCSE Mathematics, you are usually asked to solve a pair of equations in two unknowns, x and y, where every equation must be true at the same time. This topic is not only a key examination skill but also a gateway to modelling real situations such as pricing, mixing solutions, and motion problems.

联立方程是一组包含相同未知变量的两个或多个方程。在 IGCSE 数学中，通常要求解含有两个未知量 x 和 y 的方程组，并且每个方程必须同时成立。这一主题不仅是关键的应试技能，也是建模现实情境（如价格计算、溶液混合和运动问题）的重要基础。

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

A pair of simultaneous equations involves two unknown quantities, usually x and y. The word ‘simultaneous’ literally means ‘at the same time’: a valid solution must satisfy every equation in the system. If a pair of values works for only one equation but not the other, it is not a solution of the system.

一对联立方程涉及两个未知量，通常为 x 和 y。”同时”一词的字面意思是”在同一时刻”：一个有效的解必须满足方程组中的每一个方程。如果某一组值只满足其中一个方程而不能满足另一个，那么它就不是该方程组的解。

For example, consider the linear system below:

例如，考虑下面的线性方程组：

2x + 3y = 7

x − y = 1

If x = 2 and y = 1, then 2(2) + 3(1) = 7 and 2 − 1 = 1, so this ordered pair is indeed a solution. Always check your answer in both original equations before moving on.

如果 x = 2，y = 1，那么 2(2) + 3(1) = 7 且 2 − 1 = 1，因此这个有序对确实是方程组的解。在继续之前，务必把答案代回两个原方程进行检验。

Linear equations have the form ax + by = c and produce straight-line graphs.

线性方程的形式为 ax + by = c，其图像为直线。
When both equations are linear, the system usually has exactly one solution, which is the point of intersection of the two lines.

当两个方程均为线性时，方程组通常有唯一解，即两条直线的交点。
When one equation is quadratic, the system can have two, one, or zero solutions, depending on whether the line and curve intersect.

当其中一个方程为二次方程时，方程组可能有两个、一个或零个解，具体取决于直线与曲线的相交情况。

2. The Elimination Method | 消元法

Elimination is one of the most reliable algebraic methods. The goal is to eliminate one variable by adding or subtracting the equations, after adjusting the coefficients if necessary. This method works best when the coefficients are simple integers, such as 2x + 3y = 8 combined with 3x − 2y = −1.

消元法是最可靠的代数方法之一。其目标是通过相加或相减方程来消去一个变量，必要时先调整系数。当系数是简单的整数时，这种方法最为高效，例如 2x + 3y = 8 与 3x − 2y = −1 的组合。

2x + 3y = 8 (Equation 1)

3x − 2y = −1 (Equation 2)

To eliminate y, multiply Equation 1 by 2 and Equation 2 by 3. This makes the y coefficients +6 and −6:

为了消去 y，将方程 1 乘以 2，将方程 2 乘以 3。这样 y 的系数分别变为 +6 和 −6：

4x + 6y = 16

9x − 6y = −3

Now add the two equations vertically. The y terms cancel, giving 13x = 13, so x = 1. Substituting x = 1 back into Equation 1 gives 2(1) + 3y = 8, so 3y = 6 and y = 2.

现在将两个方程纵向相加。y 项相互抵消，得到 13x = 13，因此 x = 1。将 x = 1 代回方程 1，得到 2(1) + 3y = 8，即 3y = 6，所以 y = 2。

As a final check, substitute both values into Equation 2: 3(1) − 2(2) = 3 − 4 = −1. The solution is x = 1, y = 2.

作为最后检验，将两个值代入方程 2：3(1) − 2(2) = 3 − 4 = −1。因此解为 x = 1，y = 2。

If the coefficients are already equal, simply add or subtract directly.

如果两个方程中某个变量的系数已经相同，直接相加或相减即可。
If the signs are opposite, add; if the signs are the same, subtract.

如果符号相反，则相加；如果符号相同，则相减。
Multiply every term in the equation by the chosen number, not just one term.

乘以选定数字时，要乘以方程中的每一项，而不是只乘一项。

3. The Substitution Method | 代入法

Substitution is especially useful when one equation already has a variable isolated, such as y = 2x + 1. You replace the isolated variable in the other equation with its algebraic expression, then solve a single equation in one variable.

代入法特别适用于其中一个方程已经将某个变量单独表示的情况，例如 y = 2x + 1。你将另一个方程中的该变量替换为它的代数表达式，然后求解一个只含一个变量的方程。

y = 2x + 1

x + y = 10

Substitute y = 2x + 1 into the second equation: x + (2x + 1) = 10, which simplifies to 3x + 1 = 10. Solving gives 3x = 9, so x = 3. Now substitute x = 3 back into y = 2x + 1: y = 2(3) + 1 = 7.

将 y = 2x + 1 代入第二个方程：x + (2x + 1) = 10，化简为 3x + 1 = 10。解得 3x = 9，因此 x = 3。再将 x = 3 代回 y = 2x + 1：y = 2(3) + 1 = 7。

Check in the second equation: 3 + 7 = 10. The solution is x = 3, y = 7.

在第二个方程中检验：3 + 7 = 10。因此解为 x = 3，y = 7。

Elimination 消元法	Substitution 代入法
Best when coefficients are small and similar in size. 最适合系数较小且相近的情形。	Best when one equation has a variable already isolated. 最适合已经有一个变量被单独表示的情形。
Avoids fractions until the final step. 在最后一步之前通常避免分数。	Requires careful expansion of brackets. 需要小心地展开括号。

4. One Linear and One Quadratic | 线性与二次方程联立

When the system contains one linear equation and one quadratic equation, the substitution method is generally the best approach. Solve the linear equation for one variable, then substitute into the quadratic equation. This produces a quadratic equation in one variable, which you can solve by factorisation or the quadratic formula.

当方程组包含一个线性方程和一个二次方程时，代入法通常是最佳选择。先从线性方程解出一个变量，再代入二次方程。这会得到一个只含一个变量的二次方程，可通过因式分解或求根公式求解。

y = x + 1

x² + y² = 25

Substitute y = x + 1 into x² + y² = 25:

将 y = x + 1 代入 x² + y² = 25：

x² + (x + 1)² = 25

x² + x² + 2x + 1 = 25

2x² + 2x − 24 = 0

Divide through by 2 to simplify: x² + x − 12 = 0. Factorising gives (x + 4)(x − 3) = 0, so x = −4 or x = 3. Use the linear equation to find the corresponding y values: when x = −4, y = −3; when x = 3, y = 4.

两边同时除以 2 化简：x² + x − 12 = 0。因式分解得 (x + 4)(x − 3) = 0，所以 x = −4 或 x = 3。利用线性方程求出对应的 y 值：当 x = −4 时，y = −3；当 x = 3 时，y = 4。

There are therefore two solutions: (−4, −3) and (3, 4). Geometrically, these are the intersection points of the straight line y = x + 1 and the circle x² + y² = 25 with radius 5 centred at the origin.

因此方程组有两个解：(−4, −3) 和 (3, 4)。从几何角度看，这两个解就是直线 y = x + 1 与圆心在原点、半径为 5 的圆 x² + y² = 25 的交点。

5. The Graphical Method | 图象法

Solving equations graphically involves drawing all equations on the same set of axes and reading off the coordinates of the intersection points. For linear equations, rearrange each equation into the form y = mx + c, then plot the lines. The x and y coordinates of the intersection point give the solution.

图解法是指在同一坐标系中画出所有方程的图像，然后读出交点的坐标。对于线性方程，先将每个方程变形为 y = mx + c 的形式，再画出直线。交点的 x 坐标和 y 坐标就是方程组的解。

y = 2x − 1

x + y = 5

Rewriting the second equation gives y = 5 − x. Plotting y = 2x − 1 and y = 5 − x on the same axes, the two lines intersect at x = 2, y = 3. Verify: 2(2) − 1 = 3 and 2 + 3 = 5.

将第二个方程改写为 y = 5 − x。在同一坐标系中画出 y = 2x − 1 和 y = 5 − x，两条直线相交于 x = 2，y = 3。验证：2(2) − 1 = 3 且 2 + 3 = 5。

Advantage: the method is visual and helps you understand the meaning of a solution.

优点：图象法直观，有助于理解解的实际含义。
Disadvantage: it only gives approximate answers unless the intersection has integer coordinates.

缺点：除非交点坐标为整数，否则只能得到近似解。
The graphical method is useful for checking algebraic answers quickly.

图象法可用于快速检验代数方法得到的答案。

6. Solving Word Problems | 应用题建模

Many examination questions translate real-world situations into simultaneous equations. Read the problem carefully, choose two variables, and write two independent equations from the information given. Then use substitution or elimination to solve the system.

许多考试题将现实情境转化为联立方程。要仔细读题，设定两个变量，并根据题目信息写出两个相互独立的方程，然后使用代入法或消元法求解。

Example: In a fruit shop, 2 apples and 3 bananas cost $7, while 1 apple and 4 bananas cost $6. Find the cost of one apple and one banana.

示例：在一家水果店中，2 个苹果和 3 根香蕉共需 7 美元，而 1 个苹果和 4 根香蕉共需 6 美元。求一个苹果和一根香蕉的单价。

Let x be the cost of one apple and y be the cost of one banana. The equations are:

设 x 为一个苹果的单价，

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Simultaneous Equations | 联立方程