📚 Simultaneous Equations | 联立方程
Simultaneous equations are a set of two or more equations that share the same unknown variables. In IGCSE Mathematics, solving these systems is a core skill tested in both Paper 2 and Paper 4, and it appears in many real-world problem-solving contexts.
联立方程组是包含两个或更多方程、共享相同未知变量的一组方程。在 IGCSE 数学中，解联立方程是 Paper 2 和 Paper 4 均会考查的核心技能，也广泛出现在实际应用问题中。
This revision guide will walk you through the key methods: substitution, elimination, and graphical approaches. We will also cover linear-quadratic systems, special cases, and common exam traps.
本复习指南将带你掌握三种关键方法：代入法、消元法和图像法。我们还会讨论线性与二次方程联立、特殊情况以及常见考试陷阱。
1. What Are Simultaneous Equations? | 什么是联立方程
When two equations contain the same two unknown variables (usually x and y), we call them simultaneous equations. A solution is an ordered pair (x, y) that satisfies both equations at the same time.
当两个方程包含相同的两个未知变量（通常为 x 和 y）时，我们称之为联立方程。解就是同时满足这两个方程的有序数对 (x, y)。
For example, consider the system:
例如，考虑以下方程组：
x + y = 10
x − y = 4
ld always verify your answer by substituting the values back into both original equations.
这里 x = 7，y = 3，因为 7 + 3 = 10 且 7 − 3 = 4。你应该始终将数值代回原方程进行检验。
2. Setting Up Equations from Word Problems | 从应用题建立方程
Many exam questions require you to translate a word problem into a pair of simultaneous equations. Follow this three-step strategy.
许多考试题目要求你将文字应用题转化为联立方程组。请遵循以下三步策略。
Step 1: Define the unknown quantities using clear letters, such as p for the cost of a pencil.
第一步：用清晰的字母定义未知量，例如用 p 表示一支铅笔的价格。
Step 2: Write one equation for each piece of given information.
第二步：根据每一条已知信息写出一个方程。
Step 3: Solve the system and interpret the answers in the original context.
第三步：解方程组，并将答案放回原问题情境中进行解释。
Worked example: The total cost of 2 pencils and 3 pens is $13. The cost of 1 pencil and 2 pens is $8. Find the cost of each item.
示例：2 支铅笔和 3 支钢笔的总价为 13 美元。1 支铅笔和 2 支钢笔的总价为 8 美元。求每件物品的价格。
Let p represent a pencil and q represent a pen. The equations are:
设 p 表示铅笔，q 表示钢笔。可列方程组：
2p + 3q = 13
p + 2q = 8
13, which simplifies to −q = −3, so q = 3. Then p = 8 − 2(3) = 2. Therefore a pencil costs $2 and a pen costs $3.
由第二个方程得 p = 8 − 2q。代入第一个方程：2(8 − 2q) + 3q = 13，即 16 − 4q + 3q = 13，化简得 −q = −3，所以 q = 3。于是 p = 8 − 2(3) = 2。因此铅笔每支 2 美元，钢笔每支 3 美元。
3. The Substitution Method | 代入法
The substitution method is especially useful when one variable is already isolated, or can easily be isolated. It works well in linear-quadratic systems too.
代入法特别适用于某个变量已经被单独表示出来，或很容易被表示出来的情况。它也适用于线性与二次方程的联立系统。
Step 1: Rearrange one equation to make one variable the subject.
第一步：将一个方程变形，使一个变量作为被表示对象。
Step 2: Substitute this expression into the other equation.
第二步：将这个表达式代入另一个方程。
Step 3: Solve the resulting linear equation.
第三步：解出所得的一元一次方程。
Step 4: Substitute the value back to find the other variable.
第四步：代回原式，求出另一个变量的值。
Worked example: Solve
示例：解方程组
y = 2x + 1
3x + y = 11
Since y is already the subject, substitute y = 2x + 1 into 3x + y = 11:
由于 y 已被单独表示，将 y = 2x + 1 代入 3x + y = 11：
3x + (2x + 1) = 11
so 5x = 10 and x = 2. Then y = 2(2) + 1 = 5. The solution is (2, 5). Check: 3(2) + 5 = 11 ✓.
由此得 5x + 1 = 11，所以 5x = 10，x = 2。于是 y = 2(2) + 1 = 5。解为 (2, 5)。检验：3(2) + 5 = 11 ✓。
4. The Elimination Method | 消元法
The elimination method involves adding or subtracting the equations to remove one variable. It is often the fastest approach when the coefficients are convenient.
消元法通过将两个方程相加或相减来消去一个变量。当系数比较整齐时，这往往是最快捷的方法。
Step 1: Align the equations so that like terms are in the same vertical columns.
第一步：将方程对齐，使同类项位于同一竖直列。
Step 2: Multiply one or both equations by a constant so that the coefficient of one variable matches (or is the opposite sign) in both equations.
第二步：将其中一个或两个方程乘以适当的常数，使某个变量的系数在两个方程中相同（或互为相反数）。
Step 3: Add or subtract the equations to eliminate that variable.
第三步：将两方程相加或相减，消去该变量。
Step 4: Solve the resulting equation, then substitute back to find the other variable.
第四步：解出所得方程，再代回求出另一个变量。
Worked example: Solve
示例：解方程组
2x + 5y = 19
2x − y = 1
The coefficient of x is 2 in both equations, so subtract the second equation from the first:
两个方程中 x 的系数都为 2，因此用第一个方程减去第二个方程：
(2x + 5y) − (2x − y) = 19 − 1
2x − 3 = 1, so 2x = 4 and x = 2. The solution is (2, 3). Check: 2(2) + 5(3) = 4 + 15 = 19 ✓.
由此得 6y = 18，所以 y = 3。将 y = 3 代入 2x − y = 1：2x − 3 = 1，所以 2x = 4，x = 2。解为 (2, 3)。检验：2(2) + 5(3) = 4 + 15 = 19 ✓。
If the coefficients have opposite signs, add the equations instead of subtracting. Always line up the equations carefully to avoid sign errors.
如果系数互为相反数，则应把两个方程相加而不是相减。书写方程时务必对齐，避免符号错误。
5. The Graphical Method | 图像法
Graphically, each linear equation in two variables represents a straight line. The solution to the simultaneous equations is the point where the two lines intersect.
从图像上看，二元一次方程对应一条直线。联立方程组的解就是两条直线的交点。
Step 1: Rearrange each equation into the form y = mx + c.
第一步：将每个方程化为 y = mx + c 的形式。
Step 2: Draw both lines on the same coordinate axes.
第二步：在同一坐标系中画出两条直线。
Step 3: Read the coordinates of the intersection point.
第三步：读出交点坐标。
Step 4: Verify the coordinates by substituting into both equations.
第四步：将坐标代回两个方程进行验证。
Worked example: Solve graphically the system
示例：用作图法解方程组
y = 2x + 1
y = −x + 4
Both lines are already in slope-intercept form. The first line has slope 2 and y-intercept 1; the second line has slope −1 and y-intercept 4.
两条直线都已化为斜截式。第一条直线的斜率为 2，y 截距为 1；第二条直线的斜率为 −1，y 截距为 4。
2x + 1 = −x + 4. This gives 3x = 3, so x = 1, and y = 3. The intersection point is (1, 3).
为了用代数方法求出交点，令 2x + 1 = −x + 4。得 3x = 3，故 x = 1，y = 3。交点坐标为 (1, 3)。
In the exam, the graphical method is often used to estimate solutions. Read the intersection as accurately as possible, then check it algebraically if the question permits.
在考试中，作图法通常用于估算解。请尽可能准确地读出交点，如果题目允许，再用代数方法验证。
6. Special Cases: No Solution and Infinite Solutions | 特殊情况：无解与无穷多解
Not every pair of simultaneous linear equations has a unique solution. You should recognise the two special cases below.
并非每一对联立线性方程都有唯一解。你需要识别下面两种特殊情况。
| Case Type | Graphical Meaning | Algebraic Sign |
| Unique solution 唯一解 |
Two lines intersect at one point 两条直线交于一点 |
Different gradients 斜率不同 |
| No solution 无解 |
Parallel lines never meet 两条平行线永不相交 |
Same gradient, different y-intercept 斜率相同，y 截距不同 |
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